Page I INTRODrUCTION TO THE NATIONAL ARITHMETIC → , ON THE INDUCTIVE SYSTEM, COMIBINING THE ANALYTIC AND SYNTHETIC METHODS; TN WHICH THIE PRINCIPLES OF THE SCIENCE ARE FULLY EXPLAINED AND ILLUSTRATED. DESIGNED FOR COMMON SCHOOLS AND ACADEMIES. BY BENJANIIN GREENLEAF, A. I. AUTHOR OF THE "NATIONAL ARITHMETIC → ," "ALGEBRA," ETC. NEW STEREOTYPE EDITION, WITH ADDITIONS AND IMPROVEMENTS. BOSTON: PUBLISHED BY ROBERT S. DAVIS & CO. NEW YORK: D. APPLETON & CO., AND MASON BROTHERS. PHILADELPHIA: J. B. LIPPINCOTT AND COMPANY. CHICAGO; WILLIAMI B. KEEN. 1860.

Page II OFFICE OF THE CONTROLLERS OF PUBLIC SCHOOLS. FIRST SCHOOL DISTRICT OF PENNSILVANIA. PHILADELPIIIA, December 14, 1859. At a Meeting of the Controllers of Public Schools, First District of Pennsylvania, held at the CONTROLLERS' CHIAMBER, on Tuesday, December 13th, 1859, the following Resolution was adopted: — Resolved: That GREENLEAF'S COMMON SCHOOL AND NATIONAL ARITHMETICS be introduced to be used in the Public Schools of this District. ROBERT J. HEMPIHILL, Secretary. GREENLEAF'S SERIES OF MATHE"MATICS. 1. NEW PRIMARY ARITHMETIC → ; OR, MENTAL ARITHMETIC → , upon the Ins ductive Plan; with Easy Exercises for the Slate. Designed for Primary Schools. 72 pp. 2. INTELLECTUAL ARITHMETIC → , upon the Inductive Plan; being an advanced Intellectual Course, for Common Schools and Academies. Improved edition. 154 pp. 3. COMMON SCHOOL ARITHMETIC → ; OR, INTRODUCTION TO THE NATIONAL ← ARITHMETIC → . Improved stereotype edition. 324 pp. 4.'THE NATIONAL ← ARITHMETIC → , being a complete course of Higher ← Arithmetic → , for advanced scholars in Common Schools, High Schools, and Academies. New electrotype edition, with additions and improvements. 444 pp. 5. PRACTICAL TREATISE ON ALGEBRA, for Academies and High Schools, and for advanced Students in Common Schools. Improved stereotype edition. 360 pp. 6. ELEMENTS OF GEOMETRY; with Practical Applications to Mensuration. Designed for Academies and IIigh Schools. Electrotype edition. 320 pp. COMPLETE KEYS TO TItE INTELLECTUAL, COMMON SCHOOL, ANf NATIONAL ARITIHMETICS, THE PRACTICAL TREATISE ON ALGEBRA, AND GEOMETRY, containing Solutions and Explanations, for Teachers only. In 5 volumes. D?' Two editions of the NATIONAL ARTTHICETrC, and also of the COMIHON SCHOOL ARITHIEWTIC, one containing the ANSWERS to the examples, and the other -without them, are published. Teachers are requested to state in their orders which edition they prefer. Entered according to Act of Congress, in the year 1842, by BENJAMIN GREENLEAF, in the Clerk's Office of the District Court for the District of Massachusetts. Entered according to Act of Congress, in the year 1848, by BENJAMIN GREENLEAF, in the Clerk's Office of the District Court of the District of Massachusetts. Entered according to Act of Congress, in the year 1856, by BENJAMIN GREENLEAF, in the Clerk's Office of the District Court of the District of Massachusetts.

Page III PRE FA C E. TIrE present edition of this work has been thoroughly revised and re-written, and also improved by the addition of much valuable new material, rendering it a sufficiently complete practical treatise for the majority of learners. The arrangement is strictly progressive; the aim having been to introduce subjects in an order most in accordance with the laws governing the proper development of mind. The rules have generally been deduced from the analysis of onie or more questions, so that the reasons for the methods of solution adopted are rendered intelligible to the pupil; no knowledge of a principle being required, that has not been previously illustrated and explained. In this respect, it is believed the work will be found to differ from most other arithmetics. In preparation of the rules, definitions, and illustrations, the utmost care has been taken to express them in language simple, precise, and accurate. The examples are of a practical character, and adapted not only to fix in the mind the principles, which they involve, but also to interest the pupil, exercise his ingenuity, and inspire a love for mathematical science. The reasons for the operations are explained, and an attempt is made to secure to the learner a knowledge of the philosophy of the subject, and prevent the too prevalent practice of merely performing, mechanically, operations, which he does not understand. Analysis has been made a prominent subject, and employed in the solution of questions under most of the rules, in which it could be used with any practical advantage; and it cannot be too strongly recommended to the pupil to make use of this mode of operation, where it is recommended by the author. All the most important methods of abridging operations, applicable to business transactions, have been given a place in the work, and, so introduced, as not to be regarded as mere blind mechanical expedients, but as rational labor-saving processes. Old rules and distinctions, which modern improvements have rendered unnecessary, and which, deservedly, are becoming obsolete, have been avoided.

Page IV IV PREFACE. Rules for finding the greatest common divisor of fractions, and for finding the least common multiple of fractions; methods of equating accounts; division of duodecimals; exchange, foreign and inland; and several important tables, are among the new features of this edition, which will be found, it is believed, very valuable. The articles on money, weights, measures, interest, and duties are the results of extensive correspondence and much laborious research, and are strictly conformable to present usage, and recent legislation, state and national. Questions have been inserted at the bottom of each page, de. signed to direct the attention of teachers and pupils to the most important principles of the science, and fix them in the mind. It is not intended, however, nor is it desirable, that the teachershould servilely confine himself to these questions; but vary their form, and extend them at pleasure, and invariably require the pupil thoroughly to understand the subject, and give the reasons for the various steps in the operation, by which he arrives at any result in the solution of a question. The object of studying mathematics is not only to acquire a knowledge of the subject, but also to secure mental discipline, to induce a habit of close and patient thought, and of persevering and thorough investigation. For the attainment of this object, the examples for the exercise of the pupil are numerous, and variously diversified, and so constructed as necessarily to require careful thought and reflection for the right application of principles. The author would respectfully suggest to teachers, who may use this book, to require their pupils to become familiar with each rule before they proceed to a new one; and, for this purpose, a frequent review of rules and principles will be of service, and will greatly. facilitate their progress. If the pupil has not a clear idea of the principles involved in the solution of questions, he will find but little pleasure in the study of the science; for no scholar can be pleased with what he does not understand. BENJAMIN GREENLEAF. BRADFORD, MAss., August 1st, 1856. NOTICE. ~g' Two editions of this work, and also of the NATIONAL ← ARITHMETIC → , one containing the ANSWERS to the examples, and the other without them, are now published.

Page V CONTENTS. SECTION I. Page Page NOTATION AND NUMERATION,..... 7 Avoirdupois Weight, Table,..... 87 Notation......... 7 Cloth Measure, Table,....... 89 Table of Roman Letters....... 8 Long Measure, Table,....... 90 Exercises in Roman Notation,.... 9 Surveyors, Measure, Table,..... 93 Numeration,............11 Square 2Measure, Table,..... 94 French Numeration Table,...... 11 Cubic or Solid Pleasure, Table,... 96 Exercises in French Numeration,. 12 Wine or Liquid Measure, Table,.. 98 Exercises in French Notation and Beer Measure, Table,....... 99 Numeration............ 13 Dry Measure, Table........ 100 English Nunleration Table,..... 14 Measure of Time, Table...... 102 Exercises in English Numeration,.. 15 Circular Measure, Table,......105 Exercises in English Notation and Miscellaneous Table,........ 106 N umeration.......... 15 Miscellaneous Exercises in Reduction, 107 SECTION II. SECTION XI. ADDITION. —Mental Exercises,... 16 ADDITION OF COMPOUND NUMBERS. - Addition Table............16 English Money..........110 Examples for Practice in the different SECTION III. Weights and Measures,. 111 SUBTRACTION. - Mental Exercises,... 25 Subtraction Table......... 25 SECTIO SUBTRACTION OF COMPOUND NUMBERS.SECTION IV7 T English Money,..........114 MULTIPLICATION. - Mental Exercises,. 33 Multiplication Table,........ 33 SECTION XIII. SECTION V MISCELLANEOUS EXERCISES IN ADDITION AND SUBTRACTION OF COMDIVISION. — Mental Exercises,.... 44 POUND NUMBERS,.......119 Division Table,........... 44SECTION XIV. SECTION VI. MULTIPLICATION OF COMPOUND NUMQUESTIONS INVOLVING FRACTIONS,.. 57 BERS.............121 SECTION VII. SECTION XV. CONTRACTIONS IN MULTIPLICATION AND DIvSION OP COMPOUND NUMBERs,.. 125 DIV18I0s,....... 61 Contractions in Multiplication,....61 SECTION XVI. Contractrons in Division,......63 Contractions i DivisionMISCELLANEOUS EXAMIPLES IN MULTISECTION VIII. PLICATION AND DIVISION OF COm0MISCELLANEOUS EXAMPLES INVOLVING POUND NUMBERS)....... 129 THE FOREGOING RULES...... 65 SECTION XVII. SECTION IX. PROPERTIES AND RELATIONS OF NUmUNITED STATES MONEY,....... 69 BERS,............. 130 Reduction of United States Money,.70 Table of Prime Numbers,.131 Addition of United States Money,. 71 A Prime Factor of a Number,... 131 Subtraction of United States Money, 73 Cancellation,..........133 Multiplication of U. States Money,. 74 A Common Divisor,........ 136 Division of United States Money,.. 75 The Greatest Common Divisor,... 136 Practical Questions by Analysis,.. 76 A Common Multiple,........138 Bills, Exercises in,....... 79 Ledger Accounts,.. 81 SECTION XVIII. FRACTIONS. - CoMMNoN FRACTIONS,.. 140 SECTION X. Reduction of Common Fractions,.. 142REDUCTION,............. 82 A Common Denominator...... 146 English Money, Table,.......82 Addition of Common Fractions,... 148 Troy Weight, Table.8,,. 84 Subtraction of Common Fractions,.. 150 Ap6tihearie' Weight, Table, 86 MultiplieatlOn of Common Fradtions, 165

Page VI VI CONTENTS. Page Page Division of Comimon Fractions,... 160 SECTION XXXII. Complex Fractions,..... 165 PROFIT AND LOSS,..........248 Greatest Common Divisor of Fractions, 167 Miscellaneous Examples in Profit and Least Common Multiple of Fractions,. 167 Loss,.............253 Miscellaneous Exercises in Fractions, 169 SECTION XXXIII Reduction of Fractions of Compound Numbers,............170 PARTNERSHIP, OR COMPANY BUSINESS,. 254 Addition of Fractions of Compound Numbers,...........174 SECTION XXXIV. Subtraction of Fractions of Compound.258 Numbers,... 15 Reduction of Currencies......... 25 Questions to be performed by Analysis 176 Reducon of Currencies, Miscellaneous Questions by Analysis, 179 SECTION XXXV. SECTION XIX. ExCnANAE,.........261 DECIMAL FRACTIONS.......... 181 Inland Bills............ 262 Numeration of Decimal Fractions,.. 182 Foreign Bills... 263 Notation of Decimal Fractions,... 183 Exchange on England,.......263 Addition of Decimals....... 184 Exchange on France,....... 265 Subtraction of Decimals,..... 185 Multiplication of Decimals,.... 186 SECTION XXXVI. Division of Decimals,........188 DUODECIAMALS............266 Reduction of Decimals........ 190 Addition and Subtraction of DuodeciMiscellaneous Exercises in Decimals, 193 mals,.....266 SECTION XX. Multiplication of Duodecimals,... 267 PERCENTAGE,....... 194 SECTION XXXVII. SECTION XXI. INVOLUTION............269 SIMPLE INTEREST.......... 196 SECTION XXXVIII. Mliscellaneous Exercises in Interest,. 204 EVOLUION,.............. 271 Partial Payments,......... 205 Extraction of the Square Root,... 272 Problems in Interest,.......210 Application of the Square Root,... 276 SECTION XXII. Extraction of the Cube Root,...281 Application of the Cube Root,....285 COMfPOUND INTEREST,......... 212 Table,.............. 214 SECTION XXXIX. SECTION XXIII. ARITHMETICAL PROGRESSION,...... 287 ISCOUNT.. 216 Annuities at Simple Interest by Arithmetical Progression,.......292 SECTION XXIV. SECTION XL. COMMISSION, BROKERAGE, AND STOCKS, 218 GEOMETRIL PROGRESSION.. 294 SECTION XXV. Annuities at Compound Interest by BBANKfINGO....... 2.21 Geometrical Progression,.... 298 Bank Discount,.......... 222 SECTION XLI. SECTION XXVI. ALLIGATION............3. 00 INSURANCE.........224 Alligation Medial,..........300 SECTION XXVII. Alligation Alternate,........ 301 SECTION XXVII. CUSTOM-HOUSE BUSINESS.......225 SECTION XLII. PERMUTATION,........... 305 SECTION XXVIII. ASSESS.ENT OF TAxEs........ 227 SECTION XLIII. SECTION XXIX. MENSURATION OF SURFACES,.. 3..06 SECTION XXIX. EQUATION OF PAYMENTS,....... 230 SECTION XLIV. SECTION XXX. MENSURATION OF SOLIDS,....... 312 RATIO,...........237 SECTION XLV. SECTION XXXI. MENSURATION OF LUMBER AND TIMBER,. 318 PROPORTION;....239 Simple Proportion,.240 SECTION XLVI. Compound Proportion....... 245 MISCELLANEOUS QUESTIONS......819

Page 7 ARIT -H[ ETIC. ARTICLE 1. QUANTITY is anything that can be measured. A unit is a single thing, or one. A number is either a unit or a collection of units. An abstract number is a number, whose units have no reference to any particular thing or quantity; as two, five, seven. A concrete number is a number, whose units have reference to some particular thing or quantity; as two books, five feet, seven gallons. ← ARITHMETIC → is the science of numbers, and the art of computing by them. A rule of ← arithmetic → is a direction for performing an operation with numbers. The introductory and principal rules of ← arithmetic → are Notation and Numeration, Addition, Subtraction, Multiplication, and Division. The last four are called the fundamental rules, because upon them depend all other arithmetical processes. ~ I. NOTATION AND NUMERATION. NOTATION. ART. 2. NOTATION is the art of expressing numbers by figures or other symbols. There are two methods of notation in common use; the Roman and the Arabic. QUESTIONS. -Art. 1. What is quantity? What is a unit? What is a number? What is an abstract number? What is a concrete number? What is ← arithmetic → ? What is a rule? Which are the introductory rules? What are the last four called? - Art. 2. What is notation? How many kinds of notation in common use? What are they?

Page 8 8 NOTATION. [SECT. L ART. 3. The Roman notation, so called from its originating with the ancient Romans, employs in expressing numbers seven capital letters, viz.: I for one; V for five; X for ten; L for fifty; C for one hundred; D for five hundred; M for one thousand. All the other numbers are expressed by the use of these letters, either in repetitions or combinations; as, II expresses two; IV, four; VI, six, &c. By a repetition of a letter, the value denoted by the letter is represented as repeated; as, XX represents twenty; CCC, three hundred. By writing a letter denoting a less value before a letter denoting a greater, their difference of value is represented; as, IV represents four; XL, forty. By writing a letter denoting a less value after a letter denoting a greater, their sum is represented; as, VI represents six; XV, fifteen. A dash (-) placed over a letter increases the value denoted by the letter a thousand times; as, V represents five thousand; IV, four thousand. TABLE or ROMAN LETTERS. I one. LXXX eighty. II two. XC ninety. III three. C one hundred. IV four. CC two hundred. V five. CCC three hundred. VI six. CCC( four hundred. VII seven. D five hundred. VIII eight. DC six hundred. IX nine. DCC seven hundred. X ten. DCCC eight hundred. XX twenty. DCCCC nine hundred. XXX thirty. M one thousand. XL forty. MD fifteen hundred. L fifty. MM two thousand. LX sixty. X ten thousand. LXX seventy. M one million. QUESTIONS. - Art. 3. Why is the Roman notation so called? By what are numbers expressed in the Roman notation? What effect has the repetition of a letter? What is the effect of writing a letter expressing a less value before a letter denoting a greater? What of writing the letter after another denoting a greater value? How many times is the value denoted by a letter increased by placing a dash over it? Repeat the mole.

Page 9 bECT. I.] NOTATION. 9 The Roman notation is now but little used, except in numbering sections, chapters, and other divisions of books. EXERCISES IN ROMAN NOTATION. The learner may write the following numbers in letters: 1. Ninety-six. Ans. XCVI. 2. Eighty-seven. 3. One hundred and ten. 4. One hundred and sixty-nine. 5. Two hundred and seventy-five. 6. Five hundred and forty-two. 7. One thousand three hundred and nineteen. 8. One thousand eight hundred and fifty-eight. ART. 4, The Arabic notation, so called from its having been made known through the Arabs, employs in expressing numbers ten characters or figures, viz.: 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. one, two, three, four, five, six, seven, eight, nine, cipher. The first nine are sometimes called digits, from digitus, the Latin signifying a finger, because of the use formerly made of the fingers in reckoning. The cipher, also, has sometimes been called naught, or zero, from its expressing the absence of a number, or nothing, when standing alone. ART. Se The particular position a figure occupies with regard to other figures is called its PLACE; as in'32, counting from the right, the 2 occupies the first place, and the 3 the second place, and so on for any other like arrangement of figures. The digits have been denominated significant figures, because each expresses of itself a positive value, always representing so many units, or ones, asits name indicates. But the size or value of the units represented by a figure differs with the place occupied by the figure. For example, there are written together to represent a number three figures, thus, 366 (three hundred and sixty-six). Each of the figures, without regard to its place, expresses units, or ones; but these units, or ones, differ in value. The 6 occupying the first place represents 6 single units; the 6 occupying the second place repreQUESTIONS. - What use is now made of Roman notation? - Art. 4. How many characters are employed in the Arabic notation? What are the first nine called, and why? What is the cipher sometimes called? What does it represent when standing alone? - Art. 5. What is meant by the place of a figure? What have the digits been denominated? Why? How does the size or value of units represented by figures differ?

Page 10 10 NOTATION. [SECT. I sents 6 tens, or 6 units each ten times the size or value of a unit of the first place; and the 3 occupying the third place represents 3 hundreds, or 3 units each one hundred times the size or value of a unit of the first place. ART. 6. The cipher becomes significant when connected with other figures, by filling a place that otherwise would be vacant; as in 10 (ten), where it occupies the vacant place of units; in 120 (one hundred and twenty), where it also occupies the vacant place of units; and in 304 (three hundred and four), tvhere it fills the vacant place of tens. ART. 7, The simple value of a unit is the value expressed by a figure standing alone; or, in a collection, when standing in the right-hand place. Thus 6 alone, or in 26, expresses a simple value of six single units, or ones. The local value of a unit is the value expressed by a figure when it is used in combination with another figure or figures, and depends upon the place the figure occupies. The local values expressed by figures will be made plain by the following table and its explanation.,0 Xn M' S ei 4 - Zarc: z AThe figures in this table are read thus 9 Nine. 9 8 Ninety-eight. 9 8 7 Nine hundred eighty-seven. 9 8 7 6 Nine thousand eight hundred seventy-six. 9 8 7 6 5 Ninety-eight thousand seven hundred sixty-five.?9 8 7 6 5 4 PNine hundred eighty-seven thousand six hundred fifty-four. 98 7 6 5 4 3 Nine millions eight hundred seventy-six thousanti five hundred forty-three. QUESTrONS.- Art. 6. When does a cipher become significant?-Art. 7.'What is the simple value of a unit? What is the local value of a unit? What is the design of this table 7

Page 11 SBcT. I.] NUMERATIO0. 11 It will be noticed in the preceding table, that each figure in the right-hand or units' place expresses the local value of so many units; but when standing in the second place, it expresses the local value of so many tens, each of the value of ten ones; when in the third place, the local value of so many hundreds, each of the value of' ten tens; when in the fourth place, the local value of so many thousands, each of the value of' ten hundreds; the value expressed by any figure being always made tenfold by each removal of it one place to the left hand. NUMERATION. ART. 80 NUMIERATION is the art of reading numbers when expressed by figures. ART. 9, There are two methods of numeration in common use: the French and the Ezglish. ART. 10. The French method is that in general use on the continent of Europe and in the United States. It separates figures into groups, called periods, of three places each, and gives a distinct name to each period. FRENCH NUMERATION TABLE..-~..', * - O X 3 14 ~ rs. Period of Period of Period of Period of Period of Period of Period of Perod o ~~3~ -c~.f -j. lions. lions. lions. QUESTIONS. - Art. 7. What value is expressed by a figure standing in the right-hand or units' place? What in the second place? What in the third? How do figures increase from the right towards the left? -Art. 8. What is numeration? What are the two methods of numeration in common use? Where is the French method mole generally used? Repeat the French Numeration Table. What are the names of the different periods in the table? What is the value of the numbers represented in the table expressed in words.?

Page 12 12 N UMERATION, [SECT. L The value of the numbers represented in this table, expressed in words, is, One hundred twenty-seven sextillions, eight hundred ninety-four quintillions, two hundred thirty-seven quadrillions, eight hundred sixty-seven trillions, one hundred twenty-three billions, six hundred seventy-eight millions, four hundred seventyeight thousand, six hundred thirty-eight. The names of the periods above Sextillions, in their order, are, Septillions, Octillions, Nonillions, Decillions, Undecillions, Duodecillions, Tredecillions, Quatuordecillions, Quindecillions, Sexdecillious, Septendecillions, Octodecillions, Novemdecillions, Vigintillions, &c. ART. 11, The successive places occupied by figures are often called orders. Hence, a figure in the right-hand or units' place is called a figure of the first order, or of the order of units; a figure in the second place is a figure of the second order, or of the order of tens; in the third place, of the order of hundreds, and so on. Thus, in the number 1847, the 7 is of the order of units, 4 of the order of tens, 8 of the order of hundreds, and 1 of the order of thousands, each figure expressing as many units as its name indicates of that order to which it belongs; so that we read the whole number, one thousand eight hundred and forty-seven. ART. 12, From the preceding table and explanation, we deduce the following rule for numerating and reading numbers expressed by figures according to the French method. RULE. — Begin at the. right hand, and point off the figures into periods of TmREE places each. Then, commencing at the left hand, read the figures -of each period, adding the name of each period excepting that of units. ExERCISES IN FRENCH NUMERATION. The learner may read orally, or write in words, the numbers represented by the following figures: 1. 152 5. 2254 9. 84093 13. 610711 2. 276 6. 4384 10. 98612 14. 3031671 3. 998 7. 7932 11. 592614 15.,4869021 4. 1057 8. 42198 12. 400619 16. 637313789 QlESTIONS. - Art. 10. What are the names of the periods above sextillions? -Art. 11. What are the successive places of the figures in the table called? Of what order is the first or right-hand figure? The second? The third? &c. - Art. 12. What is the rule for numerating and reading numbers according to the French method?

Page 13 SECT. I.] NUMERATION~. 13 17. 39461928 24. 3761700137706717 18. 427143271 25. 242173562357421 19. 6301706716 26. 870037637471078635 20. 143776700333 27. 8216243812706381 21. 20463162486135 28. 2403172914376931 22. 63821024711802 29. 3761706137706167138 23. 44770630147671 30. 610167637896430607761607 ART. 130. To write numbers by figures according to the French method, we have the following RULE. - Beyin at the left hand, and write in each successive order the figure belonging to it. if any intervening order would otherwise be vacant, fill the place by a cipher. EXERCISES IN FRENCE NOTATION AND NUMERATION. The learner may represent by figures, and read, the following numbers: 1. Forty-seven. 2. Three hundred fifty-nine. 3. Six thousand five hundred seventy-five. 4. Nine hundred and eight. 5. Nineteen thousand. 6. Fifteen hundred and four. 7. Twenty-seven millions five hundred. 8. Ninety-nine thousand ninety-nine. 9. Forty-two millions two thousand and five. 10. Four hundred eight thousand ninety-six. 11. Five thousand four hundred and two. 12.,Nine hundred seven millions eight hundred five thousand and seventy-four. 13. Three hundred forty-seven thousand nine hundred and fifteen. 14. Eighty-nine thousand forty-seven. 15. Fifty-one thousand eighty-one. 16. Seven thousand three hundred ninety-five. 17. Fifty-seven billions fifty-nine millions ninety-nine thousand and forty-seven. QUESTIONS. - Art. 13., What is the rule for writing numbers according to the French method? At which hand do you begin to numerate figures? Where do you begin to read them? At which hand do you begin to write numbers? Why? 2

Page 14 14 NUMERATION. iSECT. 1 ART. 14. The following table exhibits the English method of numeration, in which it will be observed that the figures are separated by commas into periods of six figures each. The first or right-hand period is regarded as units and thousands of units, the second, as millions and thousands of millions; and so on, six places being assigned to each period designated by a distinct name, ENGLISH NUMERATION TABLE. _-'~i ] E~. accrin to h Egi m th i, E e h red thi tosn eigh Ehundc-~red int i io Esee Ere eleven 137890, 711716, 371712, 456 11. Period of Tril- Period of Bil- Period of Mil- Period of lions. lions. lions. Units. The value of the figures in the above table, expressed in words according to the English method, is, One hundred thirty-seven thousand eight hundred ninety trillions; seven'hurdred eleven thousand seven hundred sixteen billions; three hundred seventyone thousand seven hundred twelve millions; four hundred fiftysix thousand seven hundred eleven. Although there is the same number of figures in the English and in the French table, yet it will be observed that in the French table we have the names of three periods other than in the English. It will also be observed that the variation commences after the ninth place, or the place of hundreds of millions. If, therefore, we would know the value of numbers QUESTIONS.- Art. 14. How many figures in each period in the English method of numeration? What orders are found in the English method that are not in the French? Give the names of the periods in the English Numeration Table, beginning with the period of units. Repeat the table, giving the names of all the orders or places. What is the value of the nunmbers in the table expressed in words-? How do the figures in the English and French table compare as to numbers? How as to periods? Why is this difference? Has a million the same value reckoned by the Frenoh table as when reckoned by the English?

Page 15 bECT. J..] NUMERATION. 15 higher than hundreds of millions, when we see them written in words, or hear them read, we need to know whether they are expressed according to the French or the English method of numeration. The English method of numeration is that generally used in Great Britain, and in the British Provinces. ART. 150 To numerate and read numbers expressed by figures according to the English method, we have the following RULE. -Begin at the right hand, and point off the figures into periods of six places each. Then, commencing at the left hand, read the figures of each period, adding the name of each period, excepting that of units. EXERCISES IN ENGLISII NUMERATION. The learner may read orally, or write in words, the following numbers: 1. 125 5. 27306387903 2. 1063 6. 531470983712 3. 25842 7. 4230578032765038 4. 90435.7 8. 716756378807370767086389706473 ART. 16, To write numbers in figures, according to the English method, we have the following RULE. - Begin at the left hand, and write in each successive order the figure belonging to it. if any intervening order'would otherwise be vacant, fill the place by a cipher. ExERcISES IN ENGLISH NOTATION AND NUMERATION. The learner may write in figures, and read, the following numbers: 1. Three hundred twenty-five thousand four hundred and twelve. 2. Two hundred fourteen thousand, one hundred sixty-five millions, seventy-eight thousand and fifty-six. 3. Forty-two billions, six hundred seventeen thousand thirtyone millions, forty-one thousand three hundred forty-two. 4. Two thousand eight billions, nine thousand eighty-two millions, seven hundred one thousand, nine hundred and eight. QUESTIONS. - Has the billion the same value as that by the French table? Why not? By whioh table has it the greater value? Where is the English method of numerating in use?- Art. 15. What is the rule for numerating and reading numbers by the English method? - Art. 16. What is the rule for writing numbers according to the English method?

Page 16 16 ADDITION. [SECT. II ~ II. ADDITION. MjENTAL EXERCISES. ART.,7. WIIEN it is required to find a single number to express the sum of the units contained in several smaller numbers, the process is called Addition. Ex. 1. James has 3 pears, and his younger brother has 4, how many have both? ILLUSTRATION.- TO solve this question, the 3 pears and 4 pears must be added together; thus, 3 added to 4 makes 7. ~Therefore James and his brother have 7 pears. ADDITION TABLE. 2 and O are 2.3 and 0 are 3 4 and 0 are 4 5 and 0 are 5 2 and 1 are 3 3 and 1 are 4 4 and 1 are 5 5 and 1 are 6 2 and 2 are 4 3 and 2 are 5 4 and 2 are 6 5 and 2 are 7 2 and 3 are 5 3 and 3 are 6 4 and 3 are 7 5 and 3 are 8 2 and 4 are 6 3 and 4 are 7 4 and 4 are 8 5 and 4 are 9 2 and 5 are 7 3 and 5 are 8 4 and 5 are 9 5 and 5 are 10 2 and 6 are 8 3 and 6 are 9 4 and 6 are 10 5 and 6 are 11 2 and 7 are 9 3 and 7 are 10 4 and 7 are 11 5 and 7 are 12 2 and 8 are 10 3 and 8 are 11 4 and 8 are 12 5 and 8 are 13 2 and 9 are 11 3 and 9 are 12 4 and 9 are 13 5 and 9 are 14 2 and 10 are 12 3 and 10 are 13 4 and 10 are 14 5 and 10 are 15 2 and 11 are 13 3 and 11 are 14 4 and 11 are 15 5 and 11 are 16 2 and 12 are 14 3 and 12 are 15 4 and 12 are 16 5 and 12 are 17 6 and 0 are 6 7 and 0 are 7 8 and 0 are 8 9 and 0 are 9 6 and 1 are 7 7 and 1 are 8 8 and I are 9 9 and 1 are 10 6 and 2 are 8 7 and 2 are 9 8 and 2 are. 10 9 and 2 are 11 6 and 3 are 9 7 and 3 are 10 8 and 3 are 11 9 and 3 are 12 6 and 4 are 10 7 and 4 are 11 8 and 4 are 12 9 and 4 are 13 6 and 5 are 11 7 and 5 are 12 8 and 5 are 13 9 and 5 are 14 6 and 6 are 12 7 and 6 are 13 8 and 6 are 14 9 and 6 are 15 6 and 7 are 13 7 and 7 are 14 8 and 7 are 15 9 and 7 are 16 6 and 8 are 14 7 and 8 are 15 8 and 8 are 16 9 and 8 are 17 6 and 9 are 15 7 and 9 are 16 8 and 9 are 17 9 and 9 are 18 6 and 10 are 16 7 and 10 are 17 8 and 10 are 18 9 and 10 are 19 6 and 11 are 17 7 and 11 are 13 8 and 11 are 19 9 and 11 are 20 6 and 12 are 18 7 and 12 are 19 8 and 12 are 20 9 and 12 are 21 10 and 0 are 10 11 and 0 are 11 12 and 0 are 12 13 and 0 are 13 10 and 1 are 11 11 and I are 12 12 and 1 are 13 13 and 1 are 14 10 and 2 are 12 1'1 and 2 are 13 12 and 2 are 14 13 and 2 are 15 10 and 3 are 13 11 and 3 are 14 12 and 3 are 15 13 and 3 are 16 10 and 4 are 14 11 and 4 are 15 12 and 4 are 16 13 and 4 are 17 10 and 5 are 15 11 and 5 are 16 12 and 5 are 17 13 and 5 are 18 10 and 6 are 16 11 and 6 are 17 12 and 6 are 18 13 and 6 are 19 10 and 7 are 17 11 and 7 ale 18 12 and 7 are 19 13 and I are 20 10 and 8 are 18 11 and 8 are 19 12 and 8 are 20 13 and 8 are 21 10 and 9 are 19 11 and 9 are'20 12 and 9 are 21 13 and 9 are 22 10 and 10 are 20 11 and 10 are 21 12 and 10 are 22 13 and 10 are 23 10 and 11 are 21 11 and 11 are 22 12 and 11 are 23 13 and 11 are 24 10 and 12 are 22 11 and 12 are 23 12 and 12 are 24 13 and 12 are 25 QUESTION. - Art. 17. 1W1hat is the process called by which we find the suen of several numbers?

Page 17 SECT. II.] ADDITION. 17 2. Ilow many are 2 and 3? 2 and 5? 2 and 7? 2 and 9? 2and4? 2and2? 2and 8? 2and6? 3. How many are3and 3? 3 and 5? 3-and 7? 3 and 9? 3 and4? 3 and6? 3 and 8 7 3 and 3? 4. How many are 4 and 3? 4 and 5? 4 and 8? 4 and 9? 4 and 1? 4 and 2? 4 and 4?. 4 and 7? 5. How many are 5 and 3? 5 and 4? 5 and 7? 5 and 8? 5 and 9? 5 and 2? 5 and 5 7 5 and 6? 5 and i? 6. How many are 6 and 2? 6 and 4? 6 and 3? 6 and 5? 6 and 7? 6 and 9? 6 and I? 6 and 6? 6 and 8? 7. How many are 7 and 3? 7 and 5? 7 and 7? 7 and 6? 7 and 8? 7 and 9? 7 and 2? 7 and 4? 7 and 10? 8. How many are 8 and 2? 8 and 4? 8 and 5? 8 and 7? 8 and 9? 8 and S? 8 and 1? 8 and 3? 8 and 6? 9. How many are 9 and 1? 9 and 3? 9 and 5? 9 and 4? 9 and 6? 9 and 8? 9and9? 9and2? 10. James had 4 apples, Samuel gave him 5 more, and John gave him 6; how many had he in all? 11. Gave 7 dollars for a barrel of flour, 5 dollars for a hundred weight of sugar, and 8 dollars for a tub of butter; what did I give for the whole? 12. Paid 5 dollars for a pair of boots, 12 dollars for a coat, and 6 dollars for a vest; what was the whole cost? 13. Gave 25 cents for an ← arithmetic → , and 67 for a geography; what was the cost of both? ILLUSTRATION. - We may divide the cents into tens and units. Thus, 25 equals 2 tens and 5 units; 67 -equals 6 tens and 7 units; 2 tens and 6 tens are 8 tens; and 5 units and 7 units are 12 units, or 1 ten and 2 units; 1 ten and 2 units added to 8 tens make 9 tens and 2 units, or 92. Therefore the ← arithmetic → and geography cost 92 cents. 14. On the fourth of July 20 cents were given to Emily, 15 cents to Betsey, 10 -cents to Benjamin, and none to Lydia; what did they all receive? 15. Bought four loads of hay; the first cost 15 dollars, the second 12 dollars, the third 20 dollars, and the fourth 17 dol lars; what was the price of the whole? 16. Gave 55 dollars for a horse, 40 dollars for a wagon, and 17 dollars for a harness-; what did they all cost? 17. Sold 3 loads of wood for 17 dollars, 6 tons of timber for 19 dollars, and a pair of oxen for 60 dollars; what sum did I receive? 2*

Page 18 18 ADDITION. [SECT. II. ART. IS, From the solution of the preceding questions, the learner will perceive, that ADDITION is the process of finding the sum of two or more numbers. The result obtained,, is called their amonunt. Addition is commonly represented by this character, +-, which signifies plus, or added to. The expression 7+-5 is read, 7 plus 5, or 7 added to 5. This character, =, is called the sign of equality, and signifies equal to. The expression 7+-5=12 is read, 7 plus 5, or 7 added to 5, is equal to 12. EXERCISES FOR THE SLATE. ART. 9., The method of operation when the numbers are large, and the sum of each column is less than 10. Ex. 1. A man bought a watch for 42 dollars, a coat for 16 dollars, and a set of maps for 21 dollars; what did he pay for the whole? Ans. 79 dollars. OPERATION. Having arranged the numbers so that Dollars. all the units of the same order shall stand 42 in the same column, we first add the col1 6 umn of units; thus, 1 and 6 are 7, and 2 2 1 are 9 (units), and write down the amount under the column of units. VWe then add Amount 7 9 the column of tens; thus, 2 and 1 are 3, and 4t'are 7 (tens), which we write under the column of tens, and thus find the amount of the whole to be 79 dollars. ART. 20@ First Method of Proof. — Begin at the top and add the columns downward in the same manner as they were before added upward, and if the two sums agree the work is presumed to be right. The reason of this proof is, that, by adding downward, the order of the figures is inverted; and, therefore, any error made in the first addition would probably be detected in the second. NoTE.- This method of proof is generally used in business. QUESTIONS. —Art. 18. What is addition? What is the sign bf addition, and what does it signify? What is the sign of equality, and what does it signify? —Art. 19. How are numbers arranged for addition?'Which column must first be added? Why? Where do you place its sum? Where must the sum &o each column be placed? What is the whole sum called? - Art. 20. How is addition proved? What is the reason for this method of proof? Is this method in common use?

Page 19 SECT ILl ADDITION. 19 EXAMPLES FOR PRACTICE. 2. 3. 4. 5. -Miles. Furlongs. Days. Weeks. 151 234 472 121 212 423 315 516 S21 321 102 361 Ans. 6 8 4 6. What is the sum of 231, 114, and 324? Ans. 669. 7. Required the sum of 235, 321, and 142. Ans. 698. 8. What is the sum of 11, 22, 505, and 461? Ans. 999. 9. Sold twelve ploughs for 104 dollars, two wagons for 214 dollars, and one chaise for 121 dollars; what was the amount of the whole? Ans. 439 dollars. 10. A drover bought 125 sheep of one man, 432 of another, and of a third 311; how many did he buy? Ans. 868 sheep. ART. 21.l Method of operation when the sum of any column is equal to or exceeds 10. Ex. 1. I have three lots of wild land; the first contains 246 acres, the second 764 acres, and the third 918 acres. I wish to know how many acres are in the three lots. Ans. 1928 acres. OPERATION. Having arranged the numbers as in Acres. the preceding examples, we first add 2 4 6 the units; thus, 8 and 4 are 12' and 6 7 6 4 are 18 units, equal 1 ten and 8 units. 9 1 8 We write the 8 units under the column of units, we carry or add the 1 ten to Amount 1 9 2 8 the column of tens; thus, 1 added to 1 makes 2, and 6 are 8, and 4 are 12 (tens), equal to 1 hundred and 2 tens. We write the 2 tens under the column of tens, and add the I hundred to the column of hundreds; thus, 1 added to 9 makes 10, and 7 are 17, and 2 are 19 (hundreds), equal to 1 thousand and 9 hundreds. We write the 9 under the column of hundreds; and there being no other column to be added, we set down the 1 thousand in thousands' place, and find the amount of the several numbers to be 1928. NOTE. -A more concise way, in practice, is to omit calling the name of each figure as added, and name only results. QUESTIONS. - Art. 21. When the sum of any column exceeds ten, where are the units written? What is done with the tens? Why do you carry, or add, one for every 10? How is the sum of the last column written?

Page 20 20 ADDITION. [SEer. II. ART. 22, From the preceding examples and illustrations in addition, we deduce the following general RULE. - Write the numbers so that all the figures of the same order shall stand in the same column. Add, upward, all the figures in the column of units, and, if the amount be less than ten, write it underneath. But, if the amount be ten or more, write down the unit figure only, and add in the figure denotling the ten or tens with the next column. Proceed in this way with each column, until all are added, observing to write down the whole amount of the last column. ART. 23. Second Method of Proof. -Separate the numbers to be added into two parts, by drawing a horizontal line between them. Add the numbers below the line, and set down their sum. Then add this sum and the number, or numbers, above the line together; and, if their stun is equal to the first amount, the work is presumed to be right. The reason of this proof depends on the principle, That the sum of all the parts into which any number is divided is equal to the whole. EXAIMPLES FOR PRACTICE. 2. 2. 3. 3. OPERATION. OPERATION AND PROOF. OPERATION. OPERATION AND PROOF. 526 526 241 241 317 532 529- 317 207' 532 132 529 913 207 132 913 Ans. 1504 Ans. 1893 First am't 1 5 0 4 First am't l 8 9 3 978 1652 Ans. 1504 Ans. 1893 4. 5. 6. 7. 8. 9. Dollars. Miles. Pounds. Rods. Inches. Feet. 11 47 127 678 789 1769 23 87 396 971 478 7895 97 58 787 -147 719 7563 86 83 456 716 937 8765 217 275 1766 2512 2923 25992 QUESTIONs.- Art 22. What is the general rule for addition? —Art. 23. What is the second method of proving addition? What is the reason of this method of proof?

Page 21 SECT. 1I.j ADDITION. 21 10. 11. 12. 13. 14. Ounces. Drams. Cents. Eagles. Degrees. 876 789 123 471 1234 376 567 478 617 3456 715 743 716 871 6544 678 435 478 317 7891 910 678 127 899 8766 15. 16. 17. 18. Feet. Inches. Hours. Minutes. 78956 71678 71123 98765 37667 12345 45678 12345 12345 6-7890 346 80 671 11 67890 34567 56777 33333 78999 89012 67812 71345 13579 78917 71444 99999 19. 20. 21. 22. Days. Years. Months. Hogsheads 17875897 789567 37 30176 7167512 7613 1378956 31 876567 123123 700714 8601 98765 700.71 367 11 7896 475 76117 9911 789 1069 4611779 89120 78 374176 9171 710 7 761 131765 4325 23. Add 1001, 76, 10078, 15, 8761, 7, and 1678. Ans. 21616. 24. Add 49, 761, 3756, 8, 150, 761761, and 18. Ans. 766503. 25. Required the sum of 3717, 8, 7, 10001, 58, 18, and 5. Ans. 13814. 26. Add 19, 181, 5, 897156, 81, 800, and 71512. Ans. 969754. 27. What is the sum of 999, 8081, 9, 1567, 88, 91, 7, and 878? Ans. 11720. 28. Add 71, 18765, 9111, 1471, 678, 9, 1446, and 71. Ans. 31622. 29. Add 51, 1, 7671, 89, 871787, 61, and 70001. Ans. 949661.

Page 22 22 ADDITION. [SECT. 11 30. What is the sum of 71, 8956, 1, 785, 587, and 76178? Ans. 86578. 31. Add 9999, 8008, 8, 81, 4777, and 516785. Ans. 539658. 32. Add 5, 7, 8911, 467, 47895, and 87. Ans. 57372. 33. Add 123456, 71, 8005, 21, and 716787. Ans. 848340. 34. Add 47, 911111, 717, 81, 88767, and 56. Ans. 1000779. 35. What is the -sum of 71, 8899, 4, 7111, and 678679? Ans. 694764. 36. Add 81, 879, 41, 76789, 42, 1, and 78967. Ans. 156800. 37. Add 917658, 75, 876789, 46, and 8222. Ans. 1802790. 38. Add 91, 76756895, 76, 14, 3, and 76378. Ans. 76833457. 39. Add 10, 100, 1000, 10000, 100000, and 1000000. Ans. 1111110. 40. What is the sum of 9, 99, 99, 1111, 8000, and 5? Ans. 9323. 41. Add 41, 7651, 7678956, 43, 15, and 6780. Ans. 7693486. 42. Add 1234, 7891, 3146751, 27, 9, and 5. Ans. 3155917. 43. What is the sum of 19, 91, 1, 1, 1478, 1007, and 46? Ans. 2643. 44. Add four hundred seventy-six, seventy-one, one hundred five, and three hundred eighty-seven. Ans. 1039. 45. Add fifty-six thousand seven hundred eighty-five, seven hundred five, thirty-six, one hundred seventy thousand and one, and four hundred seven. Ans. 227934. 46. Add fifty-six thousand seven hundred eleven, three thousand seventy-one, four hundred seventy-one, sixty-one, and three thousand and one. Ans. 63315. 47. What is the sum of the following numbers: seven hundred thousand seven hundred one, seventeen thousand nine, one million six hundred thousand seven hundred six, forty-seven thousand -six hundred seventy-one, seven thousand forty-seven, four hundred one, and nine? Ans. 2373544. 48. Gave 73 dollars for a watch, 15 dollars for a cane, 119 dollars for a horse, 376 dollars for a carriage, and 7689 dollars for a house; how much did they all cost? Ans. 8272 dollars.

Page 23 SECT. II.] ADDITION. 23 49. In an orchard, 15 trees bear plums, 73 trees bear apples, 29 trees bear pears, and 14 trees bear cherries; how many trees are there in the orchard? Ans. 131 trees. 50. The hind quarters of an ox weighed 375 pounds each, the tbre quarters 315 pounds each; the hide weighed 96 pounds, and the tallow 87 pounds. What was the whole weight of the ox? Ans. 1563 pounds. 51. A man bought a farm for 1728 dollars, and sold it so as to gain 375 dollars; how much did he sell it for? Ans. 2103 dollars. 52. A merchant bought five pieces of cloth. For the first he gave 376 dollars, for the second 198 dollars, for the third 896 dollars, for the fourth 691 dollars, and for the fifth 96 dollars. How much did he give for the whole? Ans. 2257 dollars. 53. A merchant bought five hogsheads of molasses for 375 dollars, and sold it so as to gain 25 dollars on each hogshead; for how much did he sell it? Ans. 500 dollars. 54. John Smith's farm is worth 7896 dollars; he has bank stock valued at 369 dollars, and he has in cash 850 dollars. How much is he worth? Ans. 9115 dollars. 55. Required the number of inhabitants in the New England States. By the census of 1850 there were in Maine 583,169, in New Hampshire 317,976, in Massachusetts 994,514, in Rhode Island 147,545, in Connecticut 370,792, and in Vermont 314,120. Ans. 2,728,116. 56. Required the number of inhabitants in the Middle States, including the District of Columbia. In 1850 there were in New York 3,097,394, in New Jersey 489,555, in Pennsylvania 2,311,786, in Delaware 91,532, in Maryland 583,034, and in the District of Columbia 51,687. Ans. 6,624,988. 57. Required the number of inhabitants in the Southern States. In 1850 there were in Virginia 1,421,661, in North Carolina 869,039, in South Carolina 668,507, in Georgia 906,185, and in. Florida 87,445. Ans. 3,952,837. 58. Required the number of inhabitants in the South-Western States. In 1850 there were in Alabama 771,623, in Mississippi 606,526, in Louisiana 517,762, in Texas 212,592, in Arkansas 209,897, and in Tennessee 1,002,917. Ans. 3,321,317. 59. Required the number of inhabitants in the North-Western States and Territories. In 1850 there were in Missouri 682,044, in Kentucky 982,405, in Ohio 1,980,329, in Indiana 988,416, in Illinois 851.470, in Michigan 397,654, in Wisconsin 305,391,

Page 24 24 ADDITION. [SECT. I in Iowa 192,214, in California 92,597, and in the Territories 92,298. Ans. 6,564,818. ART. 24, Method of adding two or more columns at a single operation. Ex. 1. Washington lived 68 years; John Adams, 91 years Jefferson, 83 years; Madison, 85 years. What is the sum of the years they all lived? Ans. 327. OPERATION. Years. Beginning with the number last written down, we add the units and tens, thus: 85 9 1 and 3 equal 88, and 80 equal 168, and 1 8 3 equal 169, and 90 equal 259, and 8 equal 8 5 267, and 60 equal 327, the sum sought. In like manner may be added more than Amount 3 2 7 two columns at one operation. NOTE. — The examples that follow can be performed as the above, or by the common method, or by both, as the teacher may advise. 2. 3. 4. 5. 6. Ounces. Yards. Feet. Inches. Chaldrons. 12834 2345 3456 7891 5678 5678 6789 7891 1356 3215 9012 1023 3456 7891 6789 3456 4456 7891 2345 3214 7890 7890 3456 6789 1 234 1345 1234 7890 1234 3789 6789 5678 1378 5678 1379 3216 9012 8123 9123 9008 7890 3456 4567 4567 1071 1030 7890 8912 8912 7163 7055 1345 3456 3456 6781 5678 6789 7891 7812 1780 1234 3456 3456 3456 3007 5678 7890 7891 7812 5617 9001 5678 3783 3713 4456 2345 9012 1237 7891 3456. 6789 3456 7891 1357 7891 1030 7890 1007 9009 3070 7816 1234 5670 8765 4567 1781 5678 1234 4321 3456

Page 25 3:c'r6. TIIo] SUBTRACTION. 2 ~ III. SUBTRACTION. MIENTAL EXERCISES. ART. 25, WurE it is required to find the difference between two numbers, the process is called Subtraction. The operation is the reverse of addition. Ex. 1. John has 7 oranges, and his sister but 4; how many more has John than his sister? ILLUSTRATION. - We first inquire what number added to 4 will make 7. From addition we learn that 4 and 3 are 7; consequently, if 4 oranges be taken from 7 oranges, 3 will remain Hence John has 3 oranges more than his sister. SUBTRACTION TABLE. 1 from 1 leaves 0 2 from 2 leaves 0 3 from 3 leaves 0 4 from 4 leaves 0 1 from 2 leaves 1 2 from 3 leaves 1 3 from 4 leaves 1 4 from 5 leaves 1 1 from 3 leaves 2 2 from 4 leaves 2 3 from 5 leaves 2 4 from 6 leaves 2 1 from 4 leaves 3 2 from 5 leaves 3 3 from 6 leaves 3 4 firom 7 leaves 3 1 from 5 leaves 4 2 from 6 leaves 4 3 from 7 leaves 4 4 from 8 leaves 4 1 from 6 leaves 5 2 from 7 leaves 5 3 from 8 leaves 5 4 from 9 leaves 5 1 from 7 leaves 6 2 from 8 leaves 6 3 from 9 leaves 6 4 from 10 leaves 6 1 from 8 leaves 7 2 from 9 leaves 7 3 from 10 leaves 7 4 from 11 leaves 7 1 firom 9 leaves 8 2 from 10 leaves 8 3 from 11 leaves 8 4 from 12 leaves 8 1 from 10 leaves 9 2 from 11 leaves 9 3 from 12 leaves 9 4 from 13 leaves 9 1 from 11 leaves 10 2 from 12 leaves 10 3 from 13 leaves 10 4 firom 14 leaves 10 1 from 12 leaves 11 2 from 13 leaves 11 3 from 14 leaves 11 4 from 15 leaves 11 1 from 13 leaves 12 2 from 14 leaves 12 3 from 15 leaves 12 4 from 16 leaves 12 5 fiom 5 leaves 0 6 from 6 leaves 0 7 from 7 leaves 0 8 from 8 leaves 0 5 fiom 6 leaves 1 6 from 7 leaves 1 7 from 8 leaves 1 8 fiom 9 leaves 1 5 from 7 leaves 2 6 from 8 leaves 2 7 from 9 leaves 2 8 from 10 leaves 2 5 from 8 leaves 3 6 from 9 leaves 3 7 from 10 leaves 3 8 from 11 leaves 3 5 from 9 leaves 4 6 fiom 10 leaves 4 7 from 11 leaves 4 8 from 12 leaves 4 5 fiom 10 leaves 5 6 from 11 leaves 5 7 from 12-leaves 5 8 from 13 leaves 5 5 from 11 leaves 6 6 from 12 leaves 6 7 from 13 leaves 6 8 from 14 leaves 6 5 firom 12 leaves 7 6 from 13 leaves 7 7 from 14 leaves 7 8 from 15 leaves 7 5 from 13 leaves 8 6 from 14 leaves 8 7 from 15 leaves 8 8 from 16 leaves 8 5 from 14 leaves 9 6 from 15 leaves 9 7 from 16 leaves 9 8 from 17 leaves 9 5 fiom 15 leaves 10 6 fiom 16 leaves 10 7 from 17 leaves 10 8 from 18 leaves 10 5 fiom 16 leaves 11 6 from 17 leaves 11 7 from 18 leaves 11 8 from 19 leaves 11 5 from 17 leaves 12 6 from 18 leaves 12 7 from 19 leaves 12 8 from 20 leaves 12 9 fiom 9 leaves 0 10 fiom 10 leaves 0 11 from 11 leaves 0 12 fiom 12 leaves 0 9 fiom 10 leaves 1 10 from 11 leaves 1 11 fiom 12 leaves 1 12 from 13 leaves 1 9 from 11 leaves 2 10 from 12 leaves 2 11 from 13 leaves 2 12 from 14 leaves 2 9 fiom 12 leaves 3 10 from 13 leaves 3 11 from 14 leaves 3 12 from 15 leaves 3 9 fiom 13 leaves 4 10 fi'om 14 leaves 4 11 from 15 leaves 4 12 fiom 16 leaves 4 9 from 14 leaves 5 10 from 15 leaves 5 11 from 16 leaves 5 12 fiom 17 leaves 5 9 from 15 leaves 6 10 from 16 leaves 6 11 from 17 leaves 6 12 from 18 leaves 6 9 from 16 leaves 7 10 fiom 17 leaves 7 11 fiom 18 leaves 7 12 fiom 19 leaves 7 9 from 17 leaves 8 10 fiom 18 leaves 8 11 from 19 leaves 8 12 from 20 leaves 8 9 from 18 leaves 9 10 from 19 leaves 9 11 from 20 leaves 9 12 from 21 leaves 9 9 from 19 leaves 10 10 fiom 20 leaves 10 11 fiom 21 leaves 10 12 from 22 leaves 10 9 fron 20 leaves 11 i 10 from 21 leaves 11 11 from 22 leaves 11 12 from 23 leaves 11 9 from 21 leaves 12! 10 from 22 leaves 12 11 from 23 leaves 12 12 from 24 leaves 12 QUESTIONS. - Art. 25. What does subtraction teach? Of what is it the reverse? 3

Page 26 )2b SUBTRACTION. [SECT. III. 2. Thomas had five oranges, and gave two of them to John how many had he left? 3. Peter had six marbles, and gave two of them to Samuel' how many had he left? 4. Lydia had four cakes; having lost one, how many had she left? 5. Daniel, having eight cents, gives three to Mary; how many has he left? 6. Benjamin had ten nuts; he gave four to Jane, and three to Emily; how many had he left? 7. Moses gives eleven oranges to John, and eight to Enoch: how many more has John than Enoch? 8. Paid seven dollars for a pair of boots, and two dollars for shoes; how much did the boots cost more than the shoes? 9. How many are 4 less 2? 4 less 1? 4 less 4? 10. How many are 4 less 3? 5 less 1? 5 less 5? 11. How many are 5 less 2? 5 less 3? 5 less 4? 12. How many are 6 less 1? 6 less 2? 6 less 4? 6 less 5? 13. How many are 7 less 2? 7 less 3? 7 less 4? 7 less 6? 14. How many are 8 less 6? 8 less 5? 8 less 2? 8 less 4? 8 less 1? 15. How many are 9 less 2? 9 less 4? 9 less. 5? 9 less 71 9 less 3? 16. How many are 10 less 8? lOless7? leOlss5? 10 less 3? 10 less 1? 17. How many are 11 less 9? 11 less 7? 11 less 5? 11 less 3? 11 less 4? 18. How many are 12 less T0? 12 less 8? 12 less 6? 12 less 4? 12 less 7? 19. How many are 13 less 11? 13 less 10? 13 less 7? 13 less 9? 13 less 5? 20. How many are 14 less 11? 14 less 9? 14 less 8? 14 less 6? 14 less 7? 14 less 3? 21. I-Jow many are 15 less 2? 15 less 4? 15 less 5? 15 less 7? 15 less 9? 15 less 13? 22. How many are 16 less 3? 16 less 4? 16 less 7? 16 less 9? 16 less 11? 16 lless 15? 23. How many are 17 less 1? 17 less 3? 17 less 5? 17 less 7? 17 less 8? 17 less 12? 24. How many are 18 less 2 18 less 4? 18 less 7? 18 less8? 18 less 10? 18 less 12? 25. I-ow many are 19 less I? 19 less 3? 19 less 5? 19 less 7? 19 less 9? 19 less 16?

Page 27 SECT. II.I SUBTRACTION. 27 26. How many are 20 less 5? 20 less 8? 20 less 9? 20 less 12? 20 less 15? 20 less 19? 27. Bought a horse for 60 dollars, and sold him for 90 dollars; h:w much did I gain? ILLUSTRATION. -- We may divide the two prices of the horse into tens, and subtract the greater from the less. Thus 60 equals 6 tens, and 90 equals 9 tens; 6 tens from 9 tens leave 3 tens, or 30. Therefore I gained 30 dollars. 28. Sold a wagon for 70 dollars, which cost me 100 dollars; how much did I lose? 29. John travels 30 miles a day, and Samuel 90 miles; what is the difference? 30. I have 100 dollars, and after I shall have given 20 to Benjamin, and paid a debt of 30 dollars to J. Smith, how many dollars have I left? 31. John Smith, Jr., had 170 dollars; he gave his oldest daughter, Angeline, 40 dollars, his youngest daughter, Mary, 50 dollars, his oldest son, James, 30, and his youngest son, William, 20 dollars; he also paid 20 dollars for his taxes; how many dollars had he remaining? AnT. 20, The pupil, having solved the preceding questions, will perceive, that SuBTRACTION is the taking of one number from another to find the difference. When the two numbers are unequal, the larger is called the Minuend, and the less number the Suhtrahend. The answer, or number found by the operation, is called the Difference, or ~Remainder. NOTE. - The words minuend and subtrahend are derived from two Latin words; the former from minuendunm, which signifies to be diminished or made tess, and the latter from subtrahendunm, which means to be subtracted or taken away. ART. 27 SIGNs. - Subtraction is denoted by a short horizontal line, thus -, signifying minus, or less. It indicates that the number following is to be taken from the one that precedes it. The expression 6 —2 = 4 is read, 6 minus, or less, 2 is equal to 4. QUESTIONS. - Art. 26. What is subtraction? What is the greater number called? What is the,tnb-S number called? What the answer? —Art. 27. What is the sign of subtraction? What does it signify and indicate?

Page 28 '2b SUBTRACTION. [SIc. HI EXERCISES FOR THE SLATME. ART. 28, Method of operation, when the numbers are large and each figure in the subtrahend is less than the figure above iA in the minuend. Ex. 1. Let it be required to take 245 from 468, and to find their difference. Ans. 223. OPERATION. We place the less number under the Minuend 4 6 8 greater, units under units, tens under Subtrahend 2 4 5 tens, &c., and draw a line below them. We then begin at the right hand, and Remainder 2 2 3 say, 5 units fi'om 8 units leave 3 units, and write the 3 in units' place below. We then say, 4 tens from 6 tens leave 2 tens, and write the 2 in tens' place below; and proceed with the next figure, and say, 2 hundreds from 4 hundreds leave 2 hundreds, which we write in hundreds' place below. We thus find the difference to be 223. ART. 29, First Ilethod of Proof. - Add the remainder and the subtrahend together, and their sum will be equal to the minuend, if the work is right. This method of proof depends on the principle, Thzat the greater of any two numbers is equal to the less added to the difference between them. EXAMPLES FOR PRACTICE. 2. 2. 3. 3. OPERATION. OPERATION AND PROOF. OPERATION. OPERATION AND PROOF. Minuend 5 4 7 5 4 7 9 8 6 9 8 6 Subtrahend 235 235 7 63 763 Remainder 312 312 223 223 Min. 5 4 7 Min. 9 86 4. 5. 6. 7. From 684 735 8 6 4 9 4 8 Take 4 6 2 5 2 3 6 5 1 7 4 6 8. A farmer paid 539 dollars for a span of fine horses, and sold them for 425 dollars; how much did he lose? Ans. 114 dollars. 9. A farmer raised 896 bushels of wheat, and sold 675 bushels of it; how much did he reserve for his own use? Ans. 221 bushels. QUESTIONS. -Art. 28. How are numbers arranged for subtraction? Where do you begin to subtract? Why?'Where do you write the difference? Art. 29. What is the first method of proving subtraction? What is the reason of this proof, or on what principle does it depend?

Page 29 SEC. 111.] SUBTRACTi()N. 29 10. A gentleman gave his son 3692 dollars, and his daughter 1212 dollars less than his son; how much did his daughter receive? Ans. 2480 dollars. ART. 306 Method of operation when any figure in the subtrahend is greater than the figure above it in the minuend. Ex. 1. If I have 624 dollars, and lose 342 of them, how many remain? Ans. 282. OPERATION. We first take the 2 units from the 4 units, Minuend 6 2 4 and find the difference to be 2 units, which we Subtrahend 3 4 2 write under the figure subtracted. We then proceed to tLake the 4 tens from the 2 tens Remainder 2 8 2 above it; but we here find a difficulty, since the 4 is greater than 2, and cannot be subtracted from it. WVe therefore add 10 to the 2 tens, which makes 12 tens, and then subtract the 4 from 12, and 8 tens remain, which we write below. Then, to compensate for the 10 thus added to the 2 in the minuend, we add one to the 3 hundreds in the next higher place in the subtrahend, which makes 4 hundreds, and subtract the 4 from 6 hundreds, and 2 hundreds remain. The remainder, therefore, is 282. The reason of this operation depends upon the self-evident truth, That, if any two numbers are equally increased, their difference remains the same. In this example 10 tons, equal to 1 hundred, were added:o the 2 tens in the upper number, and 1 was added to the 2 hunIreds in the lower number. Now, since the 3 stands in the hundreds' place, the 1 added was in fact 1 hundred. Hence, the two numbers being equally increased, the difference is the same. NonE. - This addition of 10 tb the minuend is sometimes called borrowing 10, and the addition of 1 to the subtrahend is called carrying 1. ART. 3o Froem the preceding examples and illustrations in subtraction, we deduce the following general RuLE. - Place the less number under the greater, so that units of the same order shall stand in the same column. Commencing at the right hand, subtract each figure of tie subtrahend from the figure above it..f any figure of the subtrahend is large,- than the figure above it in the minuend, add 10 to that figure of the minuend before subtracting, and then add 1 to the next figure of the subtrahend. QUESTIONs. - Art. 30. How do you proceed when a figure of the subtrahend is larger than the one above it in the minuend? How do you compensate for the 10 which is added to the minuend? What is the reason for this addition to the minuend and subtrahend? How does it appear that the 1 added to the subtrahend equals the 10 added to the minuend? What is the addition of 10 to the minuend sometimes called? The addition of 1 to the subtrahend? - Art. 31. What is the general rule for subtraction? Ssi.

Page 30 30 SUBTRACTION. [SEC. IIt. ART. 32, Second Method of Proof. - Subtract the remainder or difference from the minuend, and the result will be like the subtrahend if the work is right. This method of proof depends on the principle, That the smnaller of any two numbers is equal tb the remainder obtainea by subtracting their difference from the greater. EXAMPLES FOR PRACTICE. 2. 2. 3. 3. OPERATION. OPERATION AND PROOF. OPERATION. OPERATION AND PROOF Minuend 3 7 6 3 7 6 5 3 1 5 311 Subtrahend 1 6 7 16 7 3 8 9 3 8 9 Remainder 2 0 9 2 0 9 1 4 2 1 4 2 Sub. 167 Sub. 389 4. 5. 6. 7. Tons. Gallons. Pecks. Feet. From 978 67158 14711 1 00000 Take 199 14339 9197 90909 Ans. 779 52819 5 5514 9091 8. 9. 10. 11. Miles. Dollars. Minutes. Seconds. From67895 456798 765321 555555 Take 19999 19 0 899 9 1 7 7 777 1 7 7 7 7 7 12. 13. Rods. Acres. From100200300400500 1000000000000 Take 90807060504030 999999999999 14. From 671111 take 199999. Ans. 471112. 15. From 1789100 take 808088. Ans. 981012. 16. From 1000000 take 999999. Ans. 1. 17. From 9999999 take 1607. Ans. 9998392. 18. From 6101507601061 take 3806790989. Ans. 6097700810072. QUESTIONS. - Art. 32. What is the second method of proving subtraction? What is the reason for this method of proof, or on what principle does it depend?

Page 31 SECT. III.] SUBTRACTION. 31 19. From 8054010657811 take 76909748598. Ans. 7977100909213. 20. From 7100071641115 take 10071178. Ans. 7100061569937 21. From 501505010678 take 794090589. Ans. 500710920089. 22. Take 99'999999 from 100000000. A ns. 1. 23. Take 44444444 from 500000000. Ans. 455555556. 24. Take 1234567890 from 9987`654321. Ans. 8753086431. 25. From 800700567 take 1010101. Ans. 799690466. 26. Take twenty-five thousand twenty-five from- twenty-five millions. Ans. 24974975. 27. Take nine thousand ninety-nine from ninety-nine thousand. Ans. 89901. 28. From one hundred one millions ten thousand one hundred one take ten millions one hundred one thousand and ten. Ans. 90909091. 29. From one million take nine. Ans. 99991. 30. From three thousand take thirty-three. Ans. 2967. 31. From one hundred millions take five thousand. Ans. 99995000. 32. From 1,728 dollars, I paid 961 dollars; how many remain? Ans. 767 dollars. 33. Our national independence was declared in 1776; how many years from that period to the close of the last war with Great Britain, in 1815? Ans. 39 years. 34. The last transit of Venus was in 1769, and the next will be in 1874; how many years will intervene? Ans. 105 years. 35. The State of New Jersey contains 6851 square miles, and Delaware 2120. How many more square miles has the former State than the latter? Ans. 4731. 36. In 1840 the number of inhabitants in the United States was 17,069,453, and in 1850 it was 23,191,876; what was the increase? Ans. 6,122,423. 37. In 1850 there were raised in the State of Ohio 56,619,608 bushels of corn, and in 1853, 73,436,690 bushels; what was the increase? Ans. 16,817,082 bushels. 38. By the census of 1850, 13,121,498 bushels of wheat were raised in New York, and 15,367,691 bushels in Pennsylvania; how many bushels in the latter State more than in the former? Ans. 2,246,193 bushels.

Page 32 32 SUBTRACTION. [EECT. II1 39. The city of New York owes 13,960,856 dollars, and Boston owes 7,779,855 dollars; how much more does New York owe than Boston? Ans. 6,181,001 dollars. 40. From five hundred eighty-one thousand take three thou sand and ninety-six. Ans. 577,904. 41. It was ascertained by a transit of Venus, June 3, 1769 that the mean distance of the earth from the sun was ninety-five millions one hundred seventy-three thousand one hundred twentyseven miles, and that the mean distance of Mars firom the sun was one hundred forty-five millions fourteen thousand one hundred forty-eight miles. Required the difference of their distances from the sun. Ans. 49,841,021 miles. ART. 3. Method of subtracting when there are two or more subtrahends. Ex. 1. A man owing 767 dollars, paid at one time 190 dollars, at another time 131 dollars, at. another time 155 dollars; how much did he then owe? Ans 291 dollars. FIRST OPERATION. SECOND OPERATION. In the first opDollars. Dollars. eration, the several Minuend 7 6 7 Minuend 7 6 7 subtrahends, having been properly 1`f 9 O 1 trzthe S1 3 1 placed, are added 1 9 0 Subtrahends 1 9 0 for a single subtra15 5 155 llend, to be taken from the minuend. Subtrahend:4 7 6 Remainder 2 91 In the second In the second, the Remainder 2 9 1 subtrahends are subtracted, as -they are added, at one operation, thus: beginning with units, 5 and 1 equal 6, which from 7 units leaves 1 unit; passing to tens, 5 and 9 and 3 equal 17 tens; reserving the left-hand figure to add in with the figures of the subtrahends in the nexst column, the right-hand figure, 7, being larger than 6 tens of the minuend, we add 10 to the 6, an1, subtracting, have left 9 tens; and, passing to hundreds, we add in the left-hand figure 1 reserved from the 17 tens, and also add 1, equal 10 tens, to compensate for the 10 added to the minuend, and with the other figures, 1 and 1 and 1 equal 5 hundreds, which, takeirsn firom 7 hundreds, leave 2 hundreds; and 291 as the answer sought. 2. E. Webster owned 6,765 acres of land, and he gave to his oldest brother 2,196 acres, and his uncle Rollins 1,981 acres; how much has he left? Ans. 2,588 acres. 3. The real estate of James Dow is valued at 3,769 dollars, and his personal estate at 2,648 dollars; he owes John Smith 1,728 dollars, and Job Tyler 1,161 dollars; how much is Dow worth? Ans. 3,1'528 dollars.

Page 33 SECT. IV. MULTIPLICATION. 83 $ IV. MULTIPLICATION. MENTAL EXERCISES. ART. 34 WIIEN any number is to be added to itself several times, the operation may be shortened by a process called M1ultiplication. Ex. 1. If a man can earn 8 dollars in 1 week, what will he earn in 4 weeks? ILLUSTiATION. -It is evident, since a man can earn 8 dollars in 1 week, in 4 weeks he will earn 4 times as much, and the result may be obtained by addition; thus, 8 + 8 -+- + 8 32; or, by a more convenient process, by setting down the 8 but once, and multiplying it by 4, the number of times it is to be repeated; thus, 4 times 8 are 32. Hence in 4 weeks he will earn 32 dollars. MULTIPLICATION TABLE. 2 times 1 are 2 3times 1 are 3 4 times 1 are 4 5 times l are 5 2 times 2 are 4 3 tiies 2 are 6 4 times 2 are 8 5 times 2 are 10 2 times 3 are 6 3 times 3 are 9 4 times 3 are 12 5 times 3 a-e 15 2times 4are 8 3 times 4 are 12 4 times 4 are 16 5 times 4 are 20 2 times S are 10 3 times 5 are 15 4 times 5 are 20 5 times 5 are 25 2 times 6 are 12 3 times 6 are 18 4 times 6 are 24 5 times 6 are 30 2 times 7 are 14 3 times 7 are 21 4 times 7 are 28 5 times 7 are 35 2 times 8 are 16 3 times 8 are 24 4 times 8 are 32 5 times 8 are 40 2 times 9 are 18 3 times 9 are 27 4 times 9 are 36 5 times 9 are 45 2 timles 10 are 20 3 times 10-are 30 4 times 10 are 40 5 times 10 are 50 2 times 11 are 22 3 times 11 are 33 4 times 11 are 44 5 times 11 are 55 2 times 12 are 24 3 times 12 are 36 4 times 12 are 48 5 times 12 are 60 6 times 1 are C 7 times 1 are 7 8 times 1 are 8 9 times I are 9 6times 2 are I, 7 times 2 are 14 8 times 2 are 16 9 times 2 a:re 18 6 times 3 are 18 7 times 3 are 21 8 times 3 are 24 9 times 3 are 27 6 tiles 4 are 24 i 7 times 4 iae'8 8 tilnes 4 are 32 9 timles 4 ale 36 6 tilres 5 are 30'7 times 5 are 35 8 times 5 are 40 9 times 5 are 45 6 times 6 are 36 1 ti;nes 6 are 42 8 times 6 are 48 9 times 6 are 54 6 times 7 are 42 7 timJes 7 are 4-9 8 times 7 are 56 9 times 7 are 63 6 timles 8 are 48 7 times 8 are 56 8 times 8 are 64 9 times 8 are 72 6 timens 9 are 54 7 times 9 are 63 8 times 9 are 72 9 times 9 are 81 6 times 10 are 60 7 times 10 arle 70 8 times 10 are 80 9 times 10 are 90 6 times 11 are 66 7 times 11 are 77 8 times 11 are 88 9 times 11 are 99 6 times 12 are 2 i 7times 12 are 84 8 times 12 are 96 9 times 12 are 108 10 times 1 are 10 10 times 11 are 110 11 times 8 are 88 12 times 4 lare 43 10 times 2 are 20 10 times 12 are 120 11'imnes 9 are 99 12 times 5 are 60 10 times 3 are O0 - 11 times 10 irle 110 12 times 6 arle 72 10 times 4 are 40 11 times 1 are 11 11 timnes 11 are 121 12 times 7 are 84 10 times 5 are 50 11 times 2 al'e 22 11 timtres 12 are 132 12 tines 8 are 96 10 times 6 are 60 11 timles 3 are 33 - 12 tinmes 9 are 108 10 tiiles 7 are 70 1i times 4 acr 44 12 times I are 12 12 times 10 are 120 10 times 8 si'e 80 11 times 5 are 55 12 tines 2 are 24 12 times 1i s're 1]:2 10 times 9 are 90 11 times 6 are 66 12 times 3 are 36 12 times 12 are 144 10 tilmes iO are 100 11 imes 7 are 77, QuFEsio'l -N Art. 34. lfow may the process of adding a number to itself several times be shortened?

Page 34 34 MULTIPLICATION. [SECT. IV, 2. What cost 5 barrels of flour at 6 dollars per barrel? ILLUSTRATION. - If a barrel of flour cost 6 dollars, 5 barrels will cost 5 times as much; 5 times 6 are 30. Hence 5 barrels of flour at 6 dollars per barrel will cost 30 dollars. 3. What cost 6 bushels of beans at 2 dollars per bushel? 4. What cost 5 quarts of cherries at 7 cents per quart? 5. WVhat will 7 quarts of vinegar cost at 12 cents per quart? 6. WVhat cost 9 acres of land at 10 dollars per acre? 7. If a pint of currants cost 4 cents, what cost 9 pints? 8. If in 1 penny there are 4 farthings, how many in 9 pence? In 7 pence? In 8 pence? In 4 pence? In 3 pence? 9. If 12 pence make a shilling, how many pence in 3 shillings? In 5 shillings? In 7 shillings? In 9 shillings? 10. If one pound of raisins cost 6 cents, what cost 4 pounds? 5 pounds? 6 pounds? 7 pounds? 8 pounds? 9 pounds? 10 pounds? 12 pounds? 11. In one acre there are four roods; how many roods in 2 acres? In 3 acres? In 4 acres? In 5 acres? In 6 acres? In 9 acres? 12. A good pair of boots is worth 5 dollars; what must I give for 5 patirs? For 6 pairs? For 7 pairs? For 8 pairs? For 9 pairs? 13. A cord of good walnut wood may be obtained fbr 8 dollars; what must I give for 4 cords? For 6 cords? For 9 cords? 14. What cost 4 quarts of milk at 5 cent, a quart, and 8 gallons of vinegar at 10 cents a gallon? 15. If a man earn 7 dollars a week, how much will he earn in 3 weeks? In 4 weeks? In 5 weeks? In 6 weeks? In 7 weeks? In 9 weeks? 16. If 1 thousand feet of boards cost 12 dollars, what cost 4 thousand?. 5 thousand? 6 thousand? 7 thousand? 9 thousand? 12 thousand? 17. If 3 pairs of shoes buy I pair of boots, how many pairs of shoes will it take to buy, 7 pairs of boots? 18. If 5 bushels of apples buy 1 barrel of flour, how many bushels of apples are equal in value to 12 barrels of flour? 19. If 1 yard of canvas cost 25 cents, what will 12 yards cost? ILLUSTRATION. -The number 25 is composed of 2 tens and 5 units; 12 tines 2 tens are 24 tens; and 12 times 5 units are

Page 35 bECT. IV. ] MIULTIPLCATlO.N. 35 60 units, or 6 tens, 24 tens added to 6 tens make 30 tens, or 300. Therefore, 12 yards will cost 300 cents, or 3 dollars. 20. In 1 pound there are 20 shillings; how many shillings in 3 pounds? In 4 pounds? In 6 pounds? 21. A gallon of molasses is worth 25 cents; what is the value of 2 gallons? Of 3 gallons? Of 4 gallons? Of 5 gallons? Of 6 gallons? Of 9 gallons? 22. If 1 man earn 12 dollars in 16 days, how much would 10 men earn in the same time? 23. If a steam-engine runs 28 miles in 1 hour, how far will it run in 4 hours? In 6 hours? In 9 hours? 24. If the earth turns on its axis 15 degrees in 1 hour, how far will it turn in 7 hours? In 11 hours? In 12 hours? 25. In a certain regiment there are 8 companies, in each company 6 platoons, and in each platoon 12 soldiers; how many soldiers are there in the regiment? 26. If 1 man walk 7 miles in 1 hour, how far will he walk in 8 hours? In 9 hours? In 11 hours? In 12 hours? In 20 hours? In 30 hours? ART. 35, The learner, having performed the foregoing questions, will perceive that JMULTIPLICATION is the process of taking a number as many times as there are units in another number. In multiplication three terms are employed, called the Multiplicand, the Multiplier, and the Product. The multiplicand is the number to be multiplied or taken. The multiplier is the number by which we multiply, and denotes the number of times the multiplicand is to be taken. The product is the result, or number produced by the multiplication. The multiplicand and multiplier are often called FACTORs. SIGNS. — The sign of multiplication is formed by two short lines crossing each other obliquely; thus, X. It shows that the numbers between which it is placed are to be multiplied together; thus, the expression 7 X 5 = 35 is read, 7 multiplied by 5 is equal to 35. QUESTIONS. - Art. 35. What is multiplication? What three terms are employed? What is the multiplicand? What is the multiplier? What is the product? What are the multiplicand and multiplier often called? What is the sign of multiplication? What does it show?

Page 36 36) MULTIPLICATION [SECT. IV. EXERCISES FOR THE SLATE. ART. 36. Method of operation when the multiplier does not exceed 12. Ex. 1. Let it be required to multiply 175 by 7. Ans. 1225. OPERATION. SHaving written the multiplier under BMultiplicand 1 7 5 the unit figure of the multiplicand, we Multiplier 7 multiply the 5 units by 7, obtaining 35, - and set down the 5 units directly under Product 1 2 2 5 the 7, and reserve the 3 tens for the tens' column. We then multiply the 7 tens by 7, obtaining 49, and, adding the 3 tens which were reserved, we have 52 tens, or 5 hundreds and 2 tens. Writing down the 2 tens, and reserving the 5 hundreds, we multiply 1 by 7; and, adding the reserved 5 hundreds, we have 12 hundreds, which we write down in full, and the product is 1225. EXAMPLES FOR PRACTICE. 2. 3. 4. Multiply 8756 456 7 7896 By 4 3 5 Ans. 3 5 0 2 4 1 3 7 0 1 3 9 4 8 0 5. 6. 7. 8. 56807 -47893 61657 89765 5 6 7 9 284035 287358 431599 807885 9. Multiply 767853 by 9. Ans. 6910677. 10. Multiply 876538765 by 8. Ans. 7012310120. 11. Multiply 7654328 by 7. Ans. 53580296. 12. Multiply 4976387 by 5. Ans. 24881935. 13. Multiply 8765448 by 12. Ans. 105185376. 14. Multiply 4567839 by 11. Ans. 50246(229. 15. What cost 8675 barrels of flour at 7 dollars per barrel? Ans. 60725 dollars. QUESTIONS. - Art. 36. How must numbers be written for multiplication? At which hand do you begin to multiply? Why? Where do you write the first or right-hand figure of the product of each figure in the multiplicaud? Why? What is done with the tens or left-hand figure of each product? fNow, then, do you proceed when the multiplier does not exceed 12?

Page 37 ECT. I1V M ULTIPLICATION. 37 16. What cost 25384 tons of hay at 9 dollars per ton? Ans. 228456 dollars. 17. If, on 1 page in this book, there are 2538 letters, how many are there on 11 pages? Ans. 27918 letters. ART. 37, Method of operation when the multiplier exceeds 12. Ex. 1. Let it be required to multiply 763 by 24. Ans. 18312, OPERATION. bWe write the multiplier under the Multiplicand 7 6 3 multiplicand, and proceed to multiply Multiplier 2 4 the multiplicand by 4, the unit figure of the multiplier, precisely as in Art. 36. 3 0 5 2 We then, in like manner, multiply the 1 5 2 6 multiplicand by the 2 tens in the multiplier, taking care to write the first figure Product 1 8 3 1 2 obtained by this multiplication in tens' column, directly under the 2 of the mul tiplier; and, adding together these partial products obtained by the two multiplications, and placed as in the operation, we have the full product of 763 multiplied by 24, which is 18312. ART. 38, The preceding examples sufficiently illustrate the principle and method of multiplication; hence the following general RULE. - Write the multiplier under the multsplicand, arranging units under units, tens under tens, 4-c. Multiply each figure of the multiplicand by each figure of the multiplier, beginning with the right-hand figure, writing the right-hand figure of each product under the figure multiplied, and adding the left-hand Jigure or figures, if any, to the succeeding product. If the multiplier consists of More than one figure, the right-hand figure of each partial product must be placed directly under the figure of the multiplier that produces it. The sum of the partial products will be the whole product required. NOTE. - When there are ciphers between the significant figures of the multiplier, pass over them in the operation, and multiply by the significant figures only, remembering to set the first figure of the product directly under the figure of the multiplier that produces it. QuEsTioNs. - Art. 37. How do you proceed when the multiplier exceeds 12 Where do you set the first figure of each partial product? Why'? How is the true product found? —Art. 38. What is the general rule for multiplication? When there are ciphers between the significant figures of the multiplier, how do you proceed? 4

Page 38 38 MULTTIPiLICATION. [SECT. I[V. ART. 39a First Method of Proof. - Multiply the multiplier by the multiplicand, and if the result is like the first product the work is supposed to be right. The reason of this proof depends on the principle, That, when two or more numbers are multiplied together, the product is the same, whatever the order of multiplying them. Ex. 2. Multiply 7895 by 56. Ans. 442120. OPERATION. PROOF. Multiplicand 7 8 9 5 5 6 Multiplier 5 6 7 8 9 5 47370 280 39475 504 448 Product 4 4 2 1 2 0 392 Product 44212 0 NoE. - The common mode of proof in business is to divide the product by the multiplier, and, if the work is right, the quotient will be like the multiplicand. This mode of proof anticipates the principles of division, and therefore cannot be employed without a previous knowledge of that rule. ART. 40o Second Method of Proof. - Begin at the left hand of the multiplicand, and add together its successive figures toward the right till the sum obtained equals or exceeds the number nine. If it equals it, drop the nine, and begin to add again at this point, and proceed till you obtain a sum equal to, or greater than, nine. If it exceeds nine, drop the nine as before, and carry the excess to the next figure, and then continue the addition as before. Proceed in this way till you have added all the figures in the multiplicand and rejected all the nines contained in it, and write the final excess at the right hand of the multiplicand. Proceed in the same manner with the multiplier, and write the final excess under that of the multiplicand. iMultiply these excesses together, and place the excess of nines in their product at the right. Then proceed to find the excess of nines in the product obtained by the original operation; and, if the work is right, QUESTIoNs. - Art. 39. How is multiplication proved by the first method? What is the reason for this method of proof? What is the common mode of proof in business? - Art. 4Q0. What is the second method of proving ruulti. plication?

Page 39 SECT. IV.] MULTIPLICATION. 39 the excess thus found will be equal to the excess contained in the product of the above excesses of the multiplicand and mlultiplier. Ex. 3. OPEIRATION. Multiplicand 12 3 4 5 = 6 excess. Multiplier 2 2 31 = 8 excess. 1 2345 4 8 3 37035 ______2 4 6 910 Proof. 2469-0 PI Aduct 27541695 = 3) NOTE. - This method of proof, though perhaps sufficiently sure for common purposes, is not always a test of the correctness of an operation. If two or more figures in the work should be transposed, or the value of one figure be just as much too great as another is too small1 or if a nine be set down in the place of a cipher, or the contrary, the excess of nines will be the same, and still the work may not be correct. Such a balance of errors will not, however, be likely to occur. EXAMPLES FOR PRACTICE. 4. 5. Multiply 6 7 8 9 5 78956 By 36 47 407370 552692 203685 315824 Ans. 2444220 371093 2 6. 7. TMultiply 89325 47 896 By 901 2008 89325 383168 803925 95792 Ans. 80481825 96175168 8. What cost 47 hogsheads of molasses at, 13 dollars per hogshead? Ans. 611 dollars. 9. What cost 97 oxen at 29 dollars each? Ans. 2813 dollars. QUESTIONS. - IS this method of proof always a true test of the correctness of an operation? What is the reason for this method of proof'?

Page 40 40 MULTIPLICATION. [SECT. IV. 10. Sold a farm containing 367 acres, at 97 dollars per acre what was the amount? Ans. 35599 dollars. 11. An army of 17006 men receive each 109 dollars as theix annual pay; what is the amount paid the whole army'? Ans. 1853654 dollars. 12. If a mechanic deposit annually in the Savings Bank 407 dollars, what will be the sum deposited in 27 years? Ans. 10989 dollars. 13. If a man travel 37 miles in 1 day, how far will he travel in 365 days? Ans. 13505 miles. 14. If there be 24 hours in 1 day, how many hours in 365 days? Ans. 8760 hours. 15. How many gallons in 87 hogsheads, there being 63 gallons in each? Ans. 5481 gallons. 16. If the expenses of the Massachusetts Legislature be 1839 dollars per day, what will be the amount in a session of 109 days? Ans. 200451 dollars. 17. If a hogshead of sugar contains 368 pounds, how many pounds in 187 hogsheads? Ans. 68816 pounds. 18. Multiply 675 by 476. Ans. 321300. 19. Multiply 679 by 763. Ans. 518077. 20. Multiply 899 by 981. Ans. 881919. 21. Multiply 7854 by 1234. Ans. 9691836. 22. Multiply 3001 by 6071. Ans. 18219071. 23. Multiply 7117 by 9876. Ans. 70287492. 24. Multiply 376546 by 407091. Ans. 153288487686. 25. Multiply 7001009 by 7007867. Ans. 49062139937803. 26. Multiply five hundred and eighty-six by nine hundred and eight. Ans. 532088. 27. Multiply three thousand eight hundred and five by one thousand and seven. Ans. 3831635. 28. Multiply two thousand and seventy-one by seven hundred and six. Ans. 1462126. 29. Multiply eighty-eight thousand and eight by three thousand and seven. Ans. 264640056. 30. Multiply ninety thousand eight hundred and seven by one thousand and ninety-one. Ans. 99070437. 31. Multiply ninety thousand eight hundred and seven by nine thousand one hundred and six. Ans. 826888542. 32. Multiply fifty thousand and one by five thousand eight hundred and seven. Ans. 290355807. 33. Multiply eighty thousand and nine by nine thousand and sixteen. Ans. 721361144.

Page 41 SECT. IV. j MULTIPLICATION. 41 34. Multiply forty-seven thousand and thirteen by eighty thousand eight hundred and seven. Ans. 3798979491. ART. o1. A COMPOSITE number is one produced by multiply. ing together two or more numbers greater than unity or one; thus, 12 is a composite number, since it is the product of 3 X 4; and also 24 is a composite number, since it is the product of 2X3 X4. A FACTOR of any number is a name given to one of two or more numbers, which, being multiplied together, produce that number; thus, 3 and 4 are factors of 12, since 3 X 4 = 12. ART. 42. To multiply by a composite number. Ex. 1. A merchant bought fifteen pieces of broadcloth, at 96 dollars per piece; how much did he pay for the whole? Ans. 1440 dollars. The factors of 15 are 3 and 5. Now, if we multiply the 9 6 dolls., price of 1 piece. price of one piece by the factor 3 3, we get the cost of 3 pieces; 2 8 8 dolls., price of 3 pieces. and then, by multiplying the and then, by multiplying the 2 8 8dolls., price of pieces. cost of 3 pieces by the factor 5 5, it is evident we obtain the 1 4 4 0 dolls., price of 15 pieces. cost of 15, the number of pieces bought, since 15 is equal to 5 times 3. Hence we adopt the following RULE. - 1Multiply the multiplicand by one of the factors of the multiplier, and the product thus obtained by another, and so on until each of the factors has been used as a multiplier. The last product will be the answer. NoTE. - The product of any number of factors is the same in whatever order they are multiplied. Thus, 3 X 4 = 12, and 4 X 3 = 12. EXAMPLES FOR PRACTICE. 2. Multiply 30613 by 25 = 5 X 5. Ans, 765325. 3. Multiply 1469 by 84 = 7 X 12. Ans. 123396. 4. Multiply 7546 by 81, using its factors. Ans. 611226. 5. Multiply 7901 by 125, using its factors. Ans. 987625. 6. In 1 mile there are 63360 inches; how many inches in 45 miles? Ans. 2851200 inches. 7. If in one year there are 8766 hours, how many hours in 72 years? Ans. 631152 hours. QUESTIONS. - Art. 41. What is a composite number? What is a factor of any number? - Art. 42. What are the factors of 15? How do we multiply by a composite number? Repeat the rule. In what order must the factors of a composite number be multiplied? 4*

Page 42 42 MULTIPLICATION. [SECT. IV 8. If sound moves 1142 feet in a second, how far will it move in one minute? Ans. 68520 feet. ART. 43. When the multiplier is 1 with one or more ciphers annexed to it, as 10, 100, &c. Ex. 1. In 1 day there are 24 hours; how many hours in lO days? In 100 days? Answers. 240, 2400. OPERATION. The removal of a figure one Multiplicand 2 4 2 4 place to the left increases its Multiplier 1 0 1 0 0 value ten times. (Art. 7.) If, Product 20 2 0 then, we annex a cipher to the Or 2 4th,240 240 —0 ~ right of 24, the multiplicand, Or thus, 240, 2400. each figure is removed one place to the left, and its value is increased ten times, or multiplied by 10. If two ciphers are annexed, each figure is removed two places to the left, and its value is increased 100 times, or multiplied by 100; every additional cipher increasing the value ten times. Hence the following RULE. - Annex to the multiplicand as many cilp7ers as has the multiplier. The number thus formed will be the product required. EXAMPLES FOR PRACTICE. 2. Multiply 2356 by 10. Ans. 23560. 3. Multiply 5873 by 100. Arns. 587300. 4. Multiply 7964 by 1000. Ans. 7964000. 5. Multiply 98725 by 100000. Ans. 9872500000. ART. 44. When there are ciphers on the right hand of the multiplier or multiplicand, or both. Ex. 1. What will 600 acres of land cost at 20 dollars per acre? Ans 12,000 dollars. OPERATION. The multiplicand may be resolved into Multiplicand 6 00 the factors 6 and 100, and the multiplier Multiplier 2 0 into the factors 2 and 10. Now, it is evident (Art. 42), if these several factors be Product 1 2 0 0 0 multiplied together, they will produce the same product as the original factors 600 and 20. Thus 6 X 2 =12, and 12 X 100 = 1200, and 1200 X 10 a 12000, the same result as in the operation. Hence the following QUESTIONS. - Art. 43. What is the effect of removing a figure one place to the left.? What is the effect of annexing a cipher to any figure or number? Two ciphers? &c. What is the rule when the multiplier is 1 with ciphers annexed? - Art 44. How do you arrange the figures for multiplication, when there are ciphers on the right hand of either the multiplier or multiplicand, or both? Why does multiplying the significant figures and annexing the ciphers produce the true product?

Page 43 bECT. IV.] MULTIPLICATION. 43 RULE. - Write the significant figures of the multiplier under those of the multiplicand, and multiply them together. To their product annex as many ciphers as there are on the right of both multiplicand and multiplier. EXAMPLES FOR PRACTICE. 2. 3. Multiply 8785324 713378900 By 3200 70080 17570648 57070312 26355972 49936523 Ans. 28113036800 49993593312000 4. Multiply 801070.0 by 9000909. Ans. 72103581726300. 5. Multiply 700110000 by 700110000 Ans. 490154012100000000. 6. Multiply 4070607 by 7007000. Ans. 28522743249000. 7. Mulitiply 4110000 by 1017010. Ans. 4179911100000. 8. Multiply twenty-nine millions two thousand nine hundred and nine by four hundred and four thousand. Ans. 11717175236000. 9. Multiply eighty-seven millions by eight hundred thousand seven hundred. Ans. 69660900000000. 10. Multiply one million one thousand one hundred by nine hundred nine thousand and ninety. Ans. 910089999000. 11. Multiply forty-nine millions and forty-nine by four hundred and ninety thousand. Ans 24010024010000. 12. Multiply two hundred millions two hundred by two millions two thousand and two. Ans. 400400800400400. 13. Multiply four millions forty thousand four hundred by three hundred three thousand. Ans. 1224241200000. 14. Multiply three hundred thousand thirty by forty-seven thousand seventy. Ans. 14122412100. 15. Multiply fifteen millions one hundred by two thousand two hundred. Ans. 33000220000. 16. Multiply one billion twenty thousand by one thousand one hundred. Ans. 1100022000000. QUEsTIoN. - What is the rule?

Page 44 44 DIVISION. [S CT. V ~ V. DIVISION. MENTAL EXERCISES. ART. 45, WHEN it is required to find how many times one number contains another, the process is called Division. Ex. 1. A boy has 32 cents, which he wishes to give to 8 of his companions, to each an equal number; how many must each receive? ILLUSTRATION. - It is evident that each boy must receive as many cents as the number 8 is contained times in 32. We therefore inquire what number 8 must be multiplied by to make 32. By trial, we find that 4 is the number; because 4 times 8 make 32. Hence 8 is contained in 32 4 times, and the boys receive 4 cents apiece. The following table should be studied by the learner to aid him in solving questions in division: DIVISION TABLE. 2 in 2 1 time 3 in 3 1 time 4 in 4 1 time 5 in 5 Itime 2 in 4 2 times 3 in 6 2 times 4 in 8 2 times 5 in 10 2 times 2 in 6 3 times 3 in 9 3 times 4 in 12 3 times 5 in 15 3 times 2 in 8 4 times 3 in 12 4 times 4 in 16 4 times 5 in 20 4 times 2 in 10 5 times 3 in 15 5 times 4 in 20 6times 5 in 25 5 times 2 in 12 6times 3 in 18 6 times 4 in 24 6times 5 in 30 6times 2 in 14 7 times 3 in 21 7 times 4 in 28'7 times 5 in 35 7 times 2 in 16 8 times 3 in 24 8 times 4 in 32 8times 5 in 40 8 times 2 in 18 9 times 3 in 27 9 times 4 in 36 9times 5 in 45 9times 2 in 20 10 times 3 in 30 10 times 4 in 40 10 times 5 in 50 10 times 2 in 22 11 times 3 in.33 11 times 4 in 44 11 times 5 in 55 11 times 2 in 24 12 times 3 in 36 12 times i in 48 12 times 5 in 60 12 times 6 in 6 1time 7 in 7 1 time 8 in 8 1 time 9 in 9 1 time 6 in 12 2times 7 in 14 2times 8 in 16 2times 9 in 18 2times 6 in 18 3 times 7 in 21 3 times 8 in 24 3 times 9 in 27 3 times 6 in 24 4 times 7 in 28 4 times 8 in 32 4 times 9 in 36 4 times 6-in 30 5 times 7 in 35 6 times 8 in 40 5 times 9 in 45 5 times 6 in 36 6times 7 in 42 6 times 8 in 48 6 times 9 in 54 6 times 6 in 42 7 times 7 in 49 7. times 8 in 56 7 times 9 in 63 7 times 6 in 48 8 times 7 in 56 8 times 8 in 64 8 times 9 in 72 8 times 6 in 54 9 times 7 in 63 9 times 8 in 72 9 times 9 in 81 9 times 6 in 60 10 times 7 in 70 10 times 8 in 80 10 times 9 in 90 10 times 6 in 66 11 times 7 in 77 11 times 8 in 88 11 times 9 in 99 11 times 6 in 72 12 times 7 in 84 12 times 8 in 96 12 times 9 in 108 12 times 10 in 10 1 time 10 in 110 11 times 11 in 88 8 times 12 in 48 4 times 10 in 20 2 times 10 in 120 12 times 11 in 99 9 times 12 in 60 5 times 10 in 30 3 times 11 in 110 10 times 12 in 72 6 times 10 in 40 4 times 11 in 11 1 time 11 in. 121 11 times 12 in 84 7 times 10 in 50 5 times 11 in 22 2 times 11 in 132 12 times 12 in 96 8 times 10 in 60 6 times 11 in 33 3 times 12 in 108 9 times 10 in 70 7 times 11 in 44 4 times 12 in 12 1 time 12 in 120 10 times 10 in 80 8 times 11 in 5 5 times 12 in 24 2 times 12 in 132 11 times 10 in 90 9 times 11 in 66 6 times 12 in 36 3 times 12 in 144 12 times 10 in 100 10 times 11 in 77 7 times

Page 45 BECT. V.] DIVISION. 45 2. A farmer received 8 dollars for 2 sheep; what was the price of each? ILLUSTRATION. - It is evident, since he received 8 dollars for 2 sheep, for 1 sheep he must receive as many dollars as 2 is contained times in 8. 2 is contained in 8 4 times, because 4 times 2 are 8; hence 4 dollars was the price of each sheep. 3. A man gave 15 dollars for 3 barrels of flour; what was the cost of each barrel? 4. A lady divided 20 oranges among her 5 daughters; how many did each receive? 5. If 4 casks of lime cost 12 dollars, what costs 1 cask? 6. A laborer earned 48 shillings in 6 days; what did he receive per day? 7. A man can perform a certain piece of labor in 30 days; how long will it take five men to do the same? 8. When 72 dollars are paid for 8 acres of land, what costs 1 acre? What cost 3 acres? 9. If 21 pounds of flour can be obtained for 3 dollars, how much can be obtained for 1 dollar? How much for 8 dollars? How much for 9 dollars 7 10. Gave 56 cents for 8 pounds of raisins; what costs 1 pound? What cost 7 pounds? 11. If a man walk 24 miles in 6 hours, how far will he walk in 1 hour? How far in 10 hours? 12. Paid 56 dollars for 7 hundred weight of sugar; what costs 1 hundred weight? What cost 10 hundred weight? 13. If 5 horses will eat a load of hay in 1 week, how long would it last I horse? 14. In 20, how many times 2? How many times 4? How many times 5? How many times 10? 15. In 24, how many times 3? How many times 4? How many times 6? How many times 8? 16. How many times 7 in 21? In 28? In 56? In 35? In 14? In In 77? In 70? In 84? 17. How many times 6 in 12? In 36? In 18? In 54? In 60? In 42? In 48? In 72? In 66? 18. How many times 9 in 27? In 45? In 63? In 81? In99? In 108? 19. How many times 11 in 22? In 55? In 77? In 88? In 110? In 132? 20. How many times 12 in 36? In 60? In 72? In 84? In 120 In 144?

Page 46 46 DIVISION. [SECT. V ART. 46, The pupil will now perceive that DIvisION is the process of finding how many times one number is contained in another. In division there are three principal terms': the Dividend the Divisor, and the Quotient, or answer. The dividend is the number to be divided. The divisor is the number by which we divide. The quotient is the number of times the divisor is contained in the dividend. NOTE. - Quotient is derived from the Latin word qusoties, which signi fies how often, or how many times. WVhen the dividend does not contain the divisor an exact number of times, the excess is called a renzainder, and may be regarded as a fourth term in the division. The remainder, being part of the dividend, will always be of the same' denomination or kind as the dividend, and must always be less than the divisor. ArT. 47, SIGNS. - The sign of division is a short horizontal line, with a dot above it and another below; thus,. It shows that the number before it is to be divided by the number after it. The expression 6 + 2 = 3 is read, 6 divided by 2 is equal to 3. Division is also indicated by writing the dividend above a short horizontal line and the divisor below; thus, U. The ex pression 6 - 3 is read, 6 divided by 2 is equal to 3. There is a third method of indicating division, by a curved line placed between the divisor and dividend. Thus, the expression 6 ) 12 shows that 12 is to be divided by 6. ExERCISES FOR THE SLATE. ART. 48, The method of operation by Short Division, or when the divisor does not exceed 12. Ex. 1. Divide 8574 dollars equally among 6 men. Ans. 1429 dollars. QUESTIONS. — Art. 46. What is division? What are the three principalI terms in division? What is the dividend? What is the divisor? What is the quotient? What the remainder? What will be the denomination of the remainder? I-low does it compare with the divisor? Art. 47. What is the first sign of division, and what does it show? What is the second, and what does it show? What is the third, and what does it show? - Art. 48. What is short division?

Page 47 SECT. V.] DIVISION. 47 OPERATION, We first inquire how many times Divisor 6 ) 8574 Dividend. 6, the divisor, is contained in 8, the first figure of the dividend, 1429 Quotient. which is thousands, and find it to be 1 time, and 2 thousands remaining. We wvrite the 1 directly under the 8, its dividend, for the thousands' figure of the quotient. To 5, the next figure of the dividend, which is hundreds, we regard as prefixed the 2 thousands remaining,.which equal 20 hundreds, and thus form the number 25 hundredS, in which we find the divisor 6 to be contained 4 times, and 1 hundred remaining. We write the 4 for the hundreds' figure in the quotient, and the 1 hundred remaining, equal 10 tens, we regard as prefixed to 7, the next figure of the dividend, which is tens, forming 17 tens, in which the divisor 6 is contained 2 times, and 5 tens remaining. We write the 2 for the tens' figure in the quotient, and the 5 tens remaining, equal 50 units, we regard as prefixed to 4, the last figure of the dividend, which is units, forming 54 units, in which the divisor 6 is contained 9 times. Writing the 9 for the units' figure of the quotient, we have 1429 as the entire quotient, or the number of times which the dividend contains the divisor 6. ART. 4%9 From the foregoing illustration we deduce the following RUL. - Wirite the divisor at the left hand of the dividend, with a curved line between them, and draw a horizontal line under the divideald. Then, beginning at the left, find how m9any times the divisor is contained in the fewest figures of the dividend tshat will contain it, and write the quotient under its dividend. -'f th/ere be a remainder, regard it as p2refixed to the szextic t gure of 1the dividend, and divide as before.. Should any dividend be less than the divisor, iwrite a cipher ius the quotient, ancd annex another figure, if any remnzains, Jb' a nesw dividend. NOTE 1. - When there is a remainder after dividing the last figure of the dividend, write it with the divisor underneath, with a line between them, at the right of the quotient. Nol 2. - Preyfx means to place before; znnex, to place after. ART. 0o, First Method of Proof. - Multiply the divisor by the quotient, and to the product add the remainder, if any, and, if the work is right, the stun thus obtained will be equal to the dividend. QUESTIONS.- HOw are the numbers arranged for short division? At which hand do you begin to divide? Why not begin at the right, where you begin to multiply? Where do you write the quotient? If there is a remainder after dividing a figure, what is done with it? - Art. 49. What is the rule for short division? Repeat the notes.

Page 48 48 DIVISION. [SECT. V NoE. - It will be seen, from this method of proof, that division is the reverse of multiplication. The dividend answers to the product, the divi sor to one of thefactors, and the quotient to the other. EXAMPLES FOR PRACTICEI 2. Divide 6375 by 5. OPERATION. PROOF. Divisor 5 ) 6 3 7 5 Dividend. 1 2 7 5 Quotient. 1 2 7 5 Quotient. Divisor. 6 3 7 5 Dividend. 3. 4. 5. 3)7893762 4)4763256 5)3789565 2631254 1190814 6. 7. 8. 6)8765389 7)987635 8)378532 9. 10. 11. 9)8953784 11)7678903 12)6345321 Quotients. 12. Divide 479956 by 6. 79992W 13. Divide 385678 by 7. 550966 14. Divide 438789 by 8. 54848k15. Divide 1678767 by 9. 1865296 16. Divide 11497583 by 12. 95813111 Quotients. Rem 17. Divide 5678956 by 5. 1 18. Divide 1135791 by 7. 6 19. Divide 1622550 by 8. 6 20. Divide 2028180 by 9. 3 21. Divide 2253530 by 12. 2 22. Divide 1877940 by 11. 9 Sum of the quotients, 2084732 27 QUESTIONS. - Art. 50. How is short division proved? Of what is division the reverse? To what do the three terms in division answer in multipliui.a tion? What, then, is the reason for this proof of division?

Page 49 SECT. V.] DIVISION. 49 23. Divide 944,580 dollars equally among 12 men, and what will be the share of each? Ans. 78,715 dollars. 24. Divide 154,503 acres of land equally among 9 persons. Ans. 17,167 acres. 25. A plantation in Cuba was sold for 7,011,608 dollars, and the amount was divided among 8 persons. What was paid to each person? Ans. 876,451 dollars. 26. A prize, valued at 178,656 dollars, is to be equally divided among 12 men; what is the share of each? Ans. 14,888 dollars. 27. Among 7 men, 67,123 bushels of wheat are to be distributed; how many bushels does each man receive? Ans. 9,589 bushels. 28. If 9 square feet make 1 square yard, how many yards in 895,347 square feet? Ans. 99,483 yards. 29. A township of 876,136 acres is to be divided among 8 persons; how many acres will be the portion of each? Ans. 109,517 acres. 30. Bought a farm for 5670 dollars, and sold it for 7896 dollars, and I divide the net gain among 6 persons; what does each receive? Ans. 371 dollars. 31. If 6 shillings make a dollar, how many dollars in 7890 shillings? Ans. 1315. ART. 51. The method of operation by Long Division, or, in general, when the divisor exceeds 12. Ex. 1. A gentleman divided 896 dollars equally among his 7 children; how much did each receive? Ans. 128 dollars. OPERITION. Having set down the divisor Dividend. and dividend as in short diviDivisor 7 ) 8 9 6 ( 1 2 8 Quotient. sion, we draw a curved line at 7 the right of the dividend, to mark the place for the quo1 9 tient. We then inquire how 1 4 many times 7, the divisor, is contained in the 8 hundreds 5 6 of the dividend; and, finding. 5 6 it to be 1 hundred times, we write the 1 in the quotient, QUESTIONS. - Art. 51. What is long division? What is the difference between long division and short division? How do you arrange the numbers for long division? What do you first do after arranging the numbers for long division? Where do you place the figures of the quotient? 5

Page 50 50 DIVISION. [SECT. V and multiply the divisor, 7, by it, writing the product, 7 hundreds under the 8 hundreds, from which we subtract it, and to the remainder, 1, annex the 9 tens of the dividend, making 19 tens. We now inquire how many times 7 is contained in 19 tens, and write the number, 2, at the right of the quotient figure before obtained. We then multiply the divisor by it, and place the product under the 19, and subtract as before; and to the remainder, 5, we annex 6 units, the next and last figure of the dividend, making 56 units. We proceed, as before, to find the next quotient figure, and, after subtracting the product of the divisor multiplied by it from 56, find there is no remainder left. Hence we learn that each one of the 7 children must receive 128 dollars. NOTE. - The preceding example and the four that follow are usually performed by short division, but are here introduced to illustrate more clearly the method of operation by long division. EXAMIPLES FOR PRACTICE. 2. Divide 1728 by 8. Ans. 21G. 3. Divide 987656 by 11. Ans. 897861-0. 4. Divide 123456789 by 9. Ans. 13717421. 5. Divide 390413609 by 12. Ans. 32534467 -. Ex. 6. A gentleman divided 4712 dollars equally among his 19 sons; what was the share of each? Ans. 248 dollars. OPERATION. We first inquire how Dividend. many times 19, the divisor, Divisor 19 ) 4 7 1 2 (2 4 8 Quotient. is contained in 47, the two 3 8 left-hand figures of the div8 idend; and, finding it to be 9 1 2 times, we write the 2 in 7 6 the quotient, multiply the divisor by it, and subtract 1 5 2 the product from the 47; 1 5 2 and to the remainder, 9, annex 1, the next figure of the dividend, making 91. We next inquire how many times 19 is contained in 91, place the number, 4, in the quotient, then multiply and subtract as before, and to.he remainder, 15, annex 2, the last figure of the dividend, and, proceeding as before, after finding the quotient figure, no remainder is left. Hence the share of each of the 19 sons is 248 dollars. This illustration, except in omissions, is essentially like the preceding one. QUESTIONS. - After the quotient figure is found, what is the next thing you do? Where do you place the product? What do you next do? What is the next step? How do you then proceed? Is long division the same in principle as short division?

Page 51 FECT. V.] DIVISION. 51 ART. 52. From the preceding illustrations, the pupil will perceive the propriety of the following general RULE. - Wirite the divisor and dividend as in short division, and drawl a curved line at the right hand of the dividend. Then inquire how many times the divisor is contained in the fewest filgures on the left hand of the dividend that will contain it, acld write the -result at the r ight hand of the dividend for the first quotient figzure. M:ultiply the divisor by the quotient'fyure, acz subtract the product from the figures of the dividend used, and to the remainder annex the nextfigure of the diviclendl. Find how?many timnes the divisor is contained in the number thus formed; write the figure denoting it at the ri7ght hand of the former quotient figure. Thus proceed until all the flyures of the dividend are divided. NOTE 1.- The proper remainder is in all cases less than the divisor. If, in the course of the operation, it is at any time found to be as large as, or larger than, the divisor, it wvill show that there is an error in the work, and that the quotient figure should be increased. NOTE 2. -If, at any time, the divisor, multiplied by the quotient figure, produces a product larges than the part of the dividend used, it shows that the quotient figure is too large, and must be diminished. NOTE 3. - It will often happen that, when a figure is brought down, the number will not contain the divisor; and in that case a cipher must be placed in the quotient, and another figure of the dividend brought down, and so on until the number is large enough to contain the divisor. NOTE 4. - If there is a remainder after dividing all the figures of the dividend, it must be written as directed in the preceding rule. ARlT. 53, Second JMethod of Proof. - Add together the remainder, if any, and all the products that have been produced by multiplying the divisor by the several quotient figures, and the result will be like the dividend, if the work is right. ART. 54. Third Method. - Subtract the remainder, if any, from the dividend, and divide the difference by the quotient. The result will be like the original divisor, if the work is right. NOTE. - The first method of proof (Art. 50) is usually most convenient, and is most commonly employed. QUESTIONS. - Art. 52. What is the general rule for long division? How may you know when the quotient figure is too small? How may you know when it is too large? What do you do when the part of the dividend used will not contain the divisor? - Art. 53. What is the second method of proof for division? - Art. 54. What is the third method? Can long division be proved by the first method of proof (Art. 50)?

Page 52 52 DIVISION. [SECT. V EXAMIPLES PORl PRACTICE. Ex. 7. It is required to find how many times 48 is contained in 28618. Ans. 596. OPERATION. Dividend. PROOF BY MULTIPLICATION Divisor 4 8 )2 8 6 1 8 (596Quotient 5 9 6 Quotient. 240 4 8 Divisor. 461 4768 432 2384 298 28608 2 8 8.1 0 Remainder 1 0 Remainder. 2 8 6 1 8 Dividend. 8. OPERATION. Dividend. Divisor 2 6 ) 5 6 9 8 ( 2 1 9 Quotient. PROOF BY ADDITION. *+52 52 ) 26 Products. 49 234) +2 6 4 Remainder. 238 5698 Dividend. +234 +4 Remainder. 9. OPERATION. PROOF BY DIVISION. Dividend. Dividend. Divisor 4 4)1 3 8 2 4(9 6 Quotient. 96)1 3 824(144 Divisor. 1296 96 864 422 864 384 384 384 Quotients. Rem 10. Divide 3276 by 14. 234 11. Divide 6205 by 17. 365 * This sign of addition denotes the several products to be added.

Page 53 SECT V.] DIVISION. 53 Quotients. Item. 12. Divide 3051 by 21. 145 6 13. Divide 190850 by 25. 7634 0 14. Divide 218579 by 42. 5204 11 15. Divide 9012345 by 31. 290720. 25 16. Divide 6717890 by 98. 68549 88 17. Divide 4567890 by 19. 240415 5 18. Divide 1357901 by 87. 15608 5 19. Divide 9988891 by 77. 129725 66 20. Divide 9999999 by 69. 144927 36 21. Divide 867532 by 59. 14703 55 22. Divide 167008 by 87. 1919 55 23. Divide 345678 by 379. -912 30 24. Divide 3456789567 by 987. 3502319 714 2'5, Divide 8997744444 by 345. 26080418 234 -26. Divide 4500700701 by 407. 11058232 277 27. Divide 6789563 by 1234. 95 28. Divide 78112345 by 8007. 4060 29. Divide 34533669 by 9999. 7122 30. Divide 9999999 by 3333. 0 31. Divide 47856712 by 1789. 962 32. Divide 345678901765 by 4007. 86268755 480 33. Divide 478656785178 by 56789. 8428688 22346 84. Divide 678957000107 by 10789561. 62927 2295060 85. Divide 990070171009 by 900700601. 1099 200210510 36. Divide three hundred twenty-one thousand three hundred dollars equally among six hundred seventy-five men. Ans. 476 dollars. 37. Four hundred seventy-one men purchase a township containing one hundred eighty-six thousand forty-five acres; what is the share of each? Ans. 395 acres. 38. A railroad, which cost five hundred eighteen thousand seventy-seven dollars, is divided into six hundred seventy-nine shares; what is the value of each share? Ans. 763 dollars. 39. Divide forty-two thousand four hundred thirty-five bushels of wheat equally among one hundred twenty-three nmen. Ans. -345 bushels each. 40. A prize, valued at one hundred eighty-four thousand seven hundred seventy-five dollars, is to be divided equally among four hundred seventy-five men; what is the share of each? Ans. 389 dollars. 41. A certain company purchased a valuable township for nine millions six hundred ninety-one thousand eight hundred 54

Page 54 54 DIVISION. [SECT. 5 thirty-six dollars; each share was valued at seven thousand eight hundred fifty-four dollars; of how many men did the eom pany consist? Ans. 1234 men. 42. A tax of thirty millions fifty-six thousand four hundred sixty-five dollars is assessed equally on four thousand five hundred ninety-seven towns; what sum must each town pay? Ans. 6538 12, dollars. ART. 55, Method of operation when the divisor is a com posite number. Ex. 1. A merchant bought 15 pieces of broadcloth for 1440 dollars; what was the value of each piece? Ans. 96 dollars. OPERATION. The factors of 15 are 3 3 ) 1 4 4 0 dolls., cost of 15 pieces. and 5. Now, if we divide dols.coto 5pieces. the 1440 dollars, the cost 5 ) 4 8 0 dolls., cost of 5 pieces. of 15 pieces, by 3, we obh9 6 dolls., cost of 1 piece. tain 480 dollars, which is evidently the cost of 5 pieces, because there are 5 times 3 in 15. Then, dividing 480 dollars, the cost of 5 pieces, by 5, we get the cost of 1 piece. Hence we deduce the following RULE. - Divide the dividend by one of the factors, and the quotienl thus found by another, and thus proceed till every factor has been made a divisor. The last quotient will be the true quotient required. EXAMPLES FOR PRACTICE. Quotients. 2. Divide 765325 by 25 5 X 5. 30613 3. Divide 123396 by 84 = 7 X 12. 1469 4. Divide 611226 by 81, using its factors. 7546 5. Divide 987625 by 125, using its factors. 7901 6. Divide 17472 by 96, using its factors. 182 7. Divide 34848 by 132, using its factors. 264 ART. 58. Method of finding the true remainder when there are several in the operation. Ex. 1 How many months of 4 weeks each are there in 298 days, and how many days remaining? Ans. 10 months and 18 days. QurSTIONs. -Art. 55. What are the factors of 15? What do you get the cost of, in this example, when you divide by the factor 3? What, when you divide by 5? Why? What is the rule for dividing by a composite number?

Page 55 DECT. V.] DIVISION. 5,5 OPERATION. Since there are 7 days in 1 7 ) 2 9 8 week, we first divide the 298 ~4 4 2 4 ddays by 7, and have 42 weeks 4) 4 2, 4 days 18 day. and a remainder of 4 days. 10, 2 weeks Then, since 4 weeks make 1 month, we divide the 42 weeks by 4, and have 10 months and a remainder of 2 weeks. Now, to find the true remainder in days, it is evident that we must multiply the 2 weeks by 7, because 7 days make a week, and to the product add the 4 days; thus, 2 X 7 = 14, and 14 + 4= 18 days, for the remainder. Hence the following RULE. - Multiply each remainder, except the first, by all the divisors preceding the one which produced it; and the first remainder being added to the sum of the products, the amount will be the true remainder. NoTE. — There will be but one product to add to the first remainder when there are only two divisors and two remainders. Ex. 2. Divide 789 by 36, using the factors 2, 3, and 6, and find the true remainder. Ans. 33. OPERATION. FINDING THE TRUE REMAINDER. 2 )7 8 9 5 X 3 X 2 = 30, st Product. 394~I X 2 -= 2,2dProduct. 3 )3 9 4, 1, 1st Rem. 1, 1st Remainder. 6 1, 1, 2dRem. 3, true Rem. 219 5, 3d Rem. EXAMPLES FOR PRACTICE. 3. Divide 934 by 55, using the factors 5 and 11, and find the true remainder. Ans. 54. 4. Divide 5348 by 48, using the factors 6 and 8, and find the true remainder. Ans. 20. 5. Divide 5873 by 84, using the factors 3, 4, and 7, and find the true remainder. Ans. 77. 6. Divide 249237 by 1728, using the factors 12, 6, 6, and 4, and find the true remainder. Ans. 405. ART, 57$. When the divisor is 1, with one or more ciphers at the right; as 10, 100, &c. Ex. 1. Divide 356 dollais equally among 10 men; what will each man have? Ans. 356, dollars. QUESTIONS. — Art. 56. When there are several remainders, what is the rule for finding the true remainder? Will you give the reason for this rule?

Page 56 56 DIVISION. [SECT. V. OPERATION. It will be remembered, that to mul1 0 ) 3 5 6 tiply by 10 we annex one cipher, which Q',, removes the figures one place to the Quotient 30 5, 6 em. left, and thus increases their value ten Or thus, 3 5 6. times. Now, it is obvious that if we reverse the process, and cut off the right-hand figure by a line, we remove the remaining figures one place to the right, and consequently diminish the value of each ten times, and thus divide the whole number by 10. The figures on the left of the line are the quotient, and the one on the right is the remainder, which may be written over the divisor, and, annexed to the quotient. Hence the share of each man is 35 6 dollars. EXAMIPLES FOR1 PRACTICE. Quotient. Rem. 2. Divide 6892 by 10. 689 2 3. Divide 4375 by 100. 75 4. Divide 24815 by 1000, 815 5. Divide 987654321123 by 100000000. 54321123 ART. T5, When the divisor has ciphers on the right, and is not 10, 100, &e. Ex. 1. If I divide 5832 pounds of bread equally amnong 300 soldiers, what is each one's share? Ans. 94 32 pounds. IOPERATION. The divisor, 600, may 1 0 0 ) 5 8.3 2 be resolved into the fctors 6 and 100. We first 6) 5 8, 3 2, 1st Rem. divide by the factor 100, 9~ 4, 2od Rem. by cutting off two figures at the right, and get 58 Or thus, 6 1 0 0) 5 8 1 3 2 for the quotient and 32 for a remainder. We 9, 4 3 2 then divide the quotient, 58, by the other fictoer, 6, and obtain 9 for the quotient and 4 for a remainder. The last remainder, 4, being multiplied by the divisor, 100, and 32, the first remainder, added, we obtain 432 for the true remainder (Art. 56). Hience each soldier receives 946 _ pounds. AaRT. 59 From the preceding illustrations is deduced the general RULE.- Cut off the ciphers from the divisor, and the same num/bier of figures from the right of the dividend. Then divide the remaining figures of th ividend by the remaining figures of the divisor. QUESTIONS. —Art. 57. I-low do you divide by 10? iow does it appeal that this divides the number by 10? -Art. 58. How do you divide by 00O in the example? Iow does it appear that this divides the number? —Art. 59. What is the general rule?

Page 57 bECT. VI. QUESTIONS INVOLVING FRACTIONS. 57 NOTE. - When by the operation there is a last remainder, to it must be annexed the figures cut off from the dividend to form the true remainder Should there be no last remainder, then the significant figures, if any, cut off from the dividend, will form the true remainder. EXAMPLES FOR PRACTICE. Quotients. Rem. 2. Divide 3594 by 80. 44 74 3. Divide 79872 by 240. 332 192 4. Divide 467153 by 700. 667 253 5. Divide 13112297 by 8900.. 2597 6. Divide 71897654325 by 700000000. 497654325 7. Divide 3456789123456787 by 990000. 306787 S. Divide 967231731328000 by 1020000000. 411328000 9. Divide 33166405115000 by 1600000000. 5115000 10. Divide 18191618562300 by 1000000000. 618562300 11. Divide 4766666000000 by 55550000000. 44916000000 VI. QUESTIONS INVOLVING FRACTIONS. ART. 60. IF a unit or individual thing is divided' into equal parts, each of the parts is called a Firaction of the number or thing divided. Hence a FRACTION is one or more equal parts of a unit. ILLUSTRATIONS. - 1. When any number or thing is divided into two equal parts, one of the parts is called one half, and is written thus: ~. 2. When any number or thing is divided into three equal parts, one of the parts is called one third ( t); two of the parts are called two thirds (Q). 3. When any number or thing is divided into four equal parts, one of the parts is -called one fourth (1); three of the parts, three fourths (3). 4. When any number or thing is divided into five equal parts, one of the parts is called one fifth ( ); two parts, two fifths (2); three parts, threefifths (5); and four parts,fourfifths (4). QUESTIONS. - Art. 60. What is a fraction? What is meant by one half of any number or thing? How is it written? What is meant by one third, and how is it written? What by one fourth, and how written? What by one fifth, and how written? What by four fifths, and how written? How do you find one half of any number? How one third? How one fourth? &c. How many halves make a whole one? How many thirds? How many fourths? How many fifths?

Page 58 58 QUESTIONS INVOLVING FRACTIONS. [SECT. Vi. 5. WVhen any number or thing is divided into six equal parts what is one of the parts called? Two parts? Five parts? 6. When a number or thing is divided into 7 equal parts, what is 1 part called? 2 parts? 3 parts? 4 parts? 5 parts? 6 parts? 7. When a number or thing is divided into 9 equal parts, what is I part called? 2 parts? 4 parts? 5 parts? 7 parts? 8 parts? 8. What is 1 haf of 4? Of 8? Of 16? Of 20? Of 28? Of 32? 9. What is l third of 9? Of 12-? Of 15? Of 27? Of 30? Of 36? Of 60? 10. What is 1 fourth of 8? Of 16? Of 20? Of 24? Of 40? Of 48? Of 100? 11. What is 1 fifth of 10? Of 25? Of 30? Of 35? Of 45? Of 50? Of 55? Of 65? 12. What is 1 sixth of 12? Of 18? Of 30? Of 42? Of 60? Of 72? Of 90? 13. How many fourths in 1 apple? 14. How many fourths in 2 apples? In 3 apples? In 8 apples? >; In 16 apples? 15. How many fifths in 1 barrel of flour? In 3 barrels? In 5 barrels? In 7 barrels? In 9 barrels? 16. How many sixths in 1 bushel of wheat? In 4 bushels? In 7 bushels? In 9 bushels? In 12 bushels? 17. James owns 3 fifths of a kite, and his brother Thomas the remainder. How many fifths does Thomas own? ILLUSTRATION. - Since there are 5 fifths in the kite, if Jam:es owns 3 fifths, there will remain for Thomas 5 fifths (5) less 3 fifths (-) 2 fifth. Ans. 2 fifths. 18. From a load of hay I sold 4 sevenths; how many sevenths remain? 19. John Jones found a large sum of money; he gave 5 eighths of it to the poor of the parish; how much did he reserve for himself?: 20. John Smith gave 2 ninths of his farm to his son, 3 ninths to his daughter, and the remainder to his wife; how many ninths did his wife receive? ILLUSTRATION. - Since he gave 2 ninths (2) to his son, and 3 ninths (3) to his daughter, he gave them both 29 + 9 —; and since there are 9 ninths (9) in the farm, he must have given his wife - _. 4.An. wife 9-g -g_ Ans. }..

Page 59 SECTo VYI QUESTIONS iNVOLVNG FRtACTIONSo 59 21. In a ccrtain schlool -~L or- the pupils study grammar, TI study aridthnetic, - g'eog'ra'pliy, and. the remainder philosophyo What part of tihe school staudy philosophy? 22o J. Dow spends 1 of his time in reading, 3 in labor, and 2 in visitingy How large a portion of his time remains for eating and sleeping? 23. If a yard of cloth cost $8, what cost g of a yard? What cost a of a yard? ILLUSTRATINo -- If I yard cost $8, ~ of a yard will cost: of $8 = $2; and if I of a yard cost $2, 1 will cost three times as much; 3 times $2 $ 86. Alls. G6o 24.0 f an acre of land cost $24, what will ~ of an acre cost? What will a cost? 25. When 96 cients are paid for a bushel of rye, What cost -{ of a bushel? 26 I!f I of a barrel of four cost 2 dollars, what cost 4 of barrel? iLLUmTTRATION. L If 5 cost 2 dollars, - will cost 4 times 2 dollars 8. Ans. $38 27. If -I of an acre of land cost $24 dollars, what will - of an acre cost? 280 If of a hogshead of molasses cost $11, what will a hogshead cost? 2>9. If 7 of an acre of land cost $21, what cost $ of an acre 9 What cost an acre? What cost 10 acres? iLLUsr'TRAr OXNo — If - cost $219, - will cost of 0$21, and } of $21 is 53; and 8 will cost 8 times 53 - $24, and 10 acres will cost 10 tinmes $424 -240I Ans. $24d0 3'0 If s of a hogshlead of sugar cost $1o8, what costs 1; hosheadc? WI-,,at cost 4 hogsheiads? 31. TIf o of barr'el of apples cost $1.50, what costs a barrel? IWVhat cost I 0 barrels? 3 iln R,4 —9 are paid for - - of a ton of potash, -what must be paid for 2 tons? 33o H-ow many halifbarre1s of flour are there in 2 and a half (21) barrels? ILLuiSTRATIONo - Since I barrel contains 2 halves, 2 barrels will contain 2 times 2 4 halves, and the 1 half added makels 5 halves Ans. 1 34. How many half.busheIs in 4-$ bushels of oats? 5~ bushels r? i-n 71 bosshels? In 9- biush els?

Page 60 60 QUESTIONS INVOLVING FRACTIONS. [SECT. VI 35. low many eighlths of a dollar in 21 dollars In 43 dole ars? In 7- dollars In 9- dollars? In 12' dollars? 36. How many tenths of an ounce in 4 - ounces? I1 5 7 ounces? In $8 ounces? In 10 9 ounces? 37. How many barrels of wine in 6 half (6) barrels? ILLUSTRATIO 8 — ince it takes 2 halves to make one whole one, there will be as many whole barrels in 6 halves ($6) as 2 is contained times in 6. 2 is contained in 6, 3 times, Ans. 3 barrels. 38. How many firkins of butter in $ firkins? In -O- firkins? 39. How many whole numbers in -_-? In -I 9o In 2-4? 40. How many whole numbers in -65-? In? In 32? In 41. If a skein of silk is worth 31 cents, what are 6 skeins worth? ILLusTRATrION.o If I skein is worth 34 cenJs, 6 skeins are worth 6 times as much; 6 times 31 are equal to 6 times 3 and 6 times 6; 6times 3 S 8; 6 times - = 3; 18-+ 3 =21, Ans. 21 cents. 42. Bought one pair of boots ror 86;1 what must I pay fos 4 pairs? For 8 pairs? For 10 pairs? For 12 pairs? 43. Paid 121 cents for one pound of cloves; what will 6 pounds cost? 10 pounds? 12 pounds? 443 If one pound of butter is worth 12 cents, what are 4. pounds worth? ILLUsTRnArToN - If I pound is worth 12 cents, 4$ pounds are worth 4a times as much; 4$ times 12 cents are equal to 4 times 12 and I of 12; 4 times 12 are 48, and ~ of 12 is 6; 48 cents and 6 cents are 54 cents. Ans, 54 cents. 45. When lard is sold for 9 cents per pound, what must be paid for 7-1 pounds? For 84 pounds? For 9 1 pounds? 46. Bought 1 pound of coffee at 16 cents; what will 5 -2 pounds cost? 3- pounds? 51 pounds? 6$ pounds? 47D If 1 yard of cloth is worth 20 cents, what is the value of 16, yards? 12$ yards? 8$ yards? 11I yards?,18 If 17 bushels of corn cost $1.20, what will I bushel cost? I[LLUSTRATiON.l - 1'- bushels - bushels. Now, if - cost $1.20, 1 will cost I of $1.20 = ) 0.40; and $ or a whole bushel will cost 2 times $0.40 = 0.80. Ans. 0.80. 49. If 22 pounds of coffee cost 60 -.ents, what will I pound cost3?

Page 61 SECT. VII,] CONTRACTIONS IN MULTIPLICATION. 61 ILLUSTRAYrON. — 5 15d z pounds. If J2 cost 60 cents, 5 will cost I-T of 60 cents = 5 cents; and 5, or a pound) will cost 5 times 5 cents = 25 cents. Ans. $0.25. 50. How many times will 60 contain 2?. 51. Paid $54 for 74 barrels of oil; what cost 1 barrel? Ans. $7. 52. How many times is 74 contained in 54? 53. How many cords of wood, at $51 per cord, can be bought for 866? 54. How many times will 66 contain 54? 55. Gave $40 for 62 yards of broadcloth; what cost 1 yard? 56. How many times is 64 contained in 40? 57. The distance between two places is 110 rods. I wish to divide this distance into spaces of 54 rods each. Required the number of spaces. ~ VII. CONTRACTIONS IN MULTIPLICATION AND DIVISION.* CONTRACTIONS IN MULTIPLICATION. ART. 61, To multiply by 25. Ex. 1. Multiply 876581 by 25. Ans. 21914525. OPERATION. We multiply by 100, by an4) 8 7 6 5 8 1 0 0 nexing two ciphers to the multiplicand; and since 25, the multi2 1 9 1 4 5 2 5 Product. plier, is only one fourth of 100, we divide by 4 to obtain the true product. RULE. - Annex two ciphers to the multiplicand, and divzde it by 4. If the principles on which these contractions depend are considered too difficult for the young pupil to understand at this stage of his progress, they may be omitted for the present, and attended to when he is further advanced. QUESTIONS.- Art. 61. What is the rule for multiplying by 25? What is the reason for the rule? 6

Page 62 (52 CONTRACTIONS IN MULTIPLICATION. (SECT. VII EXAMPLES FOR PRACTICE, 2. Multiply 76589658 by 25. Ans. 1914741450. 3. Multiply 567898717 by 25. Ans. 14197467925. 4. Multiply 123456789 by 25. Ans. 3086419725. ART, 62. To multiply by 33~. Ex. 1. Multiply 87678963 by 338 Ans. 2922632100. OPERATION. We multiply by 100, as be. 3 ) 8 7 6 7 8 9 6 3 0 fore; and since 33k, the mul3)8767 896300 tiplier, is only one third of 2 9 2 2 6 3 21 0 Product, 100, we divide by 3 to obtain the true product. RULE. - Annex two ciphers to the multiplicand, and divide it by 3. EXAMPLES FOR PRACTICE. 2. Multiply 356789541 by 33~. Ans. 11892984700. 3. Multiply 871132182 by 33-. Ans. 29037739400. 4. Multiply 583647912 by 33j. Ans. 19454930400. ART. 63. To multiply by 125. Ex. 1. Multiply 7896538 by 125. Ans. 987067250. OPERATION. We multiply by 1000, by 8) 7 8965 38000 annexing three ciphers to the 987067250 Product. multiplicand; and since 125, the multiplier, is only one ezghth of 1000, we divide by 8 to obtain the true product. RULE. - Annex three ciphers to the multiplicand, and divide it by 8 EXAMPLES FOR PRACTICE. 2. Multiply 7965325 by 125. Ans, 995665625. 3. Multiply 1234567 by 125, Ans. 154320875. 4. Multiply 3049862 by 125. Ans. 381232750. QUESTIONS. - Art. 62. What is the rule for multiplying by 33-? What is the reason for this rule?-Art. 63. What is the rule for multiplying by 125? Give the reason for the rule.

Page 63 bECT. VII.] CONTRACTIONS IN DIVISION. 63 ART. 64, To multiply by any number of 9's. Ex. 1. Multiply 4789653 by 99999. Ans. 478960510347. OPERATION. By adding 1 to any num478965300000 ber composed of nines, we 4 7 8 9 6 5 3 obtain a number expressed by 1 with as many ciphers 4 7 8 9 6 0 5 1 0 3 4 7 Product. annexed as there are nines in the number to which I is added. Thus, 999 + 1 -- 1000. Therefore, annexing to the multiplicand as many ciphers as there are nines in the multiplier is the same thing as multiplying the number by a multiplier too large by 1, and subtracting the number to be multiplied from this enlarged product will give the true product. RULE. - Annex as many ciphers to the multiplicand as there are 9's in the multiplier, and from this number subtract the number to be multiplied. EAXMPLES FOR PRACTICE.'2. Multiply 1234567 by 999. Ans. 1233332433. 3. Multiply 876543 by 999999. Ans. 876542123457. 4. Multiply 999999 by 999999. Ans. 999998000001. CONTRACTIONS IN DIVISION. ART. 65. To divide by 25. Ex.. Divide 1234567 by 25. Ans. 49382?3. OPERATION. Multiplying the dividend by 4 makes 12 3 4 5 6 7 it four times too great; therefore, to 4 obtain the true quotient, we must divide by 100, a divisor four times greater 4 9 3 82 268 Quotient. than the true one. This we do by cutting off two figures on the right. RULE. - Multiply the dividend by 4, and divide the product by 100: EXAMPLES FOR PRACTICE. 2. Divide 9876525 by 25. Ans. 395061. 3. Divide 1378925 by 25. Ans. 551574. Divide 899999 by 25. Ans. 35999 -9, QUESTIONS. - Art. 64. What is the rule for multiplying by any number of 9's? What is the reason for the rule? —Art. 65. What is the rule for dividing by 25? Give the reason for the rule.

Page 64 64 CONTRACTIONS IN DIVISION. [SECT. Vii ART. T6o TO divide by 33-. Ex. 1. Divide 6789543 by 333. Ans. 203686-29 OPERATION. IMultiplying the dividend by 3 makes 6 7 8 9 5 4 3 it three times too great; therefore, to obtain the true quotient, we must divide -3 by 100, a divisor three times greater 2 0 3 6 8 612 9 Quotient. than the true one. This is done by cutting off two figures on the right. RULE. - Multiply the dividend by 3, and divide the product by 100 EXAMPLES FOR PRACTICE. 2. Divide 987654321 by 33~. Ans. 296296291 63 3. Divide 8712378 by 33}. Ans. 261371T 3. 4. Divide 4789536 by 33.. Ans. 1436e8610. 5. Divide 89676 by 33~. Ans. 269 028-. 6. Divide 17854 by 333. Ans. 53512.06 ART. 67. To divide by 125. Ex. 1. Divide 9874725 by 125. Ans. 78997-3. OPERATION. Multiplying the dividend by 8 makes 9 8 7 4 7 2 5 it eight times too great; therefore, to 8 obtain the true quotient, we must divide by 1000, a divisor eighlt times grea,ter 8 9 9 718 0 0 Quotient. than the true one. We do this by cut'I Quotient. ting off three Efigures on the right. RULE. - Multiply the dividzaend by 8, and divide the product by 1000. EXAMPLES FOR IPRACTICE. 2. Divide 1728125 by 125. Ans. 13825. 3. Divide 478763250 by 125. Ans. 3830106. 4. Divide 591234875 by 125. Ans. 472.9879. 5. Divide 489648 by 125. Ans. 3917- 1s,,4u 6. Divide 836184 by 125. Ans. 66894(4o.2D QUESTIONS. - Art. 66. What is the rule for dividing by 331? Give ths reason for the rule. - Art. 67. What is the rule for dividing by 125? 12 at is the reason for the rule

Page 65 SECT. VIII. ] MISCELLANEOUS EXAMPLES. 60 ~ VIII. MISCELLANEOUS EXAMPLES, INVOLVING THE FOREGOING RULES. 1. A BOUGHT 73 hogsheads of molasses at 29 dollars per hogshead, and sold it at 37 dollars per hogshead; what did he gain? Ans. 584 dollars. 2. B bought 896 acres of wild land at 15 dollars per acre, and sold it at 43 dollars per acre; what did he gain? Ans. 25088 dollars. 3. N. Gage sold 47 bushels of corn at 57 cents per bushel, which cost him only 37 cents per bushel; how many cents did he gain? Ans. 940 cents. 4. A butcher bought a lot of beef weighing 765 pounds at 11 cents per pound, and sold it at 9 cents per pound; how many cents did he lose? Ans. 1530 cents. 5. A taverner bought 29 loads of hay at 17 dollars per load and 76 cords of wood at 5 dollars a cord; what was the amount of the hay and the wood? Ans. 873 dollars. 6. Bought 17 yards of cotton at 15 cents per yard, 46 gal. lons of molasses at 28 cents per gallon, 16 pounds of tea at 76 cents a pound, and 107 pounds of coffee at 14 cents a pound; what was the amount of my bill? Ans. 4257 cents. 7. A man travelled 78 days, and each day he walked 27 miles; what was the length of his journey? Ans. 2106 miles. 8. A man sets out from Boston to travel to New York, the distance being 223 miles, and walks 27 miles a day for 6 days in succession; what distance remains to be travelled? Ans. 61 miles. 9. What cost a farm of 365 acres at 97 dollars per acre? Ans. 35405 dollars. 10. Bought 376 oxen at 36 dollars per ox, 169 cows at 27 dollars each, 765 sheep at 4 dollars per head, and 79 elegant horses at 275 dollars each; what was paid for all? Ans. 42884 dollars. 11. J. Barker has a fine orchard, consisting of 365 trees, and each tree produces 7 barrels of apples, and these apples will bring him in market 3 dollars per barrel; what is the incomt of the orchard? Ans. 7665 dollars. 12. J. Peabody bought of E. Ames 7 yards of his best broadcloth at 9 dollars per yard, and in payment he gave Ames a 6*

Page 66 66 MISCELLANEOUS EXAMPLES. [Sacr. VIII. one hundred-dollar bill; how many dollars must Ames return to Peabody? Ans. 37 dollars. 13. Bought of P. Parker a cooking-stove for 31 dollars, 7 quirntals of his best fish at 6 dollars per quintal, 14 bushels of rye at 1 dollar per bushel, and 5 mill-saws at 16 dollars each, in part payment for the above articles, I sold him eight thousand feet of boards at 15 dollars per thousand; how much must I pay him to balance the account? Ans. 47 dollars. 14. In i day there are 24 hours; how many in 57 days? Ans. 1368 hours. 15. In one pound avoirdupois weight there are 16 ounces; how many ounces are there in 369 pounds? Ans. 5904 ounces. 16. In a square mile there are 640 acres; how many acres are there in a town, which contains 89 square miles? Ans. 56960 acres. 17. What cost 78 barrels of apples at 3 dollars per barrel? Ans. 234 dollars. 18. Bought 500 barrels of flour at 5 dollars per barrel, 47 hundred weight of cheese at 9 dollars per hundred weight, and 15 barrels of salmon at 17 dollars per barrel; what was the amount of my purchase? Ans. 3178 dollars. 19. Bought 760 acres of land at 47 dollars per acre, and sold J. Emery 171 acres at 56 dollars per acre, J. Smith 275 acres at 37 dollars per acre, and the remainder I sold to J. Kimball at 75 dollars per acre; how much did I gain by mry sales? Ans. 75c1 dollars. 20. Bought a hogshead of oil containing 184 gallons, at 75 cents per gallon; but 28 gallons having leaked out, I sold the rm1cainder at 98 cents per gallon; did I gain or lose by my bargain? Ans. 1488 cents, gain. 21. Bought a quantity of flour, for which I gave 1728 dollars, there being 288 barrels; I sold the same at 8 dollars ype barrel; how much did I gain? Ans. 576 dollars. 22. Purchased a cargo of molasses for 9212 dollars, there being 196 hogsheads; I sold the same at 67 dollars per liogs. head; how much did I gain on each hogshead? Ans. 20 dollars. 23. A farmer bought 5 yoke of oxen at 87 dollars a yo'ke 37 cows at 37 dollars each; 89 sheep at 3 dollars apiece. He sold the oxen at 98 dollars a yoke; for the cows he received 40 dollars each; and for the sheep he lhad 4 dollars apiece. How m uch did he gain by his trade?' Ans. 255 do0llars

Page 67 SECT. VIII.] MISCELLANEOUS EXAMPLES. 67 24. The sum of two numbers is 5482, and the smaller number is 1962; what is the difference? Ans. 3520 25. The difference between two numbers is 125, and the smaller number is 1482; what is the greater? Ans. 1607. 26. The difference between two numbers is 1282, and the greater number is 6958; what is the smaller? Ans. 5676. 27. If the dividend is 21775, and the divisor 871, what is the quotient? Ans. 25. 28. If the quotient is 482, and the divisor 281, what is the dividend? Ans. 135442. 29. If 144 inches make 1 square foot, how many square feet in 20736 inches? Ans. 144 feet. 30. An acre contains 160 square rods; how many rods in a farm containing 769 acres? Ans. 123040 rods 31. A gentleman bought a house for three thousand fortyseven dollars, and a carriage and span of horses for five hundred seven dollars. He paid at one time two thousand seventeen dollars, and at another time nine hundred seven dollars. How much remains due? Ans. 630 dollars. 32. The erection of a factory cost 68,255 dollars; supposing this sum to be divided into 365 shares, what is the value of each? Ans. 187 dollars. 33. Bought two lots of wild land; the first contained 144 acres, for which I paid 12 dollars per acre; the second contained 108 acres, which cost 15 dollars per acre. I sold both lots at 18 dollars per acre; what was the amount of gain? Ans. 1188 dollars. 34. Sold 17 cords of oak wood at 6 dollars per cord, 36 cojrds of maple at 3 dollars per cord, and 29 cords of walnut at 7 dollars per cord. What was the amount received? Ans. 413 dollars. 35. Daniel Bailey has a fine farm of 300 acres, which cost him 73 dollars per acre, He sold 83 acres of this farm to Mlinot Thayer, for 97 dollars per acre; 42 acres to J. Russel, for 87 dollars per acre; 75 acres to J. Dana, at 75 dollars per acre; and the remainder to J. Webster, at 100 dollars per acre. What was his net gain? Ans. 5430 dollars. 36. J. Gale purchased 17 sheep for 3 dollars each, 19 cows at 27 dollars each, and 47 oxen at 57 dollars each. He sold his purchase for 3700 dollars. What did he gain? Ans. 457 dollars. 37. Purchased 17 tons of copperas at 32 dollars per ton. 1 sold 7 tons at 29 dollars per ton, 8 tons at 36 dollars per ton.

Page 68 68 MISCELLANEOUS EXAMPLES. [SECT. VII1 and the remainder at 25 dollars per ton. Did I gain or lose, and how much? Ans. 3 dollars, loss. 38. John Smith bought 28 yards of broadcloth at 5 dollars per yard; and, having lost 10 yards, he sold the remainder at 9 dollars per yard. Did he gain or lose, and how much? Ans. 22 dollars, gain. 39. Which is of the greater value, 386 acres of land at 76 dollars per acre, or 968 hogsheads of molasses at 25 dollars per hogshead? Ans. The land, by 5136 dollars. 40. Bought of J. Low 37 tons of hay at 18 dollars per ton. I paid him 75 dollars, and 12 yards of broadcloth at 4 dollars per yard. How much remains due to Low? Ans. 543 dollars. 41. A purchased of B 40 cords of wood at 5 dollars per cord, 9 tons of hay at 17 dollars per ton, 19 grindstones at 2 dollars apiece, 37 yards of broadcloth at 4 dollars per yard, and 16 barrels of flour at 6 dollars per barrel; what is the amount of A's bill? Ans. 635 dollars. 42. John Smith, Jr., bought of R. S. Davis 18 dozen of National Arithmetics at 6 dollars per dozen, 23 dozen of Mental Arithmetics at I dollar per dozen, 17 dozen Family Bibles at 3 dollars per copy; what is the amount of the bill? Ans. 743 dollars. 43. R. Hasseltine sold to John James 169 tons of timber at 7 dollars per ton, 116 cords of oak wood at 6 dollars per cord, and 37 cords of maple wood at 5 dollars per cord; James has paid Hasseltine 144 dollars in cash, and 23 yards of cloth at 4 dollars per yard; what remains due to HIasseltine? Ans. 1828 dollars. 44. J. Frost owes me on account 375 dollars, and he has paid me 6 cords of wood at 5 dollars per cord, 15 tons of hay at 12 dollars per ton, and 32 bushels of rye at 1 dollar per bushel. How much remains due to me? Ans. 133 dollars. 45. Gave 169 dollars for a chaise, 87 dollars for a harness, and 176 dollars for a horse. I sold the chaise for 187 dollars, the harness for 107 dollars, and the horse for 165 dollars. What sum have I gained? Ans. 27 dollars. 46. Bought a farm of J. C. Bradbury for 1728 dollars, for which I paid him 75 barrels of flour at 6 dollars per barrel, 9 cords of wood at 5 dollars a cord, 17 tons of hay at 25 dollars a ton, 40 bushels of wheat at 2 dollars a bushel, and 65 bushels of beans at 3 dollars a bushel; how many dollars remain due to Bradbury? Ans. 533 dollars.

Page 69 SECT. IX.] UNITED STATES MIONEY. 69 ~ IX. UNITED STATES MONEY. ART. CT, UNITED STATES MONEY, established by Congress in 1796, is the legal currency of the United States. TABLE. 10 Mills make 1 Cent, marked c. 10 Cents " 1 Dime, " d. 10 Dimes' n 1 Dollar, $ 10 Dollars' 1 Eagle, 6' E. Cents. Mills. Dimes. 1 10 Dollars. 1 = 10 100 Eagle. 1.10 _ 100 = 1000 1 = 10 100 = 1000 = 10000 SIMPLE NUMBERS, that is, numbers whose units are all of a single denomination, have thus far, in this work, been made use of alone in the operations. But as the units or denominations of United States money in crease from right to left, and decrease friom left to right, in the same manner as do the units of the several orders in simple numbers, they may, therefore, be added, subtracted, nmultiplied, and, divided, according to the same rules. Dollars are separated firom cents by a point, called a separatrix or decimal point; the first two places at the right of the point being cents; and the third place, mills. Thus, $ 1G.253 is read, sixteen dollars, twenty-five cents, three mills. Since cents occupy two places, the place of dimes and of cents, when'the number of cents is less than 10, a cipher must be written before them in the place of dimes; thus,.03,.07, &c. The coins of the United States consist of the double-eagle, eagle, half-eagle, quarter-eagle, three dollars, and dollar, made of gold; the dollar, half-dollar, quarter-dollar, dime, half-dime, and threecent piece, made of silver; the cent and half-cent, made of copper. NOTE 1. - The word IMILL is from the Latin word mnille (one thousand); the word CENT, fi'om the Latin censtlm (one hundred); the word D)mE, from a French word signifying a tithe or tenth; and the reason of these QuEs'rTNs. - Art. 68. What is United States money? Repeat the Table of United States Money. What is a simple number? AWhat are the denominations of United States money? lowv do they increase from rigoht to left? HIow are they added, subtracted, moltiplied, and divided? Ilow are dollars, cents, and mills, separated'? Why must a cipher be placed befire cents, when the number is less than 10? Why are two places. allowed for cents, while only one is allowed for mills? Name the coins of the United States.

Page 70 70 UNITED STATES MONEY. [SECT. IX. names, as applied to our coins, is found in the proportion which they respectively bear to the dollar. The term DOLLAR is said to be derived from the Danish word Daler and this from Dale, the name of a town, where it was first coined. The symbol $ represents, probably, the letter U written upon an S, denoting U. S. (United States). NoTE 2. — All the gold and silver coins of the United States are now made of one purity, nine parts of pure metal, and one part alloy. The alloy for the silver is pure copper; and that for the gold, one part copper and one part silver. The cent is now made of pure copper and nickel. The standard weight, as fixed by present laws, of the eagle, is 258 grains, Troy; the silver dollar, 412k grains; half-dollar, 192 grains; quarterdollar, 96 grains; dime, 38S grains; half-dime, 19- grains; three-cent piece, 11_1-5 grains; and the cent, new coinage, 72 grains. REDUCTION OF UNITED STATES MIONEY. ART. 69. REDUCTION of United States Money is changing the units of one of its denominations to the units of another, either of a higher or lower denomination, without altering their value. ART. 70. To reduce units from a higher denomination to a lower. Ex. 1. Reduce 25 dollars to cents and mills. Ans. 2500 cents, 25000 mills. OPERATION. 2 5 dollars. 1 00 We0 multiply the 25 by 100, be2 - c cen. cause 100 cents make 1 dollar; 00 cents. and multiply the 2500 by 10, be1 0 cause 10 mills make i cent. 2 5 0 0 0 mills. Or thus, 2 5 0 0 0 mills. RULE. - To reduce dollars to cents, annex Two ciphers; to reduce dollars to mills, annex THREE ciphers; and to reduce cents to mills, annex ONE cipher. NoTE. - Dollars, cents, and mills, expressed by a single number, are reduced to mills by merely removing the separating point; and dollars and cents, by annexing one cipher and removing the separatrix. ART. 71. To reduce units from a lower denomination to a higher. Ex. 1. Reduce 25000 mills to cents and dollars. Ans. 2500 cents, $25. QUESTIONS. - Art. 69. What is reduction of United States Money? - Art. 70. What is the rule for reducing dollars to-cents and mills? Give the reason for the rule. How do you reduce dollars and cents to cents, or dollars, cents, anld mills, to mills? What is the reason for this rule?

Page 71 SECT. IX.] UNITED STATES MONEY, 71 OPERATION. 1 0 ) 2 5 0 0 0 mills. We divide the 25000 by 10, be1 0 0 ) 2 5 0 0 cents. cause 10 mills make 1 cent; and divide the 2500 by 100, because 2 5 dollars. 100 cents make 1 dollar. Or thas, 2 510 010 mills. RULE. - To reduce mills to cents, cut of ONE figure on the right; to reduce cents'o dollars, point ofi TWO figures; and to reduce mills to dollars, poEint off THREE figures. EXAMPLES FOR PRACTICE. 1. Reduce $125 to cents. Ans. 12500 cents. 2. Reduce $345 to mills. Ans. 345000 mills. 3. Reduce 297 mills to cents. Ans. $0.297. 4. Reduce 2682 mills to dollars. Ans. $2.682. 5. Reduce 4123 cents to dollars, Ans. $41.23. 6. Reduce $156.29 to cents. Ans. 15629 cents. 7. Reduce $16.428 to mills. Ans. 16428 mills. 8. Reduce $9.87 to mills. Ans. 9870 mills. ART. 72. ADDITION OF UNITED STATES MONEY. RULE. - Write dollars, cents, and mills, so that units of the same denomination shall stand in the same column. Add as in addition of simple numbers, and place the separating point directly under that above. Proof. - The proof is the same as in addition of simple num bers. EXAMPLES FOR PRACTICE. 1. 2. 3. 4. $. cts. m. $. cts. m. $. cts. m. $. cts. 45.2 4 3 75.64 3 16.7 05 1 47.8 6 13.896 16.897 14.003 789.58 9 3.5 1 6 4 3.8 16 1 8.7 19 4 9 6.3 7 52.343 58.313 97.009 91 1.34 Ans, 2 0 4.9 9 8 19 4.6 6 9 14 6.4 3 6 2 3 4 5.15 QUIJESTIONS. - Art. 71. What is the rule for reducing mills to cents? For reducing cents to dollars? For reducing mills to dollars? Give the reason for each. - Art. 72. How must the numbers be written down in addition of United States money? How added? How pointed off? Repeat the rule.

Page 72 72 IUNITED STATES MONEY. [ESECT. IX 5. 6. 7. 8. $. cts. m. $. M.. cts. m. S. cts. m,, 86.7 1 3 87.05 9 9 1.7 63 7 86.713 1 7 6.071 37.810 84.1 6 1 345.67 8 567.819 81.475 1 0.070 907.01 7 1 2 3.456 4 0.07 8 5 3.615 861.090 7 89.0 1 2 2 1.15 6 81.1 76 1 2 3.47 6 345.67- 8 81.177 3 2.817 9 7.016 90 1.234 33.621 53.1 96 345.7 05 7 1 8.905 28.093 4 1.5 70 3 5 7.091 9. Bought a coat for $17.81, a vest for $3.75, a pair of pantaloons for $2.87, and a pair of boots for $7.18; what was the amount? Ans. $31.61. 10. Sold a load of wood for seven dollars six cents, five bushels of corn for four dollars seventy-five cents, and seven bushels of potatoes for two dollars six cents; what was received for the whole? Ans. $13.87. 11. Bought a barrel of flour for $6.50, a box of sugar for $9.87, a ton of coal for $12.77, and a box of raisins for $2.50; what was paid for the various articles? Ans. $31.64. 12. Paid $4.62 for a hat, $9.75 for a coat, $5.75 for a pair of boots, and $1.50 for an umbrella; what was paid for the whole? Ans. $21.62. 13. A grocer sold a pound of tea for $0.625; 4 pounds of butter for $0.75; 4 dozen of lemons for $0.875; 9 pounds of sugar for $0.80; and 3 pounds of dates for $0.375. What was the amount of the bill? Ans. $3.425. 14. A student purchased a Latin grammar for $0.75, a Virgil for $3.75, a Greek lexicon for $4.75, a Homer for $1.25, an English dictionary for $3.75, and a Greek Testament for $0.75; what was the amount of the bill? Ans. $15. 15. Bought of J. H. Carleton a China tea-set for ten dollars eighty-two cents, a dining-set for nine dollars sixty-two cents five mills, a solar lamp for ten dollars fifty cents, a pair of vases for four dollars sixty-two cents five mills, and a set of silver spoons for twelve dollars seventy-five cents; what did the whole cost? Ans. $48.32. 16. Bought three hundred weight of beef at seven dollars seven cents per hundred weight, four cords of wood at six dollars four cents per cord, and a cheese for three dollars nine cents; what was the amount of the bill? Ans. $48.46.

Page 73 S8ct. IX.] UNITED STATES MONEY. 73 ART. 73o SUBTRACTION OF UNITED STATES MONEY. RULE. - Write the several denominations of the subtrahend under the corresponding ones of the minuend. Subtract as in subtraction of simple numbers, and place the separatrix directly under that above. Proof. - The proof is the same as in subtraction of simple numbers. EXAMPLES FOR PRACTICE.'1. 2. 3. 4. $. cts. m. $. ets. cts. m. cts. m. $. cts. 5Min. 6 1.5 8 5 4 7 1.8 1 1 5 6.0 0 3 1 41.70 Sub. 19.197 1 5 8.1 9 1 9.0 0 9 9 0.9 91 Item. 4 2.3 8 8 3 1 3.6 2 1 3 6.9 9 4 5 0.7 9 5. 6. 7. 8. $. cts. m. $. ct... ts.. cts. m. From 71.8 61 91.0 71 815.7 01 10781.3 0 3 Take 1 9.1 9 7 1 9.0 9 5 9 0.8 0 3 9 9 9 9.0 9 7 9. From $71.07 take $5.09. Ans. $65.98. 10. From $100 take $17.17. Ans. $82.83. 11. From one hundred dollars there were paid to one man seventeen dollars nine cents, to another twenty-three dollars eight cents, and to another thirty-three dollars twenty-five cents, how much cash remained? Ans. $26.58. 12. From ten dollars take nine mills. Ans. $9.991. 13. A lady went a shopping," her mother having given her fifty dollars. She purchased a dress for fifteen dollars seven cents; a shawl for eleven dollars ten cents; a bonnet for seven dollars nine cents; and a pair of shoes for two dollars. How much money had she remaining? Ans. $14.74. 14. From one hundred dollars there were taken at one time thirty-one dollars fifteen cents seven mills; at another time, seven dollars nine cents five mills; at another time, five dollars five cents; and at another time, twenty-two dollars two cents seven mills. How much cash remained of the hundred dollars? Ans. $34.671. QUESTIONS. - Art. 73. How do you write down the numbers in subtraction of United States money? How subtract? How pointed off? Repeat the rule.

Page 74 74 UNITED STATES MONEY. [SECT. IX. ART. 74. MULTIPLICATION OF UNITED STATES MONEY. RULE. - Multiply as in multiplication of simple numbers. The product will be in the lowest denomination in the question which must be pointed off as in reduction of United States money (Art. 71.) Proof. -- The proof is the same as in multiplication of simple numbers. EXAMPLES FOR PRACTICE. 1. What will 143 barrels 2. What will 144 gallons of flour cost at $7.25 per of oil cost at $1.625 a galbarrel?, Ans. 1036.75. ion? Ans. $234. OPERATION. OPERATION. Multiplicand $7.2 5 Multiplicand $1.6 2,5 Multiplier 1 4 3 Multiplier 1 4 4 2175 6500 2900 6500 725 1625 Product $1 0 3 6.7 5 Product $2 3 4.0 0,0 3. What will 165 gallons of molasses cost at $0.27 a gallon? Ans. $44.55. 4. Sold 73 tols of timber at $5.68 a ton; what was the amount? Ans. $414.64. 5. What will 43 rakes cost at $0.17 apiece? Ans. $7.31. 6. What will 19 bushels of salt cost at $1.625 per bushel? Ans. $30.875. 7. What will 47 acres of land cost at $37.75 per acre? Ans. $1774.25. 8. What will 19 dozen penknives cost at $0.375 apiece? Ans. $85.50. 9. What is the value of 17 chests of souchong tea, each weighing 59 pounds, at $0.67 per pound? Ans. $672.01. 10. When 19 cords of wood are sold at $5.63 per cord, what is the amount? Ans. $106.97. 11. A merchant sold 18 barrels of pork, each weighing 200 pounds, at 12 cents 5-mills a pound; what did he receive? Ans. $450. QUESTIONS. - Art. 74. How do you arrange the multiplicand and multiplier in multiplication of United States money? HIow multiply? Of what denomination is the product? How must it be pointed off? Repeat the rule.

Page 75 SECT. IX. UNITED STATES MONEY. 75 12. What cost 132 tons of hay at $12.125 per ton? Ans. $1600.50. 13. A farmer sold one lot of land, containing 187 acres, at $37.50 per acre; another lot, containing 89 acres, at $137.37 per acre; and another lot, containing 57 acres, at $89.29 per acre; what was the amount received for the whole? Ans. $24 327.96. ArT. 75, DIVISION OF UNITED STATES MONEY. RUE. - Divide as in division of simple numbers. VThe quotient will be in the lowest denomination of the dividend, which must be pointed off as in reduction of United States money. (Art. 71.) NoTr. - When the dividend consists of dollars only, and is either smaller than the divisor or not divisible by it without a remainder, reduce it to a lower denomination by annexing two or three ciphers, as the case may requile, and the quotient will be cents or mills accordingly. Proof.- The proof is the same as in division of simple numbers. EXAMPLES FOR PRACTICE. 1. If 59 yards of cloth cost 2. Purchased 68 ounces of $90.27, what will 1 yard indigo for $17. What did I cost? Ans. $1.53. give per ounce? Ans. $0.25. OPERATION. OPERATION. Dividend. $. Dividend. $. Divisor 5 9 ) 9 0.2 7 ( 1.5 3 Quotient. Divisor 6 8 ) 17.0 0 ( 0.2 5 Quotient. 59 136 312 340 295 340 177 177 3. If 89 acres of land cost $12225.93, what is the value of 1 acre? Ans. $137.37. 4. When 19 yards of cloth are sold for $106.97, what should be paid for 1 yard? Ans. $5.63. QUEslTONs.- Art. 75. How do you arrange the dividend and divisor in division of United States money? How divide? Of what denomination is the quotient? HIow pointed oil? How do you proceed when the dividend is dollars only, and is either smaller than the divisor or not divisible by it without a remsainder? Repeat the rule.

Page 76 76 UNITED STATES MONEY. LSECT. IX 5. Gave $22.50 for 18 barrels of apples; what was paid for 1 barrel? For 5 barrels? For 10 barrels? Ans. $20 for all. 6. Bought 153 pounds of tea for $90.27; what was it per pound? Ans. $0.59. 7. A merchant purchased a bale of cloth, containing 73 yards, for $414.64; what was the cost of 1 yard? Ans. $5.68. 8. If 126 pounds of butter cost $16.38, what will 1 pound cost? Ans. $0.13. 9. If 63 pounds of tea cost $58.59, what will 1 pound cost? Ans. $0.93. 10. If 76 cwt. of beef cost $249.28, what will 1 cwt. cost? Ans. $3.28. 11. If 96,000 feet of boards cost $1120.32, what will a thousand feet cost? Ans. $11.67. 12. Sold 169 tons of timber for $790.92; what was received for 1 tonl? Ans. $4.68. 13. When 369 tons of potash are sold for $48910.95, what is received for I ton? Ans. $132.55. 14. Fior 19 cords of wood I paid $109.25; what was paid for 1 cord? Ans. $5.75. PRACTICAL QUESTIONS BY ANALYSIS. ART. 7ts ANALYSIS is an examination of a question by resolving it into its parts, in order to consider them separately, and thus render each step in the solution plain and intelligible. ART. 77, The price of one pound, yard, bushel, &c., being given, to find the price of any quantity. RULE. - Multiply the price by the quantity. Ex. 1. If 1 ton of hay cost $12, what will 29 tons cost? Ans. $348. ILLUSTRATION - Since 1 ton costs $12, 29 tons will cost 29 times as much: $12 X 29 $348. 2. If 1 bushel of salt cost 93 cents, what will 40 bushels cost? Vhat will 97 bushels cost? Ans. $90.21. QuEsTIoNs. - Art. 77. The price of 1 pound, &c., being given, how do you find the price of any quantity? Give the reason for this rule.

Page 77 rECT. 1X.] UNITED STATES MONEY. 7 7 3. If 1 bushel of apples cost $1.65, what will 5 bushels cost? What will 18 bushels cost? Ans. $29.70. 4. If 1 ton of clay cost $0.67, what will 7 tons cost? What will 63 tons cost? Ans. $42'21. 5. When $7.83 are paid for 1 cwt. of sugar, what will 12 cwt. cost? What will 93 cwt. cost? Ans. $728.19. 6. When $0.09 are paid for I lb. of beef, what will 12,1b. cost? What will 760 lb. cost? Ans, $68.40. 7. A gentlemanl paid $38.37 for 1 acre of land; what was the cost of 20 acres.'What would 144 acres cost? Ans. $5525.28. 8. Paid,$6.83 for 1 barrel of flour; what was the value of 9 barrels? What must be paid for 108 barrels? Ans. $737.64. ART. 780 The price of any quantity, and the quantity being given, to find the price of a unit of that quantity. RULE. - Divide the price by the quantity. 9. If 15 bushels of corn cost $10.35, what will I bushel cost? Ans. $0.69. ILLUSTRATION. - Since 15 bushels cost $10.35, 1 bushel will cost as many cents as 15 is contained times in $10.35: $10.35 15 _= 0.69. 10. Bought 65 barrels of flour for $422.50; what cost one barrel? What cost 15 barrels? Ans. $97.50. 11. For 45 acres of land a farmer paid $2025; what cost 1 acre? What 180 acres? Ans. $8100. 12. For 5 pairs of gloves a lady paid $3.45; what cost 1 pair? What cost 11 pairs? Ans. $7.59. 13. If 11 tons of hay cost $214.50, what will 1 ton cost? iWhat will 87 tons cost? Ans. $1696.50. 14. When $60 are paid for 8 dozen of arithmetics, what will 1 dozen cost? What will 87 dozen cost? Ans. $652.50. 15. Gave $5.58 for 9 bushels of potatoes; what will 1 bushel cost? What will 43 bushels cost? Ans. $26.66. 16. Bought 5 tons of hay for $85; what would I ton cost? What would 97 tons cost? Ans. $1649. QUESTIONS. - Art. 78. How do you find the price of 1 pound, &c., the price of any quantity and the quantity being given? What is the reason for this rule? 7*

Page 78 78 UNITED-, STATES MONEY. [SECT. IX 17. If J. Ladd will sell 20 lb. of butter for $3.80, what should he charge for 59 lb.? Ans..11.21. 18. Sold 27 acres of land for $472.50; what was the pric. of 1 acre? What should be given for 12 acres? Ans. $210. 19. Paid $39.69 for 7 cords of wood; what will 1 cora cost? What will 57 cords cost? Ans. $323.19. 20. Paid $10.08 for 144 lb. of pepper; what was the price of 1 pound? What cost 359 lb.? Ans. $25.13. 21. Paid $77.13 for 857 lb. of rice; what cost 1 lb.? What cost 359 lb.? Ans. $32.31. 22. J. Johnson paid $187.53 for 987 gal. of molasses; what cost 1 gal.? What cost 329 gal.? Ans. $62.51. 23. For 47 bushels of salt J. Ingersoll paid $26.32; what cost 1 bushel? What cost 39 bushels? Ans. $21.84. ART. *7, The price of any quantity and the price of a unit of that quantity being given, to find the quantity. RULE. - Divide the whole price by the price of a unit of the quantity required. 24. If I expend $150 for coal at $6 per ton, how many tons can I purchase? Ans. 25 tons. ILLUSTRATION. - Since I pay $6 for 1 ton, I can purchase as many tons with $150 as $6 is contained times in $150: $1.50 $6 = 25; therefore I can purchase 25 tons. 25. At $5 per ream, how many reams of paper can be bought for $175? Ans. 35 reams. 26. At $7.50 per barrel, how many barrels of flour call be obtained for $217.50? Ans. 29 barrels. 27. At $75 per ton, how many tons of iron can be purchased for $4875? Ans. 65 tons. 28. At $4 per yard, how many yards of cloth can be bought for $1728? Ans. 432 yards. 29. How many hundred weight of hay can be bought for $9.66, if $0.69 are paid for 1 hundred weight? Ans. 14 hundred weight. 30. If $66.51 are paid for flour at $7.39 per barirel, how many barrels can be bought? Ans. 9 barrels. 31. Paid $136.50 for wood, at $3.25 per cord; how many cords did I buy'? Ans. 42 cords. QUESTIONS. - Art. 79. How do you find the quantity, the price of 1 pound, &c., being given? Give the reason for the rule.

Page 79 SECT. IX.] UNITED STATES MONEY. 79 BILLS. ART. 80. A BILL is a paper, given by merchants, containing a statement of goods sold, and their prices. An invoice is a bill of merchandise shipped or forwarded to a purchaser, or selling agent. The date of a bill is the time and place of the transaction. The bill is against the party owing, and in favor of the party who is to receive the amount due. A bill is receipted, when the receiving of the amount due is acknowledged by the party in whose favor it is. A clerk, or any other authorized person, may, in his stead, receipt for him, as in bill 2. When the items of a bill have been rendered at different dates, the several times may be given at the left hand, as in bill 5. When the bill is in the form of an account, containing items of debt and credit in its settlement, it is required to find the difference due, or balance, as in bill 5. What is the cost of each article in, and the amount due of, each of the following bills? (1.) New York, ]May 20, 1856. Dr. JOHN SMIITH, Bought of SOMES & GRIDLEY, 82 gals. Temperance Wine, at $0.75 89 " Port do. ".92 24 pairs Silk Gloves, ".50 $155.38. Received payment, SOMES & GRIDLEY. (2.) Philadelphia, March 7, 1857. lr. LEvI WEBSTER, Bought of JAMES PRANKLAND, 6 lbs. Chocolate, at $0.18 12 " Flour, ".20 6 pairs Shoes,' 1.80 30 lbs. Candles, -.26 - 22.08. Received payment, JAMES FRANKLAND, by ENOCH OSGOOD. QUESTIONS. - Art. 80. What is a bill, in mercantile transactions? What is an invoice? When is a bill against, and when in favor of a party? How is a bill receipted?

Page 80 80 LNITED STATES M1ONEY. [SECT. IX (3.) St. Louis, March 19, 1856 Mr. WILLIAxM GREENLEAF, Bought of MiosEs ATWOOD, 86 Shovels, at $0.50 90 Spades, ".86 18 Ploughs, " 11.00 23 Haidsaws, *6 3.50 14 Hammers, 66.62 12 Mill-saws, 6 12.12 46 ewt. Iron, " 12.00 $1105.02. (4.) Boston; June 5, 1856. Mir. Amos Dow, Bought of LORD & GREENLEAF, 37 Chests Green Tea, at $23.75 42 "' Black do., " 17.50 43 Casks Wine, " 99.0'0 12 Crates Liverpool Ware, " 175.00 19 bbl. Genesee Flour, " 7.00 23 bu. Rye, " 1.52 (5.) San Francisco, May 13, 1856. iMr. JOEN WADE, 1855. To AYER, FTTTS & CO., Dr. Apr. 5. To 80 pairs Hose, at $1.20 Aug. 7. " 17 " Boots, " 3.00 " 19 " Shoes, " 1.08 Nov. 1.' 23 " Gloves, ".75 $184.77 1856. Cr. Jan. 1. 27 Young Readers, at $0.20 " " 10 Greek Lexicons, " 3.90 Feb. 10. 7 Web.ster's Dictionaries, " 4.75 Apr. 3. 19 Folio Bibles, " 2.93 " " 20 Testaments, ".37 $140.72. Balance due A., F. & Co. $44.05. Received payment, AYER, FITTS & Co.

Page 81 SECT. IX.] UNITED STATES MONEY. 81 LEDGER ACCOUNTS. ART. 81. The principal book of accounts among merchants is called a ledger. In it are brought together scattered items of account, often making long columns. As a rapid way of finding the amount of each, accountants generally add more than one column at a single operation. (Art. 24.) The examples below may be added both by the usual method and by that which is more rapid. L 2. 3. 4. $. cts. $. cts. $. cts. $. cts. 5.7 5 1.0 5 71.1 0 100.8 8 3.1 5 7.0 8 3 5.6 0 3 2 0.1 2 6.3 7 6.3 8 2 1.4 0 2 8 0.4 7 10.1 3 5.5 0 100.5 0 15 1.5 3 5.05 3.2 5 62.75.9 2 12.50 8.19 13.13 11.0 8 8.00 1.13 1.37 49.13.6 3 10.1 0 1 6.02 44.22 1.37 15.25 19.2 8 6 0.81 2 2.00 1 3.45 163.35 5 2.75 16,0 5 6,17 6 2 0.5 0 3 5.1 5 1.1L9.09 75.0 0 7 0.06.31 1.1 3 25.20 105 0.00 10.0 0 8.07 5 3.81 312 0.12 11.88 1.0 6 3 3.19 2 0 0.5 0.1 2 35.1 5 17.0 0 16.0 9 9,1 7 1 8.91 1 0.3 8 9 O 0.1 1.3 3 1 0.0 3 4 0.1 2 1 82 5.5 0 6.2 2 3 0.0 0 1 5.6 8 10 5.1 0 2.31 1.88 71.12 35.4 6 7.17 2.75 13.1 9 67.6 3 15.5 0 1.2 5 1 0.0 0 8 1.1 7 1 1.2 5 5.0 0 1 8.2 0 1 0.1 4.09 25.50 1 3.15 75.0 0 21.1 7 12.02 25.00 1 2 0.00 32.00 19.1?7 1 02.55 14.0 9 1 4.0 6 3 2.4 3 1 1.1 0 2 1 2.6 3 20.50 4 6.3 7 2 35.83 1 0 300.4 8 QUESTIONS. -Art. 81. What is a ledger? How may ledger columns bo added rapidly?

Page 82 82 REDUCTION. rSECT. I ~ X. REDUCTION. ART. 8 2 A SIMPLE number is a unit or a collection of units, either abstract, or concrete of a single kind or denomination; thus, 1 dollar, 9 apples, 12, are simple numbers. A COMPOUND number is a collection of concrete units of several kinds or denominations, taken collectively, thus, 12k. 18s. 9d., is a compound number. ART. 83, Reduction is changing numbers, either simple or compound, from one denomination to another, without altering their values. It is of two kinds, Reduction Descending, and Reduction Ascending. Reduction Descending is changing numbers of a higher de nomination to a lower denomination; as pounds to shillings, &c. It is performed by multiplication. Reduction Ascending is changing numbers of a lower denomination to a higher denomination; as farthings to pence, &c. It is the reverse of Ieduction Descending, and is performed by division. ENGLISH MONEY. ART. ~i. English or Sterling Money is the Currency of England. TABLE. 4 FPtrthings (qr. or far.) make 1 Penny, d. 12 Pence I" 1 Shilling, s. 20 Shillings " 1 Pound, X. 21 Shillings sterling Ic 1 Guinea, G. 20 Shillings " " 1 Sovereign, soy. d. far. S. 1 4 ~-.1 - 12 = 48 1 = 20 - 240 = 960 NOTE 1.- The symbol ~. stands for the Latin word libra, signifying a pound; s. for solidus, a shilling; d. for denariz s, a penny; qr. for quzadrans, a quarter. QUlESTIONS. - Art. 82. What is a simple number? What is a compound number?-Art. 83. What is reduction? How many kinds of reduction? What are they? What is reduction descending? What is reduction ascending?- Art. 84. What is English money? Repeat the table.

Page 83 SECT. X ] REDUCTION. 83 NOTE 2. — Farthings are sometimes expressed in a fraction of a penny; thus, l far. = 4 d.; 2 far. = a d.; 3 far. = 3 d. NOTE 3.- -The Pound Sterling is represented by a gold coin called a sovereign. Its legal value in United States money is $4.84. NOTE 4.- The term sterling is probably from Easterling, the popular name of certain early German traders in England, whose money was noted for the purity of its quality. MENTAL EXERCISES. 1. How many farthings in 3 pence? In 9 pence? 2. How many pence in 2 shillings? In 6 shillings? 3. How many shillings in 7 pounds? In 10 pounds? 4. Fow many pence in 8 farthings? In 24 farthings? 5. How many shillings in 24 pence? In 60 pence? 6. How many pounds in 40 shillings? In 80 shillings? EXERCISES FOR THE SLATE. ART. 85, To reduce units of a higher denomination to a lower. 1. How many farthings in 17~. 8s. 9d. 3far.? OPERATION. We multiply the 17 by 20, bc1 7~. 8s. 9d. 3far. cause 20 shillings make 1 pound, 2 0 and to this product we add the 8 shillings in the question. We then 3 4 8 shillings. multiply by 12, because 12 pence 1 2 make 1 shilling, and to the product we add the 9d. Again, we multi41 8 5 pence. ply by 4, because 4 farthings make 4 1 penny, and to this product we add the 3 far.; and we find the Ans. 1 6 7 4 3 farthings. answer to be 16743 farthings. RULE. - Multiply the highest denonmnation given by the number required of the next lower denomination to make one in the denomination maultiplied, To this product add the corresponding denomination of the multiplicand, if there be any. Proceed in this way, till the reduction is brought to the denomination required. QUESTIONS. - Art. 85. How do you reduce pounds to shillings? Why multiply by 20? How do you reduce shillings to pence? Why? Pence to farthings? Why? Guineas to shillings? What is the general rule for roeduction descenling?

Page 84 84 REDUCTION. [SECT. X ART. 8~ o To reduce the unit of a lower denomination to a higher. Ex. 2. How many pounds in 16743 farthings? OPERATION. We divide by 4, because 4 farthings 4) 1 6 7 4 3 far. make 1 penny, and the result is 4185 pence, and the remainder, 3, is far1 2 ) 4 1 8 5 d. 3far. things. We divide by 12, because 12 2 0 ) 3 4 8 s. 9d. pence make 1 shilling, and the result is 348 shillings, and the 9 remaining 1 7 X. 8s. is pence. Lastly, we divide by 20, Ans, 17C. 8s. 9d. 3fatr, because 20 shillings make 1 pound, and the result is 17~. 8s. Therefore, by annexing all the remainders to the last quotient, we find the answer to be 172. 8s. 9d. 3far. RULE. - Divide the lower denomination given by the number, which it takes of that denomination to make one of the next higher. The quotient thus obtained divide as before, and so proceed until it is brought to the denomination required. The last quotient, vith the remainders connected, will be the answer. 3. In 9~. 18s. 7d. how many pence? 4. In 2383d. how many pounds, &c.? 5. How many farthings in 142. l1s. 5d. 2frt.? 6. How many pounds in 13990far.? TROY WEIGHT. ART. 87. Troy Weight is the weight used in weighing gold, silver, and jewels. TABLE. 24 Grains (gr.) make 1 Pennyweight, pwt. 20 Pennyweights 1 Ounce, oz. 12 Ounces 1 Pound, lb. pwt. gr. oz. 1 - 24 lb. 1 = 20 - 480 1 - 12 240 3 5760 QUESTIONS. - Art. 86. How do you reduce farthings to pence? Why divide by 4? How do you reduce pence to shillings? Why? Shillings to pounds? Why? Shillings to guineas? What is the general rule for reduction ascending' What is Troy Weight used for? Repeat the Table.

Page 85 SECT. X.] RED UCTION.,8,5 NOTE 1.- The oz. stands for onza, the Spanish for ounce. NOTE 2.- A grain or corn of wheat, gathered out of the middle of the ear, was the origin of all the weights used in England. Of these grains, 32, well dried, were to make one pennyweight. But in later times it was thought sufficient to divide the same pennyweight into 24 equal parts, still called grains, being the least weight now in use, from which the rest are computed. NOTE 3. — Diamonds and other precious stones are weighed by what is called Diamond Weight, of which 16 parts nake 1 grain; 4 grains, 1 carat. 1 grain Diamond Weight is equal to A grains Troy, and 1 carat to 31 grains Troy. NOTE 4. - The Troy pound is the standard unit of weight adopted by the United States Mint, and is the same as the Imperial Troy pound of Great Britain. MENTAL EXERCISES. I. How many gr. in 2pwt.? In 10pwt.? 2. How many pennyweights in 4oz.? In 20oz.? 3. How many ounces in 21b.? In 51b.? In 101b.? 4. How many pennyweights in 48gr.? In 96gr.? 5. Hlow many ounces in 40pwt.? In 120pwt.? 6. How many pounds in 24oz.? In 60oz.? In 120oz.? ExERCISES FOnr TIIE SLATE. 1. How many grains in 721b. 2. In 419887 grains, how 10oz. 15pwt. 7gr.? many pounds? OPERATION. OPERATION. 721b. 10oz. 15pwt. 7gr. 2 4 )4 1 9887 gr. 12 -2 0)1749 5 pwt. 7gr. 8 7 4 ounces. 1 2) 8 74 oz. 15pwt. 20 1 7 4 9 5 pennyweights. 7 2 lb. loz. 2 4 Ans. 721b. 10oz. 15pwt. 7gr. 69987 34990 Ans. 4 1 9 8 8 7 grains. QUESTIONS. — What was the original of all weights in England? How many of these grains did it take to make a pennyweight? IHow many grains in a pennyweight now? By what weight are diamonds weighed? What is the standard at the mint? How do you reduce pounds to grains? Give the reason of the operation. How do you reduce grains to pounds? 8

Page 86 86 REDUCTION. [SECT. X. 3. How many grains in 76pwt. 12gr.? 4. How many pennyweights in 1836gr.? 5. In 761b. 5oz. how many grains? 6. In 440160 grains how many pounds? 7. How many pennyweights in 1441b. 9oz.? 8. How many pounds in 34740pwt.? 9. How many pounds in 17895gr.? 10. In 31b. loz. 5pwt. 15gr. how many grains? 11. A valuable gem weighing 2oz. 18pwt. 12gr. was sold for $1.37 per grain; what was the sum paid? Ans. $1923.48. APOTHECARIES' WEIGHT. ART. 88, Apothecaries' Weight is used in mixing medicine. TABLE. 20 Grains (gr.) make 1 Scruple, sc. or D 3 Scruples " 1 Dram, dr. or o 8 Drams - " 1 Ounce, oz. or 5 12 Ounces i" 1 Pound, lb. or lb sO. gr. dr. 1 - 20 OZ. 1 = 3 60 lb. 1 - 8 - 24 = 480 1 - 12 - 96 288 = 5760 NOTE 1. - In this weight the pound, ounce, and grain are the same as in Troy Weight. NOTE 2. - IMedicines are usually bought and sold by Avoirdupois Weight. NOTE 3. - In estimating the weight of fluids, 45 drops, or a common tea-spoonful, make about 1 fluid dram; 2 common table-spoonfuls, about 1 fluid ounce. MENTAL ExERCISEs. 1. In 40 grains how many scruples? In 60gr.? In 120gr.? 2. In 5 scruples how many grains? In 10sc.? In 40sc.? 3. In 3 drams how many scruples? In lOdr.? In 17dr.? 4. How many pounds in 48 ounces? In 96oz.? In 144 oz.? 5. How many ounces in 24 drams? In 64dr.? In 96dr.? QUESTIONS.- Art 88. For what is Apothecaries' VWeight used? What de. nominations of this weight are the same as those of Troy Weight? By what weight are medicines usually bonight and sold? Repeat the tatble.

Page 87 SECT. X.] REDUCTION. 87 EXERCISES FOR THE SLATE. 1. In 401b. 8oz. 5dr. Isc. 7gr. 2. How many pounds in how many grains? 234567 grains? OPERATION. OPERATION, 4 0 lb. 8oz. 5dr. isc. 7gr. 2 0 ) 2 3 4 5 6 7 gr. 12 3)117 2 8sc. 7gr. 4 8 8 ounces. 8 )3 9 0 9 dr. lsc. 3~9 O 9 drams ~ 2 ) 4 S 8 oz. 5dr. 3 4 0 lb. Soz. 1 1 7 2 8 scruples. 2 0 Ans. 401b. 8oz. 5dr. isc. 7gr. Ans. 2 3 4 5 6 7 grains. 3. How many scruples in 761b.? 4. How many pounds in 21888 D? 5. How many grains in 1441b.? 6. How many pounds in 829440gr.? 7. In 1211 8$ 33 13 18gr. how many grains? 8. In 73178 grains how many pounds? 9. How many doses are there in 75 63 23 of tartar emetic, admitting 20 grains for each dose? Ans. 188. AVOIRDUPOIS WEIGHT. ART. 89, Avoirdupois Weight is used in weighing almost every kind of goods, and all metals except gold and silver. TABLE. 16 Drams (dr.) make 1 Ounce, oz. 16 Ounces " 1 Pound, lb. 25 Pounds " 1 Quarter, qr. 4 Quarters " 1 Hundred Weight, cwt. 20 HIundred Weight " 1 Ton, T. Oz. dr. lb. 1 - 16 qr. 1 16 - 256 cwt. 1 25 = 400 6400:T. 1 - 4 100 1600 - 25600 1 = 20 = 80 = 20000 = 32000 3 51200 QUESTIONS. - How do you reduce pounds to grains? 1W1hat is the reason for the operation? How do you reduce grains to pounds? Give the reason of the operation. - Art 89. For what is Avoirdupois Weight used? Recite the table.

Page 88 88 REDUCTION. [SECT. X NOTE 1. -In cwt. the c stands for centumn, the Latin for one hundred and wt for weight. NOTE 2.- The laws of most of the States, and common practice at the present time, make 25 pounds a quarter, as given in the table. But formerly, 28 pounds were allowed to make a quarter, 112 pounds a hundred, and 2240 pounds a ton, as is still the standard of the United States government in collecting duties at the custom-houses. NOTE 3. —The term avoirdupois is from the French avoir du poid, sig nifying to have weight. NOTE 4. - 1 pound Avoirdupois = 7000 gr. Troy = 1lb. 2oz. 11 pwt. 16 gr. Troy; 1lb Troy, or Apothecary = 5760gr. Troy = 13oz. 2L~4~ dr. Avoirdupois; loz. Troy, or Apoth. = 480gr. Troy = loz. 1-97 dr. Av.; loz. Av. = 4371gr. Troy = 18pwt. 5,gr. Troy; 1dr. Apoth. = 60gr. Troy = 23 4 dr. Av.; 1dr Av. = 27.Ogr. Troy= lpwt. 3 2gr. Troy; 1 pwt. Troy = 24gr. Troy = 7 6 8 of a dr. Av.; Isc. Apoth. = 20gr. Troy Ass of a dr. Av. MENTAL ExERCISES. 1. How many drams in 30oz.? In 7oz.? In lOoz.? In 12oz.? 2. How many ounces in 1Olb.? In 151b.? In 121b.? In 100lb.? 3. How many pounds in 2 quarters? In 3qr.? In 20qr.? 4. How many quarters in 10cwt.? In 16cwt.? In 17cwt.? 5. How many tons in 80cwt.? In 100cwt.? In 600cwt.? 6. How many hundred weight in 16qr.? In 48qr.? In 96qr.? EXERCISES FOR THE SLATE. 1. How many pounds in 176T. 2. In 3537901b. how l7cwt. 3qr. 151b.? many tons? OPERATION. OPERATION. 1 7 6T. 17cwt. 3qr. 151b. 2 5) 3 5 3 7 9 0 lb. 4)14151qr. 151b 35 37 hundred weight. 2 0 ) 3 5 37 cwt. 3qr. 1415 1 quarters. 1 7 6 T. wt 2 5 Ans. 176T. 17cwt. 3qr. 151b. 3 5 3 7 9 0 pounds 28302 Ans. 35 37 9 0 pounds. QUESTIONS. - How many pounds are now allowed for a cwt., and hoT, many for a quarter of a cwt., in most of the United States, in buying and selling articles by weight? I-low many at the custom-houses? How do you reduce tons to drams? Give the reason for the operation. How do you reduce drams to tons? What is the reason for the operation?

Page 89 $cEr. X.] REDUCTION. 89 3. In 16T. 19cwt. Oqr. 10b. lloz. 5dr. how many drams? 4. In 8681141 drams how many tons? 5. In 679cwt. how many pounds? 6. In 679001b. how many cwt.? 7. What cost 17cwt. 3qr. 181b. of beef, at 7 cents per pound? Ans. $125.51. 8. What cost 48T. 17cwt. of lead, at 8 cents per pound? Ans. $7816.00. CLOTH MEASURE. ART. 90. Cloth Measure is used in measuring cloth, ribbons, lace, and other articles sold by the yard or ell. TABLE. 24 Inches (in.) make 1 Nail, na. 4 Nails 1 Quarter of a yard, qr. 4 Quarters 66 1 Yard, yd. 3 Quarters 6 1 Ell Flemish, E. F. 5 Quarters 6 1 Ell English, E. E. na. in. qr. 1 2: E. F. 1 4 9 yd. 1 = 3 12 Y 27 E. E. 1 = 1 4 16 36 1 = 1 5 = 20 45 Nor. - The Ell French is 6 quarters; the Ell Scotch, 4qr. 1Uin. MENTAL EXEICISES. i. In 2.quarters how many nails? In 5qr.? In 8qr.? In 20qr.? In25qr.? In30qr.? In 40qr.? 2. In 3 yards how many quarters? In 7yd.? In 8yd.? In 14yd.? In 19yd.? In 100yd.? In 200yd.? 3. How many quarters in 8 nails? In 20na.? In 48na.? 4. I-low many yards in 20 quarters? In 40qr.? In 100qr.? EXERCISES FOR THE SLATE. 1. How many nails in 47yd. 2. In 765 nails how many 3qr. lna.? yards? QUESTIONS.- Art. 90. For what is cloth measure used? Repeat the table. Is the ell French longer or shorter than the ell English? Whiat makes an eli Scotch?

Page 90 90 RED UCTION. [SEcT. X OPERATION. OPERATION. 4 7 yd. 3qr. lna. 4)765na. 4 ) 19 1 qr. lna. 1 9 1 quarters. Ans. 4 7 yd. 3qr. Ina Ans. 7 6 5 nails. 3. In 144yd. 3qr. how many quarters? 4. Illn 579 quarters how many.yards? 5. In 17E. E. 4qr. 3na. how many nails? 6. In 359 nails how many ells English? 7. In 126yd. Oqr. 3na. how many nails? 8. In 2019 nails how many yards? 9. What cost 49yd. 3qr. of cloth, at $2.17 per quarter of a yard? Ans. $431.83. 10. What cost 144yd. lqr. 3na. of cloth, at 25 cents per nail? Ans. $577.75. LONG MEASURE. ART. 91, Long Measure is used in measuring distances in any direction. TABLE, 12 Inches (in.) make 1Foot, ft. 3 Feet " 1 Yard, yd. 5A Yards, or 16A Feet,. " 1 Rod, )r Pole, rd. 40 Rods'; 1 Furlong, fur 8 Furlongs, or 320 Rods,'" 1 Mile, m 3 Miles " 1 League, lea. 69 Miles (nearly) " 1 Degree, deg. or o 360 Degrees " 1 Circle of the Earth. ft. in. yd. 1 = 12 rd. 1 = 3 - 36 fur. 1 = 16 1 198 m. 1 - 40 = 220 - 660 = 7920 1 = = 8 = 320 1760 5280 =6336C QUESTIoNs.- -Iow do you reduce yards to nails? How do you reduce nails to yards? What is the reason for the operation? How is long measure used? Repeat the table.

Page 91 ECT. 2..] REDUCTION. 91 NOTE. — 1. 12 lines make 1 inch; 4 inches, 1 hand; 6 feet, 1 fathom, 1 of a degree of the circumference of the earth, 1 knot, or geographical mile, equal to 1~ statute miles. NOTE. — 2. The yard is the standard unit of linear measure adopted by the United States government, and it is the same as the imperial yard of Great Britain. A mnetre, the unit of linear measure, as established by the French government, is equal to about 393u7I English inches. NOTE.- 3. The English statute mile is the same as that of the United States, but that of other countries differs in value from it; as the German short mile is equal to 6857 yards, or about 3-u9 English miles; the German long mile, to 10125 yards, or about 5a English miles; the Prussian mile, to 8237 yards, or about 4 7o English miles; the Spanish common league, to 7416 yards, or about 45 English miles; the Spanish judicial league, to 4635 yards, or about 23 English miles. NOTE.- 4. A degree of longitude is -a of any circle of latitude. As the circles of latitude diminish in length, the degrees of longitude vary in length under different parallels of latitude. Thus, under the equator, the length of a degree of longitude is about 691 statute miles; at 25~ of latitude, 62 7i miles; at 40~ of latitude, 53 miles; at 420 of latitude, 51J miles; at 49~ of latitude, 45~ miles; at 60~, 34 miles. MENTAL EXERCISES. 1. How many inches in 4 feet? In 10ft.? In 12ft.? In 20ft.? 2. How many feet in 2 yards? In yd.? n 2Oyd.? In l8yd.? 3. How many rods in 2 furlongs? In 8fur.? In lfur. r? In 30 fur.? In 100fur.? In 200fur.? In 400fur.? 4. How many leagues in 9 miles? In 21m.? In 81m.? In 144m.? In 40m.? In 50m.? In 80m.? 5. How many furlongs in 120 rods? In 360rd.? In 1440rd.? 6. How many yards in 99 feet? In 66ft.? In 144ft.? 7. How many feet in 108 inches? In 144in.? In 1728in.? ExERCISES FOR THE SLATE. 1. In 66deg. 56im. 7fur. 37rd. 12ft. 9in. how many inches? QUESTIONS. - HOW many lines make 1 inch? How many inches 1 hand? liow many feet 1 fathom? What is the standard unit of linear measure adopted by the United States? Is the value of the mile the same in all countries? Ilow much is a degree of longitude under the equator? At 40O of latitude? At 42~ of latitude? At 60~ of latitude?

Page 92 92 REDUCTION. [SECT. X OPERATION. 6 6deg. 56m. 7fur. 37rd. 12ft. 9in. 6 2. In 292849479 inches how many 6 0 0 degrees? 4 0 1 oPERATION. 11 12)292849479 4 6 21 miles. 1 6 ) 2 4 4 0 4 1 2 3 ft. 3in. 8 2 2 36975 fur, 33 )48808246 [12ft. 6in 4 0 40)147 9037 rd. 25_ 2= 1479037 rods. 8)36 9 7 5 far. 37rd. 2 9 1 ) 4 6 2 1 miles 7fur 8874224 6 6 71479038 15 7 739518J 415)27726 24404122k ft. 6 6deg. 336+ 12 [6 = 56m. 2 9 2 8 4 9 4 7 9 in. Ans. 66deg. 56m. 7fur. 37rd. 12 ft. 3in. -6in. 66 56 7 37 12 9 Ans. NoTE. - To multiply by g, we take. of the multiplicand. — To divide by 16A, we first reduce both the divisor and dividend to halves, and then divide; and the remainder being 25 half-feet, we take half of it for the true remainder = 12ft. 6in. We adopt the same principle in dividing by 69-1; the remainder being 336 sixths of miles, we divide them by 6, and the quotient is 56 miles. By adding the 6 inches to the 3 inches, we obtain the true answer. 3. In 47 miles how many feet? 4. In 248160 feet how many miles? 5. In 78dceg 50m. 7fur. 30rd. 5yd. 2ft. 10in. how many inches? 6. How many degrees in 345056794 inches? QUESTIONS. - -Iow do you reduce degrees to inches? Give the reason of the operation. HIow do you reduce inches to degrees? What is the reason for the operation? How do you multiply by? How do you divide by 161 and find the true remainder? How do you obtain the true answer in examples of this kind?

Page 93 SECT. X.] REDUCTION 93 SURVEYORS' MEASURE. ART. 92. This measure is used by surveyors in measuring land, roads, &c. TABLE. 79k 2 Inches (in.) make 1 Link, 1. 25 Links " 1 Pole, p. 100 Links, 4 Poles, or 66 Feet. " 1 Chain, cha. 10 Chains " 1 Furlong, fur. 8 Furlongs, or 80 Chains, " 1 Mile, m. 1. in. Pa. 1 = 719 cha. 1 25 198 fur. 1 - 4 = 100 _ 792 m. i 1 10 _ 40 = 1000 - 7920 1 - 8 - 80 = 320 - 8000 =- 63360 NOTE. -Gunter's chain, in length 4 poles, or 66 feet, anid divided into 100 links, is that mostly used in ordinary land surveys; but in locating roads, and like public works, an engineer's chain is usually 100 feet in length, containing 120 links, each 10 inches long. MENTAL EXERCISES. 1. In 2 poles how many links? In 4 poles? In 7 poles? 2. In 5 chains how many links? In 8cha.? In 10cha.? 3. How many poles in 50 links? In 751.? In 1251.? ExERCISES FOR THE SLATE. 1. How many links in 7m. 2. In 61630 links how many 5fur. 6cha. 301.? miles OPERATION. OPERATION. 7 m. 5fur. 6cha. 301. 1 00 ) 6 1 6 3 0 1. 10 ) 6 1 6 cha 301. 6 1 furlongs. 8 ) 6 1 fur. 6cha. 10 7 m. 5fur. 6 1 6 chains. f 1 00 0Ans. 7m. 5fir. 6cha. 301. 6 1 6 3 0 links. Ans. 3. How many miles in 4386 chains? QUESTIONS. -Art. 92. For what is surveyors' measure used? Recite the table. How do you reduce miles to links? What is the reason for the operation? How do you reduce inches to chains? To miles? Give the reason of the operation.

Page 94 94 REDUCTION. [SECT. X 4. In 54mo 66cha. how many chains? 5. In 75m, 49cha. how many poles? 6. How many miles in 24196 poles? 7. How many links in 7m. 4fur. 30rd.? 8. How many miles in 60750 links? SQUARE MEASURE. ART. 93, Square Measure is used in measuring surfaces of all kinds. TABLE. 144 Square inches make 1 Square foot, ft. 9 Square feet " 1 Square yard, yd. 304 Square yards, orr pole, 2724 S3quare feet, ~ " 1 Square rod or pole, p. 2721 Square feet, 40 Square rods or poles 6 1 Rood, R.. 4 Roods, or 160 Poles, "6 1 Acre, A. 640 Acres " 1 Square mile, S. M. ft. in. yd. 1 = 144 P. 1= 9 = 1296 R. 1 = 30 =_ 2721 = 39204 A. 1 = 40 = 1210 = 10890 = 1568160 s. ir. 1 =L 4 = 160 4840 = 43560 -- 6272640 1 = 640 = 2560 -- 102400 = 3097600 = 27878400 =a4014489600 NOTE, -- A square is a figure bounded by four equal lines, perpendicular to each other. When the four lines are each 1 foot in length, the space enclosed is I squarefoott; when 1 yard in length, I squtare yard; when 1 rod in length, 1 square rod; and so for any other dimension. 3ft. l 1yd. SC uetre In this diagram the larye square reproefoot. sents a square yard, and each of the smaller,~j —----....._-l —-- squares within it represents one squarefoot. Now, since there are three rows of small squares, and three square feet in each row, CID there wvill be 3 sq. ft. X 3 9 sq. ft. in the large square. But the largoe square is 3 ft. in length, and 3 ft. in breadth; hence, To find tze conteCts of a sqUcare, multiplY the numbers ldenotigy its lengcth Clndl breadtlh toyether. QUJESTIONS. - Art. 93. For -what is square measure used? Repeat the table. Wha.t is a square? What is a square Iot? I-ow may the contents of a square be foundc? Explain by the diagraml the reason of the operation.

Page 95 ,iECT. X.] REDUCTION. 95 MENTAL EXERCISES. 1. In 2 square feet how many square inches? 2 In 3 square yards how many square feet? In 10 sq. yd.? 3. In 5 roods how many poles? In 20 roods? In 30 roods? 4. In 7 acres how many roods? In 24 acres? In 40 acres? EXERCISES FOIR THE SLATE. 1, How many square inches in 12A. 3R. 24p. 144ft. 72in.? OPERATION. 12 A. 3R. 24p. 144ft. 72in. 4 5 1 roods. 40 2 0 6 4 poles. NOTE. -TO multiply by 4, we 2 7 2 X take 4 of the multiplicand 4132 14452 4129 516 5 6 2 0 6 8 feet. 144 2248274 2248279 5 6 2 0,6 8 Ans. 8 0 9 3 7 8 6 4 inches. 2. In 80937864 square inches how many inches? OPERATION. 144) 80 9 37 864 inches. 2 7 2 ) 5 6 2 0 6 8 ft. 72in. 4 4 1089)2248272 fourths ofa foot. 40 ) 2 06 4 poles. 576 - 4 144ft. 4)51 R. 24p. Ans. 1 2 A. 3R. 24p. 144ft. 72in. NOTE. - TO divide by the 2721-, we first reduce the divisor and dividend to fourths, and then divide. The remainder obtained, being fourths, is reduced to whole numbers by dividing by 4. QUESTIONS. - How do you reduce acres to square inches? Give the reason for the operation. HIow do you reduce square inches to acres? What is the reason for the operation? ITow do you multiply by?

Page 96 96 REDUCTION. [SECT. [E, 3. In 49A. 3R. 16p. how many square feet? 4. In 2171466 square feet how many acres? 5. What is the value of 365A. 3R. 17p. at $1.75 per square rod or pole? Ans. $102,439.75. 6. Sold a valuable piece of land, containing 3A. 1R. 30p., at $1.25 per square foot; what was received for the land? Ans. $187,171.87,5. 7. In a tract of land 12 miles square, how many square miles? How many acres? Ans. 92160 acres. 8. In 18A. OR. 16p. how many square feet? Ans. 788436 square feet. 9. Purchased 48A. 3R. 14p. of land for $2.25 per square rod, and sold the same for $3.15 per square rod; what did I gain by my bargain? Ans. $7032.60. CUBIC OR SOLID MEASURE. ART. 91, Cubic or Solid Measure is used in measuring such bodies or things as have length, breadth, and thickness; a.s timber, stone, &c. TABLE. 1728 Cubic inches (cu. in.) make 1 Cubic foot, cu. ft. 27 " feet " 1 "6 yard, cu. yd. 40 " feet " 1 Ton, T. 16 " feet " 1 Cord foot, c. ft. 8 Cord feet, or "' 1 Cord of wood, C. 128 Cubic feet, ft. in. yd. 1 - 1728 T. 1 - 27 46656 c. 1 = 14 = 40 = 69120 l - 3- = 42 = 128 221184 NOTE 1. —A pile of wood 8ft. in length, 4ft. in breadth, and 4ft. in height, contains a cord. Also, one ton of timber, as usually surveyed, contains 6052 cubic or solid feet. QUESTIONS. - How do you divide by 2721? Of what denomination is the remainder? How is the true remainder found? - Art. 94. For what is cubic measure used? Recite the table. What are the dimensions of a pile of wood containing 1 cord? How many solid feet does a ton of timber contain, as usually surveyed?

Page 97 SECT. X.J REDUCTION. 97 NOTE 2. -A cube is a solid bounded by six square and equal sides. If the cube is 1 foot long, 1 foot wide, and 1 foot high, it is called a cubic or solid foot. If the cube is 3 feet long, 3 feet wide, and 3 feet thick, it is called a cubic or solid yard. Now, since each side of a cubic yard,. as represented in the diagram, contains 9 sq. ft. of surface (Art. 93), it is plain, if a block be cut off from one 111111 1side, one foot thick, it can be divided u11;II Byl 1!111 into 9 solid blocks, with sides 1 foot in length, breadth, and thickness, and therefore will contain 9 solid feet; t!XIlIll!t!il11!' and since the whole block or cube is 3 ft. 1 yd. three feet thick, it must contain 9 solid feet X 3 27 solid feet; or 3 solid feet X 3 X - 27 solid feet.: Hence, To filnd the conteznts of a c cubic body, mult1tily toyeth7er the ucnzbers denoting the le~nyth, breadth, and thic7kness. MENTAL EXERCISES. I. In 2 cubic feet how many cubic inches? In 4 cu. ft.? 2. In 3 cubic yards how many cubic feet? In 10 cu. yd.? 3. How many cords of wood in 64 cord feet? In 96 c. ft.? 4. How many tons in 80 cu. ft. of timber? In 160 cu. ft.? EXERCISES FOR Te HE SLATE. 1. In 48 cu. yd. and 15 cu. 2. In 2265408 cubic inches f. how many cubic inches? how many cubic yards? OPERATION. OPERATION. 48 yd. 15ft. 1728)2265408 cu. in. 27 27)1311 cu. ft. 341 Ans. 4 8 yd. 15ft. 97 1 3 11 feet. 1728 10488 2622 9177 1311 Ans. 2 2 6 5 4 0 8 inches. QUESTIONS. - What is a cube? How do you find the contents of a cube? Give the reason for the operation. Describe a cubic foot. How do you reduce a ton to cubic inches? Give the reason for the operation. How do you reduce cubic inches to cubic yards? Give the reason for the operation. 9

Page 98 98 REDUCTION. [SEcT. X 3. In 45 cords of wood how many cubic inches? 4. In 9953280 cubic inches how many cords of wood? 5. How many cubic feet in a pile of wood 15ft. long, 4ft wide, and 6 ft. high? How many cords? Ans. 3C. 6 cu. ft. 6. How many cubic inches in a block of marble 4ft. long 3kft. wide, and 2ft. thick? Ans. 44928. 7. In a room 14ft. long, 12ft. wide, and 8ft. high, how many cubic feet? Ans. 1344. 8. What will 9080 cubic feet of ship-timber cost, at $11.50 per ton? Ans. $2610.50 WINE OR LIQUID MEASURE. ART. 95. Wine or Liquid Measure is used in measuring all kinds of liquids, except, in some places, beer, ale, porter, and milk. TABLE. 4 Gills (gi.) make 1 Pint, 1t. 2 Pints " 1 Quart, qt. 4 Quarts " 1 Gallon, gal. 63 Gallons 6 1 IHogshead, hhd. 2 Hogsheads 16 1 Pipe, or Butt, pi. 2 Pipes; 1 Tun, tun. pt. gi. qt. 1 - 4 gal. 1 2 8 hhd. 1 = 4 = 8 - 32 pi. i = 63 - 252 = 504 - 2016 tan. 1 2 -- 126 504 1008 = 4032 I =- = 4 = 252 1008 - 2016 - 8064 NOTE 1.- By laws of Massachusetts, 32 gallons make 1 barrel. In some states 31A gallons, and in others from 28 to 32 gallons, make 1 barrel. 42 gallons make 1 tierce, and 2 tierces, 1 puncheon. NOTE 2.- The term hogshead is often applied to any large cask that may contain from 50 to 120 gallons, or more. NOTE 3. —The Standard Unit of Liquid.kleasure adopted by the government of the United States is the Winchester Wine Gallon, which contains 231 cubic inches. It has the name Winchester, from its standard having been formerly kept at Winchester, England. The Imperial Gallon, now adopted in Great Britain, contains 277-27o4 cubic inches; so that 6 Winchester gallons make about 5 Imperial gallons. QuESTIONS. - Art. 95. For what is wine or liquid measure used? Repeat the table. How many gallons make a barrel? How many a tierce? HIow many a puncheon? How is the term hogshead often applied? What is the standard unit of liquid measure?

Page 99 SECT. X.] REDUCTION. 99 MENTAL EXERCISES 1. In 3 pints how many gills? In 5 pints? In 9 pints? 2. In 4 quarts how many pints? In 6 quarts? In 8 quarts? 3. In 5 gallons how many quarts? In 7 gallons? 4. How many gallons in 12 quarts? In 18 quarts? EXERCISES FOR THE SLATE. 1. In 47 tuns of wine how 2. In 379008 gills how many gills? many tuns? OPERATION. OPERATION. 47 tuns. 4)379008 gi. 4 2)94752 pt. 1 8 8 hogsheads. 63 4)47376 qt. 5 61 63)11844 gal. 1 8 4 4 gallons. 4 ) 1 8 8 hhd. Ans. 4 7 tuns. 4 7 3 7 6 quarts. 2 9 4 7 5 2 pints. 4 Ans. 3 7 9 0 0 8 gills. 3. Reduce 197 tuns 3hhd. 60gal. 3qu. ipt. to gills. 4. In 1596604 gills how many tuns? 5. What will 7 hogsheads of wine-cost, at 5 cents a pint? 6. What cost 18 tuns lhhd. 47gal. of oil, at $1.25 per gallon? Ans. $5807.50. BEER IMEASURE. ART. 96o Beer Measure is used in measuring beer, ale. porter, and milk. TABLE. 2 Pints (pt.) make 1 Quart, qt. 4 Quarts c' 1 Gallon, 54 Gallons 6 1 IIogshead, Rhd. qt. pt. gal. 1 2 hhd. 1 4 8 1 - -54 = 216 - 432 QUESTION. -Art. 96. Repeat the table of beer measure.

Page 100 100 REDUCTION. [SECT. X NOTE 1. -The gallon of beer measure contains 282 cubic inches; and as has been usually reckoned, 36 gallons equal 1 barrel; 2 hogsheads, or 108 gallons, 1 butt; 2 butts, or 216 gallons, 1 tun. 1 gallon beer measure -gall. ipt. 387- gi. wine measure. NOTE 2. - Beer Measure is becoming obsolete. Milk and malt liquors, at the present time, are bought and sold, very generally, by wine or liquid measure. EXERCISES FOR TIIE SLATE. 1. How many quarts in 76 2. In 16416 quarts how hogsheads? many hogsheads? OPERATION. OPERATION. 76 hhd. 4) 16416 qt. 54 5 4)_4104 gal. 3 80 4Ans. 7 6 hhd. 380 4 1 0 4 gallons. Ans. 1 6 4 1 6 quarts. 3. In 4 tuns lhhd. 17gal. lpt. how many pints? 4. How many tuns in 7481 pints? 5. What cost 7hhd. 1Sgal. of beer at 4 cents a quart? Ans. 863.36. 6X. At 15 cents per gallon, what will 18hhds. of ale cost? Ans. $145.80. DRY MEASURE. ART. 97, This measure is used in measuring grain, fruit salt, coal, &C. TABLE. 2 Pints (pt.) make 1 Quart, qt. 8 Quarts " 1 Peck, pk. 4 Pecks " 1 Bushel, bu. 8 Bushels Id 1 Quarter, qr. 36 Bushels 6 1 Chaldron, ch. qt. pt. Pk. 1 2 bu. = 8 = 16 ch. 1 = 4 _ 32 64 1 --- 36 = 144 1152 = 2304 QUESTIONS. -H fOW many cubic inches does the beer gallon contain? I-o B, do you reduce hogsheads to quarts? I-ow do you reduce quarts to hogsheads? -- Art. 97. For what is dry measure used? Repeat the table.

Page 101 SrCT. X.] REDUCTION. 101 NoTE 1. - The Standard Unit of Dry.Measure adopted by the United States government is the Wintchester bushel, which is in form a cylinder, 18~ inches in diameter, and 8 inches deep, containing 2150- 4f2 cubic inches. The Standard Iniperial bushel of Great Britain contains 2218- o'92 cubic inches, so that 32 Imperial bushels equal about 33 Winchester bushels. The gallon in Dry Measure contains 2684 cubic inches. NOTE 2. - 1gal. Dry Measure = 268-4cu. in. = Igal. lpt. 1 -gi. Wine Measure 3qt. 1l47pt. Beer Measure; Igal. W. M. = 231cu. in. = 3 qt. jpt. D. 1. = 3qt. 2 pt. B. M.; 1gal. B. M. = 282cu. in. = 1gal. ipt. 3 5~gi. W. M.= Igal. wpt. D. M.; lqt. D. M. 67'cu. in. Iqt. 155gi. W. M.; lqt. W. M. = 657cu. in. = 1-2pt. D, M.W; ipt. D. M. - 338cn. in. - ipt.J- gi. W. M.; ipt. W. M. - 28 cu. in. - pt. D. M. MENTAL ExERCISES. 1. In 2 quarts how many pints? In 5 quarts? In 7 quarts? 2. In 3 pecks how many quarts? In 6 pecks? In 9 pecks? 3. In 5 bushels how many pecks? In 10 bushels? 4. How many pecks in 16 quarts? In 25 quarts? EXERCISES FOR THIE SLATE. 1. How many quarts in 2. In 56731 quarts how 49ch. 8bu. 3pk. and 3qt.? many chaldrons? OPERATION. OPERATION. 4 9 ch. 8bu. 3pk. 3qt. 8)5 6 7 31 qt. 3006 4)7091 pk. 3qt. 302 1 4 7 3 6) 1 77 2 bu. 3pk, 1 7 7 2 bushels. 4 9 ch. 8bu. 4 7 0 9 1 peck&s Ans. 49ch. 8bu. 3pk. 3qt. 8 &ns. 5 6 7 3 1 quarts. 3. Reduce 97ch. 30bu. 2pk. to quarts. 4. In 112720 quarts how many chaldrons? 5. How many pints in 35bu. lpt.? 6. Reduce 2241 pints to bushels. 7. Reduce 18qr. 3pk. 5qt. to quarts. 8. How many quarters in 4637 quarts? 9. In 19bhu. 3pk, 7qt. Ipt. how many pints? 10. In 1279 pints how many bushels? QUESTION. - What is the standard unit of Dry Measure?

Page 102 102 REDUCTION. [SECT. A MEASURE OF TIME. ART. 98, This measure is applied to the various divisions and subdivisions into which time is divided. TABLE. 60 Seconds (sec.) make 1 IMinute, m. 60 Minutes c6 1 Hour, h 24 Hours " 1 Day, da. 7 Days " 1 Week, w 365~1 Days, or 52 weeks Julian. 4 day,1 Julian Year, y. 12 Calendar Months (mo.) " 1 Year, y. m. sec h. 1 = 60 d. 1 60 3600 wV. 1 24 1440 = 86400 Y/. 1 7 - 168 = 10080 604800 1 = 52_52 = 3654 = 8766 - 525960 -- 315'57600 NOTE 1. - The true Solar or Tropical Year is the time measured from the sun's leaving either equinox or solstice to its return to the same again, and is 365d. 5h. 48m. 49sec. nearly. The Julian Year, so called from the calendar instituted by Julius Caesar, contains 3651 days, as a medium; three years in succession containing 365 days, and the fourth year 866 days; which, as compared with the true solar year, produces an average yearly error of llm. 10-3a sec., or a difference that would amount to 1 whole day in about 120 years. The Gregorian Year, or that instituted by Pope Gregory XIII., in the year 1582, and which is now the Civil or Legal Year in use among the different nations of the earth, contains, like the Julian year, 365 days for three years in succession, and 366 days for the fourth, exceptib g centennial years whose number cannuot be exoctctl divided b/ 400. The Gregorian year is so nearly correct as to err only 1 day in 3866 years, a difference so little as hardly to be worth taking into account. A Commooz Yeccr is one of 365 days, and a Lecqp or Bissextile Year is one of 366 days. Any year is Leap Year whose number can be divided by 4 without a remainder, except years whose number can be divided without a remainder by 100, but not by 400. A Sidereal Year is the time in which the earth revolves round the sun, and is 365d. Gh. 9m. 9 6Gsec. NOTE 2. - The names of the 12 calendar months, composing the civil year, are January, February, March; April, May, June, July, August, September, October, November, December, and the number of days in each may be readily remembered by the fbllowing lines:'" Thirty days hath Septenber, April, June, and November; QUESTIONS. - Art. 98. To what is the measure of time applied? Repeat the table. How is-the true solar year measured? How long is it? Why is the Julian year so called? Who instituted the Gregorianl year? What is a Commonn ye'? a Whnt is; hidel re al 7

Page 103 IECT. X.] RPJ DUCTION. ] () And all the rest have thirty-one, Save February, which alone Bath twenty-eight.; and this, in fine, One year in four hath twenty-nine." TABLE. SHOWING THE NUMBER OF DAYS FROMI ANY DAY OF ONE MONTH TO THE SAME DAY OF ANY OTHER MONTHI IN THE SAME YEAR. FROMn ANY TO THE SAME DAY OF DAY OF Jon. Feb. Mar. Apr. May. June. July.i Aug, Sept. Oct. Nov. Dec. January 365 31 59 90 120 151 181 212 243 273 304 334 February 334 365 28 59 89 120 150 181 212 242 273 303 March 306 337 365 31 61 92 122 153 184 214 245 275 April 275 306 334 365 30 61 91 122 153 183 214 244 May 245 267 304 335 365 31 61 92 123 153 184 214 June 214 245 273 304 334 365 30 61 92 122 153 183 July 184 215 243 274 304 335 365 31 62 92 123 153 August 153 184 212 243 273 304 334 365 31 61 92 122 September 122 153 181 212 242 273 303 334 365 30 61 91 October 92 123 151 182 212 243 273 304 335 365 31 61 November 61 92 120 151 181 212 242 273 304 334 365 30 December 31 62 90 121 151 182 212 243 274 304 335 365 For example, suppose we wish to find the number of days from April 4th to November 4th, we look for April in the left-hand vertical column, and November at the top, and, where the lines intersect, is 214, the number sought. Again, if we wish the number of days from June 10th to September 16th, we find the difference between June 10th and September 10th to be 92 days, and add 6 days for the excess of the 16th over the 10th of September, so we have 98 days as the exact difference. If the end of February be included between the points of time, a day must be added in leap year. When the time includes more than one year, there must be added 365 days for each year. MENTAL EXERCISES 1. In 3 minutes how many seconds? In 5 minutes? 2. In 2 hours how many minutes? In 4 hours? 3. In 4 weeks how many days? In 6 weeks? In 9 weeks? 4. In 2 days how many hours? In 3 days? In 7 days? 5. How many weeks in 21 days? In 30 days? In 50 days? 6. How many calendar months in 2 years? In 8 years? In 10 years? In 12 years? In 20 years? QUESTIONS. - Name the months in their order. How many days has each month? How do you find by the table the number of days from April 4th to November 4th? When the time sought for is more than one year, how many days must be added?

Page 104 104 REDUCTION. [SECT. X. EXERCISES FOR THE SLATE. 1. How many seconds in 365da. 2. In 31556929 seconds 5h. 48m. 49sec., or one solar year? how many days? OPERATION. OPERATION. 3 6 5 da. 5h. 48m. 49sec. 60)3 15 5 6 9 29 2 4 60)5 2 5 9 4 8 m. 49seo 1730 24)8 76 5 h. 48m. 7 3 0 8 7 6 5 hours. 365 da. 5h. 6 0 Ans. 365da. 5h. 48m 49sec 5 2 5 9 4 8 minutes. 60 Ans. 3 1 5 5 69 2 9 seconds. 3. Reduce 296da. 18h. 32m. to minutes. 4. In 427352 minutes how many days? 5. How many seconds in 30 solar years 262da. 17h. 28m 42sec.? 6. In 969407592 seconds how many solar years? 7. How many weeks in 684592 minutes? 8. In 67w. 6d. 9h. 52m. how many minutes? 9. How many days from June 5th to Dec. 11th 10. How many days from March 17th, 1856, to May 16th 1857? Ans. 425 days. 11. How many days from December 18th, 1856, to January 30th, 1857? 12. How many days from August 30th, 1857, to June lst, 1858? 13. How many days from July 4th, 1859, to July 4th, 1860? 14. How many days from April 25th, 1855, to August 20th, 1858? Ans, 1213 days. NoTE. — The last six examples are to be performed by aid of the table on page 103. QUESTIONS. - How do you reduce years to seconds? Give the reason for the operation. Ilow do you reduce seconds to days? To years? Give the reason for the operation.

Page 105 EOCT. X.] REDUCTION. 105 CIRCULAR MEASURE. ART. C9, Circular Measure is applied to the measurement of circles and angles, and is used in reckoning latitude and longitude, and the revolutions of the planets round the sun. TABLE. 60 Seconds (/) make 1 Minute, 60 Minutes " 1 Degree, o. 30 Degrees " 1 Sign, S. 12 Signs, or 360 Degrees, " The Circle of the Zodiac, C. o 1 60. 1 60 = 3600 c. 1 = 30 = 1800 108000 1 12 = 360 = 21600 1296000 NOTE. - 1. A Circle is a plane figure A bounded by a curve line, every part of,360 which is equally distant from a point "5J~. ~6 o called its centre. li n ege, The Circumnference of a circle is the line which bounds it, as shown by the diagram. An./rc of a circle is any part of its circumference; as AB. ca\/ / \ // A Radius of a circle is a straight line drawn from its centre to its circumference; as CA, CB, or CD. 0T1 Every circle, large or small, is supposed to be divided into 360 equal parts, called degrees. A Quadrant is one fourth of a circle, or an are of 900; as AB. An tAngle, as ACB, is the inclination or opening of two lines which meet at a point, as C. The point is the vertex of the angle. If a circle be drawn around the vertex of an angle as a centre, the two sides of the angle, as radii of the circle, will include an are, which is the neasure of the angle; as the are AD = 120~ is the measure of the angle ACD, and AB - 90~, the measure of the angle ACB; hence the one is called an angle of 120~, and the other an angle of 90~. NOTE. — 2. As the earth turns on its axis from west to east every 24 hours, the sun appears to pass from east to west z-i: of 3600 of longitude every hour, or over 150 of longitude in 1 hour's time, or 10 in 4 minutes of time, and 1' in 4 seconds of time; so that, for instance, at any place, when it is noon, it is 1 hour earlier for every 15~ of longitude westward, or 1 hour later for every 150 of longitude eastward. Thus, Boston being 71~ 4' west of Greenwich, and San Francisco 510 17' west of Boston, when it is noon at Boston, it is 4h. 44m. 16sec. past noon at Greenwich, and wanting 3h. 25m. 8sec. of noon at San Francisco. QUESTIONS. - Art. 99. To what is circular measure applied? Recite the table. What is a circle? What is an angle?

Page 106 106 REDUCTION. [SECT. X EXERCISES FOR TUE SLATE. 1. How many minutes in 2. In 20937 minutes how 11S. 18" 57'? many signs? OPERATION. OPERATION. 118. 180 57' 60)20937' 3 0 3 0 ) 3 4 3~ 57' 3 4 8 degrees. 1 1. 180. 60 Ans. 2 0 9 3 7 minutes. Ans. 11S. 180 57'. 3. In 27S. 190 51' 28" how many seconds? 4. How many signs in 2987488 seconds? MISCELLANEOUS TABLE. ART. I00. This table embraces a variety of things in business important to be known. 12 units make 1 dozen. 12 dozen " 1 gross. 12 gross " 1 great gross. 20 units " 1 score. 14 pounds of Iron or Lead 4 1 stone. 60 pounds of Wheat " 1 bushel. 60 pounds of Clover-seed " 1 bushel. 60 pounds of Beans " 1 bushel. 60 pounds of Potatoes " 1 bushel. 52 pounds of Onions 6 1 bushel. 70 pounds of Corn on the Cob " 1 bushel. 56 pounds of Shelled Corn " 1 bushel. 56 pounds of Rye " 1 bushel. 56 pounds of Flax-seed " 1 bushel. 45 pounds of Tihothy-seed 6' 1 bushel. 20 pounds of Bran " 1 bushel. 48 pounds of Barley " 1 bushel. 52 pounds of Buckwheat " 1 bushel in Ky. 48 pounds of Buckwheat " 1 bushel in Mass. and Pa. 32 pounds of Oats " 1 bushel in iMs., il., O., &~. 30 pounds of Oats 6 1 bushel in Me., N.H.,Pa., [&c. QUesTIONS. - How do you reduce signs to seconds? Give the reason of the operation. How do you reduce seconds to degrees? To signs? Give the reason for the operation. IHow many degrees in a circle? Art. 100. What is embraced in the miscellaneous table?

Page 107 SECT. X.] REDUCTION. 107 196 pounds of Flour make 1 barrel. 200 pounds of Beef " 1 barrel. 200 pounds of Pork i" 1 barrel. 100 pounds of Fish " 1 iquintal. 200 pounds of Shad or Salmon " 1 barrel in N. Y., Ct. 220 pounds of Fish " 1 barrel in Md. 30 gallons of Fish 6' 1 barrel in Mass. 5 bushels of Corn "6 1 barrel in Md., Tenn., &c. OF BooKs. A sheet folded in 2 leaves forms a folio. A sheet,6 4 leaves' a quarto. A sheet " 8 leaves " an octavo. A sheet'; 12 leaves " a 12mo. A sheet " 18 leaves " an 18mo. A sheet "' 24 leaves " a 24mo. MISCELLANEOUS EXERCISES IN REDUCTION. 1. In $345.18 how many mills? 2. How many dollars in 345180 mills? 3. In 46~ 18s. 5d. how many farthings? 4. How many pounds in 45044 farthings? 5. Reduce 6lib. Ooz. 17pwt. 17gr. troy to grains. 6. In 351785 grains troy how many pounds? 7. How many scruples in 271t 38 153 1? 8. In 7852 scruples how many pounds? 9. In 83T. lcewt. 3qr. 181b. how many ounces? 10. How many tons in 2675088 ounces? 11. How many nails in 97yd. 3qr. 3na.? 12. In 1567 nails how many yards? 13. In 57 ells English how many yards? 14. How many ells English in 71yd. lqr.? 15. How many inches in 15m. 7fur. 18rd. 10ft. 6in.? 16. In 1009530 inches how many miles? 17. In 95,000,000 of miles how many inches? 18. How many miles in 6,019,200,000,000 inches? 19. In 48deg. 18m. 7fur. 18rd. how many feet? 20. In 17629557 feet how many degrees? 21. How many square feet in 7A. 311. 16p. 218flt.'t 22. In 342164 square feet how many acres? 23. How many square inches in 25 square miles? QUESTIor. — What gives name to the size or form of books?

Page 108 [08 REDUCTION. [SECT. X. 24. In 100362240000 square inches how many square miles? 25. How many cubic inches in 15 tons of timber? 26. In 1036800 cubic inches how many tons? 27. How many gills of wine in 5hhd. 17gal. 3qt.? 28. In 10648 gills how many hogsheads of wine? 29. How many quarts of beer in 29hhd. 30gal. 3qt.? 30. In 6387 quarts of beer how many hogsheads? 31. Iow many pints in 15ch. 16bu. 3pk. of wheat? 32. In 35632 pints of wheat how many chaldrons? 33. How many seconds of time in 365 days 6 hours? 34. In 31557600 seconds how many days? 35. How many hours in 1842 years (of 365da. 6h. each)? 36. In 16146972 hours how many years? 37. How many seconds in SS. 14~ 18' 17"? 38. In 915497" how many signs? 39. What will be the cost of 13 gross of steel pens, at 21 cents per pen? Ans. $46.80. 40. Bouoht 12 reams of paper at 20 cents per quire; how much did it cost? Ans. $48. 41. I wish to put 2 hogsheads of wine into bottles thatwill contain 3 quarts each; how many bottles are required? Ans. 168 bottles. 42. When $1480 are paid for 25 acres of land, what costs 1 acre? What costs 1 rood? What cost 37A. 2R 1SSp.? Ans. $2226.66. 43. John Webster bought 5cwt. 3qr. 81ib. of sugar at 9 cents per lb., for which he paid 25 barrels of apples at $1.75 per barrel; how much remains due? Ans. $9.62. 44. Bought a silver tankard weighing 21b. 7oz. for $43.50 what did it cost per oz.? How much per lb.? Ans. $18. 45. Bought 3T. lcwt. 181b. of leather at 12 cents per lb., and sold it at 9 cents per lb.; what didi I lose? Ans. $ 183.54. 46. Phineas Bailey has agreed *to grade a certain railroad at $5.75 per rod; what -will he receive for grading a road between two cities, whose distance from each other is 37mi. 7fur. 29rd.? Ans. $69856.75. 4'7. If it cost $17.29 per rod to grade a certain piece of railroad, what will be the expense of grading 15m. 6fur. 37rd.? Ans. $87,781.33. 48. What is the value of a house-lot, containing 40 square rods and.200 square feet, at $1.50 per square foot? Ans. $1 )6(5.

Page 109 SECT. X.] REDUCTION. 109 49. How many yards of carpeting, that is one yard in width, will be required to carpet a room 18ft. long and 15ft. wide? Ans. 30 yards. 50. A certain machine will cut 120 shingle-nails in a minute how many will it cut in 47 days 7 hours, admitting the machine to be in operation 10 hours per day? Ans. 3434400 nails. 51. In a field 80 rods long and 50 rods wide, how many square rods? How many acres? Ans. 25 acres. 52. How long will it take to count 18 millions, counting at the rate of 90 a minute? Ans. 138da. 21h. 20m. 53. A merchant purchased 9 bales of cloth, each containing 15 pieces, each piece 23 yards, at 8 cents per yard; what was the amount paid? Ans. $248.40. 54. Suppose a certain township is 6 miles long and 4~ miles wide, how many lots of land of 90 acres each does it contain? Ans. 192 lots. 55. The pendulum of a certain cloclk vibrates 47 times in 1 minute; how many times will it vibrate in 196 days 49m.? Ans. 13267583 times. 56. I-Iow many shingles will it take to cover a roof, each of whose equal sides is 36 feet long, with rafters 16 feet in length, supposing 1 shingle to cover 27 square indhes? Ans. 6144 shingles. 57. How many times will the large wheels of an engine turn round in going from Boston to Portland, a distance of 110 miles, supposing the wheels to be 12 feet and 6 inches in circumference? Ans. 46464 times. 58. In a certain house there are 25 rooms, in each room 7 bureaus, in each bureau 5 drawers, in each drawer 12 boxes, in each box 15 purses, in each purse 178 sovereigns, each sovereign valued at $4.84; what is the amount of the money? Ans. $135689400. 59. In 18rd. 5yd. 2ft. ilin. how many inches? Ans. 3779 inches. 60. In 3779 inches how many rods? Ans. 18rd. 5yd. 2ft. 1lin. 61. Sold ST. 17cWt. 3qr. 181b. of potash for 3 cents per pound; what was the amount? Ans. $353.79. 62. A gentleman purchased a house-lot that was 25 rods long and 16 rods wide for $100,000, and sold the same for $1.25 per square foot; what did he gain by his purchase? Ans. $36,125. 10

Page 110 110 ADDITION OF COMPOUND NUMBERS. [SECT. XI XI. ADDITION OF COMPOUND NUMBERS. ART. 101M ADDITION of Compound Numbers is the process of finding the amount of two or more compound numbers~ ENGLISH MONEY. Ex. 1. Paid a London tailor 7~. 13s. 6d. 2far. for a coat; 2t. 17s. 9d. lfhr. for a vest; 38. Ss. 3d. 3far. for pantaloons; 9~. Ils. 8d. 3f8ar. for a surtout; what was the amount of the bill? Ans. 23~. 11s. 4d. Ifar; OPEIRATION. Having written units of the same de~-. s. d. far. nomination in the same column, we find 7 13 6 2 the sum of farthings in the right-hand 2 1 7 9 1 column to be 9 farthings, equal to 2d. 3 8 3 3 and lfar. We write the ifar. Under the 9 11 8 3 column of farthings, and carry the 2d. to the column of pence; the sum of which is Ans. 2 3 1 1 4 1 28d., equal to 2s. 4d. We write the 4d. under the column of pence, and carry the 2s. to the column of shillings; the sum of which is 51s., equal to 2~. 11s. HIlaving written the 11s. under the column of shillings, we carry the 2~. to the column of pounds, and find the whole amount to be 23~. 11s. 4d. ifar. The same result can be arrived at by reducing the numbers as they are added in their respective columns. Thus, in working the example, we can, beginning with farthings, add in this way: 3far. and 3far. are 6far., equal to Id. 2far., and 1far. are Id. 3far., and 2far. are ld. 5far., equal 2d. Ifar. Writing the Ifar. under the column of farthings, carry the 2d. to the column of pence; add 2d. (carried) and 8d. are 10d., and 3d. are 13d., equal to Is. Id., and 9d. are is. 10d., and 6d. are is. 16d., equal to 2s. 4d. Writing the 4d. under the column of pence, carry the 2s. to the column of shillings; add 2s. (carried) and 11s. are 13s., and 8s. are 21s., equal to 1~. is., and 17s. are 1~. 18s., and 13s. are 1~. 31s., equal to 2X. 1ls. Writing 11s. under the column of shillings, carry the 2E. to the column of pounds, and so find the whole amount to be, as before, 23~. ls. 4d. Ifar. Thus the adding of compound numbers is like that of simple numbers, except in carrying; which difference holds also in subtracting, multiplying, and dividing of compound numbers. QUESTIONS. - Art. 101. What is addition of compound numbers? How do you arrange compound numbers for addition? Why? What is the difference between addition of compound and addition of simple numbers?

Page 111 SErCT. XI.] ADDITION OF COMPOUND NUMBERS. 111 RUTE. - Write all the given numbers so that units of the same denonlo'taliotln may stand in the samze cohlrn. Add as in a(ldition of simple numbers; and carry, from column to column, onef Jr as many units as i, takes of the denonmination added to make a unit of the denomination next higher. PRooF. -The proof is tne same as in addition of simple numbers. EXAMPLES FOR PRACTICE. TROY WEIGHT. 2. 3. lb. oz. pwt. gr. lb. oz. pwt. gr. 15 11 19 22 10 10 10 10 71 10 13 17 81 11 19 23 65 9 17 14 47 7 8 19 73 11 13 13 16 9 10 14 14 8 9 9 33 10 9 21 242 4 14 3 APOTHECARIES' WEIGHT, 4. 5. lb 5 5 D gr. l, 5 D g2'. 81 11 6 1 19 35 9 6 2 19 75 10 7 2 13 71 1 1 1 11 14 9 7 1 12 37 3 3 2 12 37 8 1 1 11 14 4 7 1 13 61 11 3 2 3 75 5 6 1 17 2 72 4 3 0 18 AVOIRDUPOIS WEIGHT. 6. 7. T. cwt. qr. lb. oz. dr. T. cwt. qr. lb. oz. dr. 71 19 3 17 14 13 14 13 2 15 15 15 14 13 1 11 13 12 13 17 3 13 11 13 39 9 3 13 9 9 46 16 3 11 13 1'0 15 17 3 16 10 14 14 15 2 7 6 9 61 16 3 13 7 8 11 17 3 10 15 11 203 17 3 23 8 8 QUESTIONS. - What is the rule? The proof?

Page 112 11] ADDITION OF COMPOUND NUMBERS. [SECT. X1 CLOTH MEASURE. 8. 9. yd. qr. na. in. E. E. qr. na. in 5 3 3 2 16 3 2 1 7 1 1 2 71 1 1 2 8 3 3 1 13 3 2 1 9 1 2 2 47 3 2 2 4 3 3 2 39 2 3 2 36 3 0 0 LONG MEASURE. 10.' 1. deg. m. fur. rd. ft. in. m. fur. rd. yd. ft. in. i8 19 7 15 11 1 12 7 35 5 2 11 61 47 6 39 10 11 13 6 15 3 1 10 78 32 5 14 9 9 16 1 17 1 2 5 17 59 7 366 16 10 13 4 13 2 1 9 28 56 1 30 16 1 17 7 36 5 2 7 205 8- 5 2 17 141 8 ==4 26 205 9. 1 17 15 2 SURVEYORS' MEASURE. 12. 13. m. fur. ch. p. 1. m. fur. ch. p. 1. 17 5 8 3 24 14 7 9 3 21 16 3 7 1 21 37 1 0 3 16 47 7 9 3 19 17 7 8 3 17 19 6 6 1 16 61 6 5 3 16 31 7 1 0 20 47 1 1 0 23 133 7 4 0 0 SQUARE MEASURE. 14. 15. A. I. p. it. in. A. R. p. yd. ft. in. 67 3 39 272 143 43 1 15 30 8 17 78 3 14 260 116 16 3 39 19 7 141 14 2 31 167 135 47 1 1 6 27 5 79 67 1 17 176 131 38 3 17 18 8 17 49 3 31. 69 117 15 1 32 11 1 117 278 3 15 1311 66 = —36 278 3 15 131 102

Page 113 SECT. XI.1 ADDITION OF COMPOUND NUMIBERS. 113 SOLID MEASURE. 16. 17. Tun. ft. in. Cord. ft. in. 17 39 1371 14 116 1169 61 17 1711 67 113 1711 47 16 1666 96 127 969 71 38 1711 19 98 1 376 47 17 167 1 37 1 414 246 11 1164 WINE MEASURE. 18. 19. Tun. hhd. gal. qt. pt. Tun. hhd. gal. qt. pt. 61 1 62 3 1 14 3 18 3 0 71 3 14 1 1 81 1 60 3 1 60 0 17 3 0 17 3 61 3 0 14 1 51 1 1 6-1 3 57 3 1 57 3 14 3 1 17 1 17 1 0 265 2 35 1 0 BEER MEASURE. 20. 21. Tun. hhd, gal qt. pt. Tun. hhcl. gal. qt. pt. 15 3 50 3 1 67 1 51 1 0 67 3 17 3 1 15 3 16 3 1 17 1 44 1 0 44 1 45 1 1 71 3 12 3 1 15 2 12 2 1 81 1 18 1 0 67 3 35 1 0 254 1 36 0 1 DRY MEASURE. 22. 23. ch. bu. pk. qt. pt. ch. bu. pk. qt. pt. 15 35 3 7 1 71 17 1 1 1 61 1 6 3 1 16 31 3 3 0 51 30 1 5 0 41 14 3 1 1 42 17 2 2 1 71 17 1 0 1 14 14 1 4 1 10 10 2 3 0 186 7 1 2 0 10-*

Page 114 114 SUBTRACTION OF COMPOUND NUMIBERS. [SECT. XI1. TIME 24. 25. y. da. h. m. S. w. da. h. m. s. 57 300 23 59'17 15 6 23 1 5 17 47 169 15 17 38 61 5 15 27 18 29 364 23 42 17 71 6 21 57 58 18 178 16 38 47 18 5 19 39 49 49 317 20 52 57 87 6 19 18 57 03 236 10 30 56 CIRCULAR MEASURE, 26. 27. S. o S i0 S 1 28 56 58 6 17 17 18 10 21 51 37 7 09 19 51 8 13 39 57 8 18 57 45 8 19 38 49 4 17 16 39 7 17 47 48 7 27 38 48 11 11 55 09 NOTE. —-The sum, of the signs, in circular motion, must always be divided by 12, and the remainder only be written down, as in Ex. 26. XII. SUBTRACTION OF COM3POUND NUMBERS. ART. 102. SUBTRACTION of Compound Numbers is the process of finding the difference between two compound numnbers. ENlGLISH MONEY. Ex, 1. From 87~. 9s. 6d. 3far., take 52~. Ils. 7d. Ifaro Ans. 34k. 17s. ld. 2far. OPERATION. HHaving placed the less nuimber under In ~. s. d.7 far. the greater, farthings under farrthings, 3in, 8 7 9 6 3 pence under pence, &c., we begin with Sub. 5 2 1 1 7 1 the farthings, thus: 1 far. from 3 far. leaves 2 far., which we set under the Renm 3 4 1 7 1 1 2 column of farthings. As we cannot QUESTIONS. Art. 102. What is subtraction of compound numbers? How do you arrange the numbers for subtraction?

Page 115 SECT. XII.] SUBTRACTION OF COMPOUND NUMBERS. 11I take 7d. from 6d., we add 12d. = Is. to the 6d., making 18d., and then subtract the 7d. from it, and set the remainder, lid., under the column of pence. We then add is. = 12d. to the Ils. in the subtrahend, making 12s., to compensate for the 12d. we added to the 6d. in the minuend. (Art. 30.) Again, since we cannot take 12s. fiom 9s., we add 20s. = 1~. to the 9s., making 29s., from which we take the 12s., and set the remainder, 17s., under the column of shillings. Having added 1~. = 20s. to the 52~., to compensate for the 20g. added to the 9s. in the minuend, we subtract the pounds as in subtraction of simple numbers, and obtain 34~. for the remainder, and as the result complete, 34~. 17s. 11d. 2far. RULE.- Write the less compound number under the greater, so that units of the same denomination shall stand in the same column. Subtract as in subtraction of simple numbers. If any number in the subtrahend is larger than that above it, add to the upper number as many units as make one of the next higher denomination before subtracting, and carry one to the next lower number before subtracting it. PROOF. - The proof is the same as in simple subtraction. EXAMPLES FOR PRACTICE. 2. 3. ~. s. d. far. ~. ~. d. far 78 11 5 2 765 16 10 1 41 13 3 3 713 17 11 3 36 18 1 3 TROY WEIGHT. 4. 5. lb oz. pwt. gr. lb. oz. pwt. gr. 15 3 12 14 711 1 3 17 9 11 17 21 19 3 18 19 5 3 14 17 APOTHECARIES' WEIGHT. 6. 7. lm g 5 gr. lb g gr. 15 7 1 2 15 161 6 3 1 17 11 9 7 1 19 97 7 1 2 18 3 9 2 0 16 QUESTIONS. - What do you do when the upper number is smaller than the lower? How many do you carry to the next denomination? What is the rule for subtraction? The proof?

Page 116 116 SUBTRACTION OF COMPOUND NUMBERS. [SECT. XII AVOIRDUPOIS WEIGHT. 8. 9. T. cwt. qr. lb. oz. dr. T. cwt. qr. lb. oz. dr 117 1 6 1 5 0 14 11 1 0 1 1 1 3 19 17 3 17 1 15 9 18 3 1 13 15 97 18 1 1214 15 CLOTH MEASURE. 10. 11, yd. qr. na. in.. E. qr. na. in. 15 1 1 2 171 2 2 1 9 3 3 1 19 3 0 2 5 1 2 1 LONG MEASURE. 12. 13. deo. m. fur. rd. yd. ft. in. deg. m. fur. rd. ft. in. 97 3 7 31 1 1 3 18 19 1 1 3 7 19 17 1 39 1 2 7 9 28 7 1 16 9 77 551 5 31 4-, 1 8 1=1 13 1 2 6 77 55 7 5 1 1 2 SURVEYORS' MIEASURE. 14. 15. m. fur. cha. p. 1. m. fur. cha. p. 1. 21 3'5 2 17 31 7 1 1 19 9 5 8 1 20 18 1 7 3 23 11 5 7 0 22 SQUARE MEASURE. 16. 17. A. R. p. ft. in. A. R. p. yd. ft. in. 116 1 13 100 113 139 1 17 18 1 30 87 3 17 200 117 97 3 18 30 1 31 28 1 35 1714 140 2 3 6 28 1 35 172 32

Page 117 ISECT. XII.] SUBTRACTION OF COMPOIJND NUMBERS. 117 SOLID MEASURE. 18. 19. T. ft. in. Cords. ft. in. 171 30 1000 571 18 1234 98 37 1234 199 19 1279 72 32 1494 WINE MEASURE. 20. 21. T. hhd. gal. qt. pt. gi. T. hhd. gal. qt. pt. gi. 171 3 8 1 1 71 1 1 1 1 1 99 1 19 3 1 3 9 3 3 3 1 3 72 1 51 1 1 2 BEER MEASURE. 22. 23. 1. hhd. gal. qt. pt. T. hhd. gal. qt. pt. 15 1 17 1 0 79 2 2 2 0 9 3 9 3 1 19 3 13 3 1 5 I 5 1 I. DRY MEASURE. 24. 25, ch. bu. pl. qt. pt. ch. bu. pk. qt. pt, 716 1 2 1 0 73 13 3 1 19 9 3 1 1 19 18 1 3 1 696 27 2 7 1 TIME. 26. 27. y. da. h. m. seQ. w. da. h. m. sec. 375 15 13 17 5 14 1 3 4 15 199 137 15 1 39 9 6 17 37 48 175 243 4 15 26 CIRCULAR MEASURE. 28. 29. S. 0 s. ~' 11 7 13 15 1 23 37 39 9 29 17 36 9 15 38 47 1 7 55 39 4 7 58 52 NoTE. - In Circular Measure, the minuend is sometimes less than the subtrahend, as in Ex. 29, in which case it must be increased by 12 signs.

Page 118 118 SUBTRACTION OF COMPOUND NUMBERS. [SECT. XII. ART. I03, To find the time between two different dates. Ex. 1. What is the difference of time between October 16th 1 852, and August 9th, 1854? Ans. ly. 9no. 23da. FIRST OPERATION. Commencing with January, the first Y. moe0. da. month in the year, and counting the M.in. 1 8 5 4 7 9 months and days in the later date up to Sub. 1 8 5 2 9 1 6 Alugust 9th, we find that 71u10. and 9 - dC. hiave elapsedc; and counting the Rem. 1 9 2 3 months and days in the earlier date, up to October 16th, we find that 9mo. and SECOND OPERATION. 16da. have elapsed. WVe, therefore, Min. 1 8 5 4 8 9 write the numbers for subtraction as in Sub. 1 8 5 2 1 0 1 6 the first operation. The same result, however, could be obtained, as some Rem. 1 9 2 3 prefer, by reckoning the number of the given months instead of the number oJ months that have elapsed since the beginning of the year, and writ. ing the numbers as in the second operation; - written either way, Tlhe earlier date being placed under the later, is subtracted, as by the preceding rule. OTI;.- Iin finding the difference'between two dates, and ill computing illtelest for less than a mnolth, 30 days are considered a llonlth. nll leyc(l transactions, a monthl is reckoned fiom any dacy in one month to the corresponding day of the following month, if it has a corresponding day, otherwise to its end. EXAMPLES FOn PRACTICE. 2. What is the time from March 21st, 1853, to Jan. 6th, 1857? Ans. 3y. 9m. 15da. 3. A note was given Nov. 15th, 1852, and paid April 25th, 1857; how lonog was it on interest? Ans. 4y. 5mno. 1Oda. 4. John Quincy Adanrr was born at Braintree, 31ass., July 11th, 1767, and died at Washington, D. -C., Feb. 23, 1848; to what age did he live? Ans. 80y. 7sno. 12da. 5. Andrew Jackson was born at Waxaw, S. C., Mlarch 15th, 1767, and died at Nashville, Tenn., June 8th, 1845; at what age did he die? Ans. 78y. 2mo. 23da. QUESTIONS. - Art, 103. From what period do you count the months and days in preparing dates for subtraction? HIow do you arrange the dates for subtraction? I-low subtract? How many days are considered a month in business transactions? What is the second method of preparing dates for subtraction?

Page 119 SECT. XIII.] MISCELLANEOUS EXERCISES. 119 X9 III. MIISCELLANEOUS EXERCISES IN ADDITION AND SUBTRACTION OF COMPOUND NUIIBEtRS. 1. 7WHAT is the amount of the following quantities of gold 41b. 8oz. 13pwt. Sgr., 51b. 11oz. 19pwt. 23gr., 81b. Ooz. 17pwt. 15gr., and 1Slb. 9oz. 14pwt. 10gr.? Ans. 371b. 7oz. 5pwt. 8gr. 2. An apothecary would mix 7hBI 33 23 23 lgr. of' rhubarb, 21R 105 03 1 3, 13go of cantharides, and 21P 35 73 2E 17gr. of opium; what is the weight of the compound? Ans. 12Th 5. 33 03 llgr. 3. Add together 17T. llcwt. 3qr. Il1b. 12oz., 11T. 17cwt. lqr. 191b. 11oz., 53T. 19cwt. lqr. 171b. 8oz., 27T. 19cwt. 3qr. 18lb. 9oz., and 16T. 3cwt. 3qr. 0lb. 13oz. Ans. 127T. 12cwt. lqr. 181b. 5oz. 4. A merchant owes a debt in London amounting to 7671~.; what remains due after he has paid 1728~. 17s. 9d.? Ans. 5942~. 2s. 3d. 5. From 731b. of silver there were made 261b. 11oz. 13pwt. l4gr. of plate; what quantity remained? Ans. 461b. Ooz. 6pwt. IOgr. 6. From 71t 8S 135 1 14gr. take 7Th 9~ 13 13 17 gr. Ans. 63Th 103 73 2E 17gr. 7. From 25T. 13cwt. take 10T. 17cwt. 191b. 14oz. Ans. 17T. 15cwt. 3qr. 51b. 2oz. 8. A merchant has 3 pieces of cloth; the first contains 37yd. 3qr. 3na., the second 18yd. lqr. 3na., and the third 31yd. lqr. 2na.; what is the whole quantity? Ans. 87yd. 3qr. Ona. 9. Sold 3 loads of hay; the first weighed 2T. 13cwt. lqr. 71b., the second 3T. 171b., and the third IT. 3qr. 1llb.; what did they all weigh? Ans. 6T. 14cwt. lqr. 201b. 10. What is the sum of the following distances: 16m. 7fur. 18rd. 14ft. 11in., 19mI. 1ifur. 13rd. 16ft. 9in., 97m. 3fur. 27rd. 13ft. 3in., and 47m. 5fur. 37rd. 13ft. lOin.? Ans. 181m. 2f1ur. 18rd. 9ft. 3in. 11. From 76yd. take 8lyd. Sqr. 2na. Ans. 57yd. Oqr. 2na. 12. From 20m. take 3m. 4fur. 18rd. 3ft. 8in. Ans. 16m. 3fur. 21rd. 2ft. 10in. 13. From 144A. 3R. take 18A. 1R. 17p. 200ft. 100in. Ans. 126A. IR. 22p. 71ft. 80in.

Page 120 120 MISCELLANEOUS EXERCISES. [SECT. XIU 14. From 18 cords take 3 cords 100ft. 1000in. Ans. 14 cords 27ft. 728in, 15. A gentleman has three farms; the first contains 169A 3R. 15p. 227ft., the second 187A. 1R. 15p. 165ft., and the third 217A. 2R. 28p. 165ft.; what is the whole quantity? Ans. 574A. 31. 20p. 12:ft. 16. There are 3 piles of wood; the first contains 18 cords 116ft. 1000in., the second 17 cords 111ft. 1600in., and the third 21 cords 109ft. 1716in.; how much in all? Ans. 58 cords 82ft, 860in. 17. From 17T. take 5T. 18ft. 765 ino Ans. 11T. 21ft. 963in. 18. From 169gal. take 76gal. 3qt. ipt. Ans. 92gal. Oqt. ipt. 19. From 17ch. 18bu. take 5ch. 20bu. ipk. 7qt. Ans. 11ch. 33bu. 2pk. lqt. 20. From 83y. take 47y. 10mo. 27d. 18h. 50m. 14s. Ans. 35y. linmo. 2d. 5h. 9m. 46s. 21. From 118. 150 36' 15" take 5S. 18~ 50' 18", Ans. 5S. 260 45' 57". 22. John Thomson has 4 casks of molasses; the first con tains 167gal. 3qt. ipt., the second 186gal. lqt. ipt., the third l08gal. 2qt. ipt., and the fourth 123ga1. 3qt. Opt.; how much is tne whole quantity? Ans. 586gal. 2qt. ipto 23. Add together 17bu. ipk. 7qt. ipt., 18bu. 3pk. 2qt. 19bu. ipk. 3qt. ipt., and 51bu. 3pk. Oqt. ipt. Ans. 107bu. ipk. 5qt. Ipt. 24. James is 13y. 4mo. 13d. old, Samuel is 12y. 11mo. 23d., and Daniel is 18y. 9mo. 29d.; what is the sum of their united ages? Ans. 45y. 2mo. 5d. 25. Add together ISy. 345d. 13h. 37m. 15s., 87y. 169d. 12h. 1Gm. 28s., 316y. 144d. 20h. 53m. 18s., and 13y. 360d. 21h. 57m. 15s. Ans. 436y. 290d. 8h. 44m. iGs. 26. A carpenter sent two of his apprentices to ascertain the length of a certain fence. The first stated it was 17rd. 16ft. 11in., the second said it was 18rd. 5in. The carpenter, finding a discrepancy in their statements, and fearing they might both be wrong, ascertained the true length himself, which was 17rd. 5yd. lift. llin.; how much did each differ from the other? 27. From a mass of silver weighing 10glb., a goldsmith made 36 spoons, weighing 51b. l1oz. 12pwt. 15gr.; a tankard, 31b. 0oz. 13pwt. 14gr.; a vase. 71b. 11oz. 14pwt. 23gr.; how much unwrought silver remains? Ans. 881b. loz. 18pwt. 20gr

Page 121 SECT. XIV.] MULTIPLICATION OF COMPOUND NUMBERS. 121 28. From a piece of cloth, containing 17yd. 3qr., there were taken two garments, the first measuring 3yd. 3qr. 2na., the second 4yd. lqr. 3na.; how much remained? Ans. 9yd. lqr. 3na. 29. Venus is 35S. 18~ 45' 15" east of the Sun, Mars is 7S. 15' 36' 18" east of Venus, and Jupiter is 5S. 210 38' 27" east of lMars; how far is Jupiter east of the Sun? Ans. 4S. 260~ 30. The longitude of a certain star is 3S. 18~ 14' 35", and the longitude of Jupiter is 11S. 250 30' 50"; how far will Jupiter have to move in his orbit to be in the same longitude with the star? Ans. 3S. 22~ 43' 45". ~ XIV. MULTIPLICATION OF COMPOUND NUMBERS. ART. 104e MULTIPLICATION of Compound Numbers is the process of taking a compound number any proposed number of times. ART. 1s5, To multiply when the multiplier is not more than 12. Ex. 1. If an acre of land cost 14~. 5s. 8d. 2far., what will 9 acres cost? Ans. 128~. 11s. 4d. 2far. OPERATION. We write the multiplier under i-. s. d-. far. the lowest denomination of the Multiplicand 1 4 5 8 2 multiplicand, and then say 9 Multiplier 9 times 2far. are 18far., equal to 4d. and 2far. We set down the Product 1 2 8 11 4 2 2far. under the number multiplied, reserving the 4d. to be added to the next product. We then say 9 times 8d. are 72d., and the 4d. make 76d., equal to 6s. and 4d., and set the 4d. under the column of pence, reserving the 6s. to be added to the next product. Then, 9 times 5s. are 45s., and 6s. make 51s., equal to 2Y~. and Ils. We place the Ils. under the column of shillings, reserving the 2f~. to be added to the next product. Again, 9 times 14~. are 126~., and 2~. make 128~. This, placed under the column of pounds, gives us 128~f. Ils. 4d. 2far. for the answer. QUEsTIoNs. -Art. 104. What is multiplication of compound numbers? — Art. 105. Explain the operation. By what do you divide the product of each denomination? What do you do with the quotient and remainders thus obtained? 11

Page 122 120 MULTIPLICATION OF COMPOUND NUMBERS. [SECT. XIV, RULE. - Multiply each denomination of the compound number as in multiplication of simple numbers, and carry as in addition of cornpouna numbers. NoE. - Going a second time carefully over the work is a good way of testing its accuracy. On learning Division of Compound Numbers, the pupil will find that rule a better method of proving multiplication of compound numbers. EXAMPLES FOPR PRACTICE, 2. 3.. 5.. s. d;I. ~. L. s. d. ~. s. d. 5 6 8 19 1 25 17 5 7 1 1 8 15 8 2 3 5 6 101 3 4 58 14 9 129 9 17 12 14 4-A 6. 7. 8 ewt. qr. lb. oz. Ton. cwt. qr. lb. ewt. qr. lb. oz. 18 3 17 10 14 15 3 12 1 9 1 8 15 6 7 8 113 2 5 12 1 03 11 0 9 154 2 21 8 9. 10. 11. lb. oz. dr. m. fur. rd. ft. deg. m. far. rd. 15 14 13 97 7 14 13 18 12 6 18 9 6 8 143 5 5 587 4 8 12 145 33 2 102 12. 13. rd. yd. ft. in. fur. rd. ft. in 23 3 2 9 9 31 16 11 9 10 213 2 0 9 98 0 4 2 NOTE. - The answers to the following questions are found in the corresponding questions in Division of Compound Numbers, p. 128. 14. What cost 7 yards of cloth at 18s. 9d. per yard? 15. If a man travel 12m. 3fur. 29rd. in one day, how far will he travel in 9 days? 16. If 1 acre produce 2 tons 13cwt. 191b. of hay, what will 8 acres produce? QUESTIONS. - What is the rule? How may the work be tested?

Page 123 SECT. XIV.] MULTIPLICATION OF COMPOUND NUMBERS. 12% 17. If a family consume 49gal. 3qt. ipt. of molasses in one month, what quantity will be suficient for one year? 1S. John Smith has 12 silver spoons, each weighing 3oz. 17pwt. 14gr.; what is the weight of all? 19. Samuel Johnson bought 7 loads of timber, each measur ing 7 tons 37ft.; what was the whole quantity? 20. If the moon move in her orbit 130 11' 35" in I day, how far will she move in 10 days? 21. If 1 dollar will purchase 21b 8g 75 13 10gr. of ipecacu anha, what quantity would 9 dollars buy? 22. If 1 dollar will buy 2A. 3R. 15p. 30yd. 8ft. 10Oin. of wild land, what quantity may be purchased for 12 dollars? 23. Joseph Doe will cut 2 cords 97ft. of wood in I day; how much will he cut in 9 days? 24. If 1 acre of land produce 3ch. 6bu. 2pk. 7qt. ipt. of corn, what will 8 acres produce? ART. 106. When the multiplier is a composite number, and none of its factors exceed 12. Ex. 1. What cost 24 yards of broadcloth at 2~. 7s. 11d. per yard? Ans. 57~. 10s. Od. OPERATION. ~. i. d. 2 pric of ard. We find the number 24 4 equal to the product of 4 and 6; we therefore multiply the 9 11 8 price of 4 yards. price first by 4, and then that 6 product by 6, and the last product is the answer. e5 7 1 0 0 price of 24 yards. Ex. 2. What cost 360 tons of iron at 17g. 16s. Id. per ton? Ans. 6409~. 10s. Od. OPERATIOIN. ~. s. d. 7 1 6 1 - price of 1 ton. We find the factors of 360 to be 6, 6, and:10. We 1 0 6 1 6 6 3 price of 6 tons. first multiply by 6, and then 6 that product by 6, and then again the last product by 6 4 0 1 9 0 =-price of 36 tons. 10. 9 10 prie f 360 tns 6409 10 0 =price of 360 tons.

Page 124 12,4 MULTIPLICATIO1N OF COMPOUNND NUMBERS. [SECT. XIV RuLE. - Multiply by the factors of the composite number an succes sion. EXAMPLES FOR PRACTICE. 3. If a man travel 3m. 7fur. 18rd. in one day, how far would he travel in 30 days? 4. If a load of hay weigh 2.tons 7cwt. 3qr. 181b., what woula be the weight of 84 similar loads? 5. When it requires 7yd. 3qr. 2na. of silk to make a lady's dress, what quantity would be sufficient to make 72 similar dresses? 6. A tailor has an order from the navy agent to make 132 garments for seamen; how much cloth will it take, supposing each garment to require 3yd. 2qr. 1na.? ART. 107. When the multiplier is not a composite number, and exceeds 12, or, if a composite number, and any of its factors exceed 12. Ex. 1. What cost 379cwt. of iron at 3~. 16s. 8d. per owt.? Ans. 1452~. 16s. 8d. OPERATION. ~. S. d. Since 379 is not a composite 3 1 6 8 X 9 units. number, we cannot resolve it 1 0 into factors; but we may sep3 8 6 8 >< 7 tens. arate it into parts, and find 3 8 61 0 X 7 tens. the value of each part sepal 0 rately; thus, 379 = 300 - 70 3 8 3 6 8 + —9. In the operation, we 3 hundreds, first multiply by 10, and then this product by 10, to get the 1 1 5 0 0 0 cost of 300cwt. cost of 1OOcwt. To find the 2 6 8 6.8 cost of 70cwt. cost of 300cwt., we multiply 3 4 1 0 0 cost of 9cwt. the last product by 3; and to find the cost of 70cwt., we 1 4 5 2 1 6 8 cost of 379cwt. multiply the cost of 10cwt. by 7; and then, to find the cost of 9cwt., we multiply the cost of lcwt. by 9. Adding the several products, we obtain 1452~. 16s. 8d. for the answer. RULE. - Having resolved the multiplier into any convenient parts, as of units, tens, 4c., multiply by these several parts, adding together the products thus obtainedjfor the required result. QUESTIONS. —Art. 106. What is the rule for multiplying by a composite number? Give the reason for the rule. - Art. 107. How do you find the cost of 300cwt. in the example? Of 70cwt.? Of 9cwt.? What is the rule when the multiplier is large, and is not a composite number?

Page 125 SECT. XV.] DIVISION OF COMPOUND NUMBERS. 125 EXAMPLES FOR PRACTICE. 2. If 1 dollar will buy 171b. 10oz. 13dr. of beef, how much may be bought for 62 dollars? 3. What cost 97 tons of lead at 2~. 17s. 9 d. per ton? 4. If' a man travel 17rn. 3fur. 19rd. 3yd. 2ft. 7in. in one day, how far would he travel in 38 days? 5. If 1 acre will produce 27bu. 3pk. 6qt. ipt. of corn, what will 98 acres produce? 6. If it require 7yd. 3qr. 2na. to make 1 cloak, what quantity would it require to make 347 cloaks? 7. One ton of iron will buy 13A. 3R. 14p. 1Syd. 7ft. 76in.of land; how many acres will 19 tons buy? 8. If 1 ton of copper ore will purchase 17T. 14cwt. 3qr. 181b l4oz. of iron ore, how much can be purchased for 451 tons? Ans. 8003T. 17cwt. lqr. 121b. 10oz. 9 XV. DIVISION OF COMPOUND NUMBERS. ART. X08, DIVISION of Compound Numbers is the process of dividing compound numbers into any proposed number of equal parts. ART. 1A09 To divide when the divisor does not exceed 12. Ex. 1. If 9 acres of land cost 128~. 11s. 4d. 2far., what is the value of 1 acre? Ans. 14. 14~. 5 d. 2faro OPERATION. ~. S. d. far. Having divided the 128~. by 9, we 9 ) 1 2 8 11 4 2 find the quotient to be 14~. and 2~. re1 4 5 8 2 maining. We place the quotient 14~. under the 128~., and to the remainder 2~., equal to 40s., we add the lls. in the question, and divide tne amount, 51s., by 9. We write the quotient, 5s. under the IUs., and to the remainder 6s., equal to 72d., we add the 4d., making 76d., which we divide by 9, and write the quotient 8d. under the 4d. To the remainder 4d., equal to 16far., we add QUESTIONS. - Art. 108. What is division of compound numbers? - Art. 109. Where do you begin to divide? Why? When there is a remainder after dividing any one denomination, what must be done with it? 11'

Page 126 126 DIVISION OF COMPOUND NUMBERS. [SECT. XV. the 2far., and divide the amount, 18far., by 9, and obtain 2far. for a quotient —which we place under the 2far. in the dividend. Thus we find the answer to be 14~ 5s. 8d. 2far. RULE. - Divide as in division of simple numbers, each denomination zn its order, beginning with the hiyhest. If there be a remainder, reduce it to the next lower denomination adding in the number already of this denomination, if any, and divida as before. PROOF. - The same as in simple numbers. NOTE. - When the divisor and dividend are both compound numbers, they must be reduced to the same denomination, and the division then is that of simple numbers. 2. 3. 4. ~. s. d. ~. s. d. ~. s. d. 2)10 13 4 3)58 14 9 5)129 9 7 5 6 8 19 11 7 25 17 11 5. 6. 7. ~. s. d. far. cwt. qr. lb. oz. ton. cwt. qr. lb, 6)112 14 4 2 6)113 2 5 12 7)103 11 0 9 18 15 8 3 18 3 17 10 14 15 3 12 8. 9. 10. cwt. qr. lb. oz. lb. oz. dr. m. fur. rd. ft. 8).154 2 21 8 9)143 5 5 6)587 4 8 12 19 1 8 15 15 14 13 I1. 12. 13. deg. m. fur. rd. rd. yd. ft. in. fur. rd. ft. in. 8)145 33 2 1 0 9)213 2 0 9 10)98 0 4 2 NoTE. - The answers to the following questions are found in the. corres. ponding numbers in Multiplication of Compound Numbers. 14. What costs 1 yard of cloth, when 7yd. can be bought for 6~. 11s. 3d.? 15. If a man, in 9 days, travel 112m. ifur. 21rd., how fir will he travel in 1 day? 16. If 8 acres produce 21T. 5cwt. 2qr. 21b. of hay, what will 1 acre produce? QrsoTIo. - What is the rule for division of compound numbers?

Page 127 SECT. XV.] DV1SION COF COMPOUND NUMBERS. 127 17. If a family consume in 1 year 598 gal. 2qt. of molasses, how much will be necessary for 1 month? 18. John Smith has 12 silver spoons, weighing 31b. 10)oz. 11pwt.; what is the weight of each spoon? 19. Samruel Johnson bought 7 loads of timber, measuring 55T. 19ft.; what was the quantity in each load? 20. If the moon, in 10 days, move in her orbit 4S. 110 55' 50", how far does she move in 1 day? 21. If $9 will buy 241t 8~ 33 19 lOgr. of ipecacuanha, how large'. quantity will $1 purchase? 22. When $12 will buy 34A. OR. 32p. 8yd. 5ft. 48in. of wild land, how much will $1 buy? 23. Joseph Doe will cut 24 cords 105 feet of wood in 9 days; how much will he cut in 1 day? 24. When 8 acres of land produce 25ch. 17bu. 3pk. 4qt. of grain, what will 1 acre produce? ART. 110. When the divisor is a composite number, and none of its factors exceed 12. Ex. 1. When 24 yards of broadcloth are sold for 57~. 10s. Od., what is the price of 1 yard? Ans. 2~. 7s. 11d. OPERATION. We find the component S. d. parts, or factors, of 24, 6) 5 7 1 0 O = price of 24 yards. are 6 and 4. We there4) 9 1 1 =fore divide the price by ) 9 1 pco yd one of these numbers, and 2 7 1 1 = price of 1 yard. the quotient by the other. RULE. - Divide by thefactors of the composite number in succession. EXAMPLES FOR PRACTICE. 2. If 360 tons of iron cost 6409~. 10S. Od., what is the cost of 1 ton? 3. If a man travel 117m. 7fur. 20rd. in 30 days, how far will he travel in 1 day? 4. If 84 loads of hay weigh 201 tons 6cwt. Oqr. 121b., what will 1 load weigh? 5. When 72 ladies require 567yd. Oqr. Ona. for their dresses, how many yards will be necessary for one lady? QUEsTIONS.- Art. 110. How does it appear that dividing by 6 in Ex. 1 gives the price of 4 yards? How do you divide by a composite number?

Page 128 128 DIVISION OF COMPOUND NUMBERS. [SECT. XV 6. WVhen 132 sailors require 470yd. lqr. of cloth to make their garments, how many yards will be necessary for 1 sailor'? ART. Il When the divisor is not a composite number, and exceeds 12, or, if'a composite number, and any of its fiactors exceed 12, the whole operation can be written down, as in the following example: Ex. 1. If 23cwt. of iron cost 171~. Is. 3d., what cost 1cwt.? Ans. 7~. 8s. 9d. OPERATION. ~. s.d. 2 3 ) 1 7 1 1 3 ( 7~. We divide the pounds by 23, and obtain 1 6 1 7 for the quotient, and 10~. remaining, which we reduce to shillings, and add the 1 0 Is., and again divide by 23, and obtain 8s. 2 0 for the quotient. The remainder, 17s., we 2 3 ) 2 0 1 (8s. reduce to pence, and add the 3d., and again I )81(, 4 d divide by 23, and obtain 9d. for the quo1 8 4 tient. Thus, the method of operation is 1 7 the same as by the general rule (Art. 109), 1 2 excepting more of the work is written down; and, by uniting the several quo2 3 ) 2 0 7 ( 9d. tients, we find the answer to be 7~. 8s. 9d. 207 2. If 862 will buy 10951b. 14oz. 6dr. of beef, how much may'oe obtained for 81? 3. Paid 280~, 5s. 9Id. for 97 tons of lead; what did it cost per ton? 4. If a man travel 662m. 4fur. 28rd. 3yd. 2ft. 2in. in 38 days, how far will he travel in I day? 5. When 98 acres produce 2739 bu. lpk. 5qt. of grain, what will 1 acre produce? 6. A tailor made 347 garments from 2732yd. 2qr. 2na. of cloth; what quantity did it take to make 1 garment? 7. When 19 tons of iron will purchase 262A. 3R. 37p. 25yd. Ift. 40in. of land, how much may be obtained for 1 ton? 8. Itf 451 tons of copper ore will purchase 8003T. 17cwt. lqr 121b. lOoz. of iron ore, how much will 1 ton purchase? Ans. 17T. 14cwt. 3qr. 181b. 14oz. QUESTION. - Art. 111. When the divisor is large, and not a composite num ber, how is the division performed?

Page 129 SECTer. XI.] MISCELLANEOUS EXAMPLES. 129 XVI. MISCELLANEOUS EXAMIPLES IN MULTIPLICATION AND DIVISION OF COMPOUND NUMBERS. 1. BOUGHiT 30 boxes of sugar, each containing Scwt. 3qr. 201b., but having lost 68cwt. 2qr. 0lb., I sold the remainder for 1a. 17s. 6d. per cwt.; what sumu did I receive? Ans. 375~. 2. A company of 144 persons purchased a tract of land containing 11067A. iR. 8p. John Smith, who was one of the company and owned an equal share with the others, sold his part of the land for is. 9Id. per square rod; what sum did he receive? Ans. 1101~. 12s. lId 3. The exact distance from Boston to the mouth of the Columbia River is 2644m. 3fur. 12rd. A man, starting from Boston, travelled 100 days, going 18m. 7fur. 32rd. each day; required his distance from the mouth of the Columbia at the end of that time. Ans. 746m. 7fur. 12rd. 4. James Bent was born July 4, 1798, at 3h. 17m. A. M.; how long had he lived Sept. 9, 1807, at llh. 19m. P. M., reckoning 365 days for each year, excepting the leap year 1804, which has 366 days? Ans. 3353da. 20h. 2m. 5. The distance from Vera Cruz, in a straight line, to the city of Mexico, is 121m. 5fur. If a man set out from Vera Cruz to travel this distance, on the first day of January, 1848, which was Saturday, and travelled 3124rd. per day until the eleventh day of January, omitting, however, as in duty bound, to travel on the Lord's day, how far would he be from the city of Mexico on the morning of that day? Ans. 43m. 4fur. 8rd. 6. Bought 16 casks of potash, each containing 7cwt. 3qr. 181b., at 5 cents per pound. I disposed of 9 casks at 6 cents per pound, and sold the remainder at 7 cents per pound; what did I gain? Ans. $182.39. 7. A merchant purchased in London 17 bales of cloth for 17g. 18s. 10l. per bale. He disposed of the cloth at Havana for sugar at 1~. 17s. 6d. per cwt. Now, if he purchased 144cwt. of sugar, what balance did he receive? Ans. 35X. Os. 2d. 8. A and B commenced travelling, the same way, round an island 50 miles in circumference. A travels 17m. 4fur. 30rd. a day, and B travels 12m. 3fur. 20rd. a day.; required how far they are apart at the end of 10 days. Ans. Im. 4fur. 20rd,

Page 130 130 PROPERTIES AND RELATIONS OE NUMBERS. [SEcr. XVI1. 9. Bought 760 barrels of flour at $5.75 per barrel, which I paid for in iron at 2 cents per pound. The purchaser aflterwards sold one half of the iron to an axe manufhcturer; what, quantity did he sell? Asl. 54T. 12cwst. 2tcqr 10. Bought 17 house-lots, each containing 44 perches, 2100 square fect. From this purchase I sold 2A. 2R. 240ft., and the remaining quantity I disposed of at Is. 2.-d. per square foot; what anlount did I receive for the last sale? Ans. 5914I~. 19s. 53d. 11. J. Spofford's farm is 100 rods square. From this he sold 1H. Spaulding a fine house-lot and garden, containing 5A. 3R. 17p., and to D. Fitts a farm 50rd. square, and to R. Thornton a farm containing 3000 square rods; what is the value of the remainder, at $1.75 per square rod? Ans. $6235.25. 12. Bought 78A. 3R. 30p. of land for $7000, and, having sold 10 house-lots, each 30rd. square, for $8.50 per square rod, I dispose of the remainder for 2 cents per square foot. How much do I gain by my bargain? Ans. $89265.35 XVII. PROPERTIES AND RELATIONS OF NUMBER8. ART. 11, AN INTEGER is a whole number; as 1, 6, 13. All numbers are either odd or even. An odd number is a number that cannot be divided by 2 without a remainder; thus, 3, 7, 11. An even number is a number that can be divided by 2 without a remainder; thus, 4, 8, 12. Numbers are also either prizme or composite. A prime number is a number which can be exactly divided only by itself or 1; as 1, 3, 5, 7. A comzposite number is a number which can be exactly divided other than by itself or 1; as 6, 9, 14. Numbers are prime to each other when they have no factor in common; thus, 7 and 11 are prime to each other, as are, also, 4, 15, and 19. QUESTIONS. - Art. 112. What is an integer? What are all numbers? What is an odd number? What is an even number? What other distinctions of numbers are mentioned.? What is a prime number? When are numbers prime to each other? What is a composite number?

Page 131 SECT. XVII.] PROPERTIES AND RELATIONS OF NUMBERS. 131 All the prime numbers not larger than 1109 are included in the following T&BLE OF PRIME NUMBERS. 1 59 139 233 337 439 557 653 1 769 883 1013 2 61 140 239 347 443 563 659 773 887 1019 3 67 151a 241 349 449 569 661 787 907 1021 5 71 157 251 353 457 571 673 797 911 1031 7 73 163 257 359 461 577 677 809 919 1033 11 79 167 263 367 463 587 683 811 929 1039 13 83. 173 269 373 467 693 691 821 937 1049 17 89 179 271 379 479 599 701 823 941 1051 19 97. 181 277 383 487 601 709 827 947 1061 23 101. 191 281 389 491 607 719 829 953 1063 29 103. 193 283 397 499 613 727 839 967 1069 31 107 197 293 401 503 617 733 853 971 1087 37 109. 199 307 409 509 619 739 857 977 1091 41 113 211 311 419 521 631 743 859 983 1093 43 127, 223 313 421 523 641 751 863 991 1097 47 131 227 317 431 541 643 757 877 997 1103 53 137 229 331 -433 547 647 761 881 1009 1109 ART. 113. A pri?,me factor of a number is a prime number that will exactly divide it; thus, the prime factors of 21 are the prime numbers 1, 3, and 7. A composite factor of a number is a composite number that will exactly divide it; thus, the composite factors of 24 are the composite numbers 4 and 6. NoTE 1. —Unity or 1 is not regarded as a material prime factor, since multiplying or dividing any number by 1 does not alter its value; it will be omitted when speaking of the prime factors of numbers. NOTE 2. —There has been discovered no direct process by which prime numbers may be found. The following facts, however, if kept in mind, will aid in ascertaining whether a number is prime or not; and, if not prime, indicate one or more of its factors: 1. 2 is the only even prime number. 2. 2 is a factor of every even number. 3. 3 is a factor of every number the sum of whose digits 3 will exactly divide; thus, 15, 81, and 546, have each 3 as a factor. 4. 4 is a factor of every number whose two right-hand figures 4 will exactly divide; thus, 316, 532, and 1724, have each 4 as a famctor. 5. 5 is the only prime number having 5 for a unit or right-hand figure. QUESTIONS. -Art. 113. What is a prime factor? What is a composite factor? How is unity or 1 regarded? Is there any direct process for de termuining prime numbers? Which is the only even prime number? Of what numbers is 2 a factor? Of what numbers is 3 a factor? Of what numbers is 4 a factor r

Page 132 132 PROPERTIES AND RELATIONS OF NUMBERS. [SECT. XVI1 6. 5 is a factor of every number whose right-hand figure is either 5 or 0; as, 15, 20, &c. 7. 6 is a factor of every even number that 3 will exactly divide; thus, 24, 108, and 830, have each 6 as a factor. 8. 7 is a factor of every numnber occupyfng four places whose two righthand figures are contained in the left-hand figure or figures exactly 3 times; thus, 602, 2107, and 3913, hlave each 7 as a factor. 9. 7 is a factor of every number occupying three or four places, when the two right-hand figurles coltai the left-lland figure or figures exactly 5 times; thus, 840, 945, and 1155, have each 7 as a factor. 10. 8 is a factor of every number -,hose three right-hand figures 8 will xsactly divide; thus, 5072, 11240, and 17128, have each 8 as a factol. 11. 9 is a factor of every number the sum of whose digits 9 will exactly divide; thus, 27, 432, and 20304, have each 9 as a factor. 12. 10 is a factor of every number whose right-hand figure is 0; as, 20, 30, &Co 13. 7, 11 and 13, are factors of any number occupying four places in which two like figures have two ciphers between them; as, 3003, 4004, 9009, &c. 14. Every prime number, except 2 and 5, has 1, 3, 7, or 9, for the right-hand figure. ART. I4 I, Method of finding the prime factors of numbers. Ex. 1. It is required to find the prime factors of 24. Ans. 2,, 2 2, 3. OPERATION. We divide by 2, the least prime number 2 2 4 greater than 1, and obtain the quotient 12 - And since 12 is a composite number, we 2 1 2 divide this also by 2, and obtain a quotient 6. We divide 6 by 2, and obtain 3 for a 2 6 quotient, which is a prime number. The 3 several divisors and the last quotient, all being prime, constitute all the prime factors of 24, which, multiplied together, 2 X 2 X 2 X 3 24 they equal. RULE. - Divide the given number by the least prime number, greater than 1, that zoill divide it, and the quotient, if a composite number, in the same manner; and continue dividing until a prime number is obtained for a quotient. The several divisors and the last quotient will be the prime factors required. NOTE. - The composite factors of any number may be found by multiplying together two or more of its prime factors. QUESTINxs. - Of what numbers is 5 a factor? Of what numbers is 6 a factor? Of what numbers is 7 a factor? Of what numbers is 8 a factor? Of what numnbers is 9 a factor? What is thbe right-hand figure of every prime number? What is the rule for finding the prime factors of numbers? Tlow may the composite factors of numbers be found?

Page 133 SECT. XVIL] PROPERTIES AND RELATIONS OF NUMBERS. 133 EXAMPLES FOR PRACTICE. 2. What are the prime factors of 36? Ans. 2, 2, 3, 3. 3. WThat are the prime factors of 48? Ans. 2, 2, 2, 2, 3. 4. WVhat are the prime factors of 56? Ans. 2, 2, 2, 7, 5. WVhat are the prime factors of 144? Ans. 2, 2, 2, 2, 3, 3. 6. What are the prime factors of 3420? Ans. 2, 2, 3, 3, 5, 19. 7. What are the prime factors of 18500? Ans. 2, 2, 5, 5, 5, 37. 8. What are the prime factors of 19965? Ans. 3, 5, 11, 11, 11. 9. What are the prime factors of 12496? Ans. 2, 2, 2, 2, 11, 271 10. What are the prime factors of 17199? Ans. 3, 3, 3, 7, 7, 13. 11. What are the prime factors of 7800? Ans. 2, 2, 2, 3, 5, 5, 13. CANCELLATION. ART. 1 5, If the dividend and divisor are both divided by the same number, the quotient is not changed. Thus, if the dividend is 20 and the divisor 4, the quotient will be 5. Now, if we divide the dividend and divisor by some number, as 2, their proportion is not changed, and we obtain 10 and 2 respectively; and 10 -- 2 - 5, the same as the original quotient. ART. 11, If a factor in any number is cancelled, the number is divided by that factor. Thus, if 15 is the dividend and 5 the divisor, the quotient will be 3. Now, since the divisor and quotient are the two factors, which, being multiplied together, produce the dividend (Art. 50), it is plain, if we cross oust or cancel the factor 5, the remaining 3 is the quotient, and by the operation the dividend 15 has been divided by 5. ART. 1f17. Cancellation is the method of shortening arithinetical operations by rejecting any factor or factors common to the divisor and dividend. QuESTIONS. —Art. 115. What is the effect on the quotient when the dividend and divisor are divided by the same number? What is the effect of cancelling a factor of any number? What is cancellation? 12

Page 134 134 PROPERTIEb AND RELATIONS OF NUMBERS. [SECT. XVII Ex. 1. A man sold 25 hundred weight of iron at 5 dollars per hundred weight, and expended the money for flour at 5 dollars per barrel; how many barrels did he purchase? Ans. 25 barrels. OPERATION. We first indicate by their signs Dividend 5 X 2 5 the multiplicatlon and division re-, Divisor 2 5. quired by the question. We then, observing 5 to be a common factor of the divisor and dividend, divide the divisor and dividend by this factor, or, which is the same thing, cancel or reject it in both, and obtain 25 for the quotient. 2. Divide the product of 12, 7, and 5, by the product of 5, 4, and 2. Ans. 10. OPERATION. Dividend I X 7 X 2 1 0 Quotient. Divisor X 2 2 Quotient Finding 4 in the divisor to be a factor of 12 in the dividend, we divide 12 by 4, cancelling these numbers, and use the 3 instead of 12. The factor 5, common to both dividend and divisor, having been cancelled, we divide the product of the remaining factors in the dividend by the product of those in the divisor, and obtain the quotient 102o 3. Divide the product of 8, 5, 16, and 21, by the product of 10, 4, 12, and 7. OPERATION. 4 Dividend $ X $ X I $ X 4, Divisor I 0 X X I X Quotient. The product of the factors 8 and 5 in the dividend is equal to the product of 10 and 4 in the divisor; therefore we cancel these factors. Finding 16 in the dividend and 12 in the divisor may be divided by 4, they are cancelled, and use made of their quotients. Again, as the product of the factors 3 and 7 of the divisor equals the 21 of the dividend, we cancel the 3, 7, and 21. The factor 4 alone remaining is the quotient. QUESTIONS. - How do you arrange the dividend and divisor for cancellation? How do you then proceed? Is the factor 5, in Ex. 1, reduced to 0 or 1 by being cancelled? How do you proceed when a number in the dividend and another in the divisor have a common factor? How do you proceed when the products of two or more factors in the dividend and divisor are alike'

Page 135 SECT. XVII.] PROPERTIES AND RELATIONS OF NUMBERS. 135 RULE. -Cancel the factor or factors common to -the dividend and divisor, and then ditvide the product of the factors remaining in the dividend by the product of those remaining in the divisor. NOrE.- 1. In arranging the numbers for cancellation, the dividend may be written above the divisor with a horizontal line between them, as in division (Art. 47); or, as some prefer, the dividend may be written on the right of the divisor, with a vertical line between them. NOTE. - 2. Cancelling a factor does not leave 0, but the quotient 1, to take its place, since rejecting a factor is the same as dividing by that factor (Art. 116). Therefore, for every factor cancelled, either in the dividend or divisor, the factor 1 remains. EXAMPLES FOR PRACTICE. 4. Divide 42 X 19 by 19. Ans. 42. 5. Divide the product of 8, 6, and 3, by the product of 6, 3, and 4. Ans. 2. 6. Divide the product of 17, 6, and 2, by the product of 6, 2, and 17. Ans. 1. 7. Sold 15 pieces of shirting, and in each piece there were 30 yards, for which I received 10 cents per yard; expended the money for 10 pieces of calico, each containing 15 yards; what was the calico per yard? Ans. 30 cents. 8. Divide the product of 12, 7, and 5, by the product of 2, 4, and 3. Ans. 17.-1 9. Divide the product of 20, 13, and 9, by the product of 13, 16, and 1. Ans. 111. 10. Divide the product of 9, 8, 2, and 14, by the product of 3, 4, 6, and 7. Ans. 4. 11. Divide the product of 16, 5, 10, and 18, by the product of 8, 6, 2, and 12. Ans. 12~1. 12. Divide the product of 22, 9,; 12, and 5, by the product of 3, 11, 6, and 4. Ans. I5. 13. Divide the product of 25, 7, 14, and 36, by the prodluct of 4, 10, 21, and 54. Ans. -1. 14. Divide the product of 26, 72, 81, and 12, by the product of 36, 13, 24, and 54. Ans. 3. 15. Divide the product of 8, 5, 3, 16, and 28, by the product of 10, 4, 12, 4, and 7. Ans. 4. 16. Divide the product of 8, 4, 9, 2, 12, 16, and 5, by the product of 4, 6, 6, 3, 8, 4, and 20. Ans. 2. 17. Divide the product of 6, 15, 16, 24, 12, 21, and 27, by the product of 2, 10, 9, 8, 36, 7, and 81. Ans. 8. QUESTIONS. -What is the rule for cancellation? How may the numbers be arranged for cancelling? What takes the place of a, cancelled factor? What remains for every factor cancelled either in the dividend or divisor?

Page 136 136 PROPERTIES AND RELATIONS OF NUMBERS. [SECT. XVI1. A COMMON DIVISOR. ART. 118, A common divisor of two or more numbers is any number that will divide them without a remainder; thus, 2 is a common divisor of 2, 4, 6, and 8. ART. 119. - To find a common divisor of two or more numbers. Ex. 1. What is the common divisor of 10, 15, and 25? Ans. 5. OPERATION. We resolve each of the given numbers into two 1 0. 5 X 2 factors, one of which is common to all of them. 1 5 = 5 X 3 In the operation 5 is the common factor, and there2 5 - 5 X 5 fore must be a common divisor of the numbers. RULE. - Resolve each of the given numbers into two factors one of which is common to all of them, and this common factor is a commlon divisor. EXAMPBLES FOR PRACTICE. 2. What is the common divisor of 3, 9, 18, 24? Ans. 3. 3. What is the common divisor of 4, 12, 16, 28? Ans. 2 or 4. ART. 120. A divisor of any factor of a number is a divisor of the number itself. Thus 3, a divisor of 9, a factor of 45, is a divisor of 45 itself. ART. 121. A common divisor of two numbers is a divisor of their-sum and of their difference. Thus 4, a common divisor of 16 and 12, is a divisor of their sum, 28, and of their difference, 4. ART. 122? A common divisor of the remtainder and the divisor is a divisor of the dividend. Thus, in a division having 12 for remainder, 36 for divisor, and 48 for dividend, 12, a common divisor of the 12 and the 36, is also a divisor of the 48. THE GREATEST COMMON DIVISOR. ART. 123. The greatest common divisor of two or more numbers is the greatest number that will divide each of them without a remainder. Thus 6 is the greatest common divisor of 12, 18, and 24. QUESTIONS. - Art. 118. What is a common divisor of two or more numbers?- Art. 119.'What is the rule?-Art. 121. Of what is the common divisor of two numbers a divisor? - Art. 122. Of what is a common divisor of the less of two numbers and of their difference a divisor? —Art. 123. What is the greatest common divisor of two or more numbers?

Page 137 OECT. XVII.] PROPERTIES AND RELATIONS OF NUMBERS. 137 ART. 124, To find the greatest common divisor of two or more numbers. Ex. 1. What is the greatest common divisor or measure of 84 and 132? Ans. 12. FIRST OPERATION. Resolving the numbers into their 8 4 2 X 2 X 3 X 7 prime factors (Art. 114), thus, 84 1 32 2 X 2 X 3 X - = 2 X 2X 3 X 7, and 132=-2X 2 X 2 X 3 - 12. 2 X 3 X 11, we find the factors 2 X 2 X 3 are common to both. Since only these common factors, or the product of two or more of such factors, will exactly divide both numbers, it follows that the product of all their common prime factors must be the greatest factor that will exactly. divide hboth of them. Therefore 2 X 2 X 3 12 is the greatest common divisor required. The same result may be obtained by a sort of trial process, as by the second operation. SECOND OPERATION. It is evident, since 84 canno 84) 1 32 ( 1 be exactly divided by a number 8 4 greater than itself, if it will also exactly divide 132, it will 4 8) 8 4 (1 be the greatest common divisor 4 8 sought. But, on trial, we find -3 6 ) 4 8 (1 84 will not exactly divide 132, 36) 48( 1 there beinof a remainder, 48 3 6 Therefore!4 is not a common 1 2) 3 6 ( 3 divisor of the two numbers. 3We know a common divisor of 48 and 84 will also be a divisor of 132 (Art. 122). We next try to find that divisor. It cannot be greater than 48. But 48 will not exactly divide 84, there ~being a remainder, 36; therefore 48 is not the greatest common div isor. Again, as the common divisor of 36 and 48 will also be a divisor of 84 (Art. 122), we try to find that divisor, knowing that it cannot be greater than 36. But 36 will not exactly divide 48, there being a remainder, 12; therefore 36 is not the greatest common divisor. As before, the common divisor of 12 and 36 will be a divisor of 18 (Art. 122); we make a trial to find that divisor, knowing that it cannot be greater than 12, and find 12 will exactly divide 36. Therefore 12 is the greatest common divisor required. RULE 1. - Resolve the given numbers into their prime factors. The product of all the factors common to the several numbers will be the greatest common divisor. Or, RULE 2. - Divide the greater number by the less, and if there be a QUESTION. - Art. 124. What are the rules for finding the greatest common divisor of two or more numbers? 12*

Page 138 138 PROPERTIES AND RELATIONS OF NUMBERS. [SECT. XVIL. remainder divide the preceding divisor by it, and so continue dividing until nothing remains. The last divisor will be the greatest common divisor. NOTE. — When the greatest common divisor is required of mnore than two numbers, find it of two of them, and then of that common divisor and of one of the other numbers, and so on for all the given numbers The last common divisor will be the greatest common divisor required. EXAMPLES FOR PRACTICE. 2. What is the greatest common divisor of 85 and 95? Ans. 5, 3. What is the greatest common divisor of 72 and 168? Ans. 24. 4. What is the greatest common divisor of 119 and 121? Ans. 1. 5. What is the greatest common divisor of 12, 18, 24, and 30? Ans. 6. 6. Having three rooms, the first 12 feet wide, the second 15 feet, and the third 18 feet, I wish to purchase a roll of the widest carpeting that will exactly fit each room without any cutting as to width. How wide must it be? Ans. 3 feet. A COMMI0ON MULTIPLE. ART. 12a5 A multiple of a number is a number that can be divided by it without a remainder; thus 6 is a multiple of 3. AnT. 12$o A common multiple of two or more numbers is a number that can be divided by each of them without a remainder; thus 12 is a common multiple of 3 and 4. ART. 127. The least common rmultipZe of two or more numbers is the least number that can be divided by each of them without a remainder; thus 30 is the least common multiple of 10 and 15. NOTE. - A multiple of a number contains all the prime factors of that number; and the common multiple of two or more numbers contains all the prime factors of each of the numbers. Therefore, the least common multiple of two or more numbers must be the least nuniber that will contain all the prime factors of them, and none others. Hence it will have each prime factor taken only the greatest number of times it is found in any of the several numbers. QUESTIONS.- Art. 125. What is a multiple of a number? - Art. 127 What is the least common multiple of a number?

Page 139 bECT. XVII.] PROPERTIES AND RELATIONS OF NUMBERS. 139 ART. 128, To find the least common multiple. Ex. 1. What is the least common multiple of 6, 9, 12? Ans. 36. FIRST OPERATION. Resolving the numbers into their 6 _ 2 X 3 prime factors, - thus, 6 = 2 X 3, and 9 3 X 3 9=3X3, and 12=2X2X3,-we 12 = 2 X 2 X 3 find their different prime factors to be 2 X 2 X 3 X 3 - 36 2 and 3. The greatest number of times the 2 occurs as a factor in any of the numbers is twice, as 2 X 2 in 12; and the greatest number of times the 3 occurs in any of the numbers is also twice, as 3 X 3 in 9. Hence 2 X 2 X 3 X 3 must be all the prime factors that are necessary in composing 6, 9, and 12; and, consequently, the product of these factors must be the least number that can be exactly divided by 6, 9, and 12. Therefore 2 X 2 X 3 X 3 = 36 is the least common multiple required. SECOND OPERATION. Another method, and one usually 3 6 9 1 2 preferred, is as by second operation. 2 2 3 4 Having arranged the numbers on a horizontal line, we divide by 3, a prime 13 2 number that will divide all of' them 3 X 2 X 3 X 2 = 3 6 without a remainder, and write the quotients in a line below. We next divide by 2, a prime number that will divide without a remainder most of them, writing down the quotients and undivided numbers as before. Then, since these numbers are prime to each other, we multiply together the divisors and the numbers on the lower line, which are all the prime factors of 6, 9, and 12, and thus obtain 36 for the least common multiple. RULE 1. - Resolve the given numbers into their prime factors. The product of these factors, taking each factor the greatest number of times it occurs in any of the numbers, will be the least common multiple. Or, RULE 2. -Having arranged the numbers on a horizontal line, divide by such a prime number as will divide most of them without a remainder, and write the quotients and undivided numbers in a line beneath. So continue to divide until no prime number greater than 1 will divide two or more of them. The product of the divisors and the numbers of the line below will be the least common multiple. NOTE 1.-When numbers are prime to each other, their product is their least common multiple. NOTE 2. -When one or more of the given numbers are factors of any one of the other numbers the factor or factors may be cancelled. QUESTION. -Art. 128. What are the rules for finding the least common multiple?

Page 140 140 FRACTIONS. [SECT. XVII1 EXAMIPLES FOR PRACTICE. 2. WVhat is the least common multiple of 7, 14, 21, and 15? Ans. 210. OPERATION. 7 r 1 4 21 15 Since 7 is a factor of 14, another of the numbers, we cancel it; and since 3 is a 2 $ 1 5 factor of 15, we also cancel that (Note 2): thus the work is rendered shorter. 7X2X15 —=10 3. What is the least common multiple of 3, 4, 5, 6, 7, and 8? Ans. 840. 4. WVhat is the least number that 10, 12, 16, 20, and 24, will divide without a remainder? Ans. 240. 5. What is the least common multiple of 9, 8, 12, 18, 24, 36, and 72? Ans. 72. 6. Five men start from the same place to go round a certain island. The first can go round it in 10 days; the second, in 12 days; the third, in 16 days; the fourth, in 18 days; the fifth, in 20 days. In what time will they all meet at the place from which they started? Ans. 720 days. ~ XVIII. FIRACTIONS. ART. 1!29. A FRACTION is an expression denoting one or more equal parts of a unit. The term fraction is derived from the Latin word frango, which signifies to break; from the idea that a number or thing is broken or separated into parts. Fractions are of two kinds, Common and Decimal. COMMON FRACTIONS. ART. 130. A COMMON FRACTION is expressed by two numbers one above the other, with a line between them. The number below the line is called the denominator; and the number above, the numerator. Thus, { Numerator 3 Three Denominator 5 Fifths. QUESTONS. - Art. 129. Whatt is a fraction? From what is the term derived, and what does it signify? Ilow nmany kinds of fractions, and what are they called?-Art. 130. How is a coummon fraction expressed? What is the number below the line ealled? The number above the line?

Page 141 BECT. XVIII.] COMMON FRACTIONS. 141 The denominator shows into how many parts the whole number is divided, and gives a name to the fraction. The numerator shows how many of these parts are taken, or expressed by the fraction. A proper fraction is one whose numerator is less than the'denominator; as, -. An inzmproper firaction is one whose numerator is equal to, or greater than, the denominator; as, 8, 9. NOTE. -A fraction, strictly speaking, is less than a unit; hence, if the numerator is equal to, or greater than, the denominator, it expresses a unit or more than a unit, and is therefore called an improper fraction. A mixed number is a whole number with a fraction; as, 7-A, A simple or single fraction has but one numerator and one denominator, and may be either proper or improper; as, -, 4. A compotend fraction is a fraction of a fraction, connected by the word of; as, 7 of 5 of a. A complex fraction is a fraction having a fraction or a mixed number for its numerator or denominator, or both; as o 7 8' 71 7T 978 11 T 9-eL ART, 131* The terrns of a fraction are its iunmerator and denominator. The unit of a fraction is the unit or whole thing from which its fractional parts, or fr~actional units, are obtained. A whole number may be expressed fractionally, by writing I for the denominator. Thus, 5 may be written 4, and read 5 ones; and 9 may be written 9, and read 9 ones. ART. 132. Fractions originate from division; thie nume - ator answers to the dividend, and the denominator to th~e divisor. Thus, when we divide 479956 by 6 (Art. 49, Ex. 12), we had a remainder of 4, which could not be divided by 6, and therefore we wrote it over the divisor, with a line between them. This expression originating from division is a fraction; the number above the line being the numerator, and the one below the deoniominator. QtESTIONS. -What does the denominator of a fraction show? What does the numerator show? What is a proper fraction? Wbhat is an improper fraction? NW1hat is a mixed number? What is a simple fraction? What is a compound fraction? What is a complex fraction? - Art. 131. Wlhat are the terms of a fraction? What is the unit of a fraction? HSow may a whole number be expressed fractionally? From what do firactions originate?

Page 142 142 REDUCTION OF COMMON FRACTIONS. [SECT. XVIII. ART. 13, From what has preceded, we perceive that the value of a frctctionz is the quotient arising from the division of the numerator by the denominator. Thus, the value of a, or 6 - 2, is 3; and the value Of 4-, or 3 4, is. REDUCTION OF COMMON FRACTIONS. AnT. 3L o Reduction of Fractions is the process of changing their formt of expression without altering their value. A fiaction is in its lowest terms, when its terms are prime to each other. (Art. 112.) ART. T35o To reduce a fraction to its lowest terms. Ex. 1. Reduce 6 to its lowest terms. Ans. W. OPERATION. We divide the terms of the fraction by 2, a factor ~2) is_ 3 common to them both, and obtain W. We divide, again, both terms of 3 by 3, a factor common to them, 3 ) i=3 and obtain!. Now, as 1 and 3 are numblers prime to each other, the fraction X is in its lowest terms. The same result would have been produced, if we had divided the terms by 6, the greatest common divisor. Since the numerator and denominator of a fraction correspond to the dividend and divisor in division (Art. 132), dividing both by the samle number, or cancelling equal factors in both (Art. 115), changes only the form of the firaction, while the value expressed ilemains the same. Therefore, Dividing the numerator and denominator of a fraction by the same number does not alter the value of the fraction. RULE. - Divide the numerator dnd denominator by any fnurmer greater than 1, that will divide them both without a remainder, and thus proceed until they are prinme to each other. Or, Divide both the numerator and denominator by their greatest common divisor. EXAMPLES FOR PRACTICE. 2. Reduce 6 to its lowest terms. Ans. 1. 3. Reduce -5 to its lowest terms. Ans. a. 4. Reduce 12 to its lowest terms. Ans. 8. 5. Reduce 1964 to its lowest terms. Ans.'2. 6. Reduce 1907 to its lowest terms. Ans. 2 7. Reduce I2- to its lowest terms. Ans. ~2' S. Reduce "8 to its lowest terms. Ans. 7. QUESTIONS. - What is the value of a fraction? - Art. 134. What is reduction of fractions? When is a fraction in its lowest terms? - Art. 135. Why does dividing both terms of a fraction by the )ame number not alter the value? Has ~ the same value as i%? Why? Repeat the rule.

Page 143 bECT. XVIII.] REDUCTION OF COMMON FRACTIONS. 143 9. Reduce 78 9 1 to its lowest terms. Ans. 9 10. WVhat is the lowest expression of 4s 4? Ans.,-s7 ART. 136. To reduce a mixed number to an improper fraction Ex. 1. In 7` how many fifths? Ans. 358. OPERATION. 7,3 5 5 Since there are 5 fifths in 1 whole one, there will be 5 times as many fifths as whole ones; therefore, 3 5 fifths. in 7 there are 35 fifths, and the 3 fifths being added 3 make 38 fifths, which are expressed thus, 358. 3 8 fifths -s — RULE. - Multiply thle whole number by the denominator of the fraction, and to the product add the numerator, and place the sum over the given denominator. NOTE. - TO reduce a whole number to a fraction of the same value, having a given denominator, we.multiply the whole number by the given denomlinator, and nmake the product the numerator; thus, 5, reduced to a fraction, having 3 for a denominator, becomes 5-. EXAMPLES FOR PRACTICE. 2. In 83 dollars how many sevenths? Ans. 79. 3. In 3} oranges how maly fourths? Ans. 1 3 4. In 9T r gallons how many elevenths? Ans. 1-~035. Reduce 8-3 to an improper fraction. Ans. 9 1 6. Reduce 1517 to an improper fraction. Ans. J —. 7. In 187 how many ninths? Ans. 169. 8. In 16l~v- how many one hundred and seventeenths? Ans. 1T848 11-yy. 9. Change t431 to an improper fraction. Ans. 5 14 10. WVhat improper fraction will express 2719x? Ans. 3GO%. 11. Change 1111:T to an improper fraction. Ans. 12 322 12. Change 125 to an improper fraction. Ans. 15. 13. Change 25 to an improper fraction, having 6 for a den6rminator. Ans. l 5 a 14. Reduce 75 to ninths. Ans. 6_7 5 15. Change 343 to the form of a fraction. Ans. 34 3 16. Reduce 84 to fifteenths. Anls. 1 2A 6 QUFSTIONS. - Art. 136. What is the rule for reducing a mixed number to en improper fraction? Give the reason. How do you reduce a whole num ber to a fraction of the same value, having a given denominator?

Page 144 144 REDUCTION OF COMMON FRACTIONS. [SECT. XVII1. ART. 137. To reduce improper fractions to whole or mixea numbers. Ex. 1. How many dollars in 37 dollars? Ans. $2-5W. OPERATION. This question may be analyzed by saying, As 16 16.) 3 7 ( 2(-5 sixteenths make one dollar, there will be as many 3 2 dollars in 37 sixteenths of a dollar as 37 contains - times 16, which is 2j-5 times. Therefore $ 2-5_ is 5 the answer. RULE.- Divide the numerator by the denominator, and the quotient will be the whole or mixed number. EXAMPLES FOR PRACTICE. 2. Reduce 96_ to a whole number. Ans. 12. 3. Change l5-17v to a mixed number. Ans. 10 8. 4. Change -#1IY- to a mixed number. Ans. 10T 1. 5. Change 17 36 - to a mixed number. Ans. 1 g 75 6. Reduce -1o7o to a mixed number. Ans. 1426. 7. Reduce 3 7 to a whole number. Ans. 1. 8. Change'67 to a whole number. Ans. 567. 9. Reduce 774 to a mixed number. Ans. 9_a 10. Reduce 8s 45 8 to a mixed number. Ans. 4 45. Ar-T. 13 8. To reduce a compound fraction to a simple fraction. Ex. 1. Reduce 4 of -7 to a simple fraction. Ans. 2 OPERATION. To show the reason of the operation, this 4 X s-7- = s2 question may be analyzed by saying, that, if'T of an apple be divided into 5 equal parts, one of these parts is I; of an apple; and, if } of -A- be 5-1, it is evident that I of T7T will be 7 times as much. 7 times 1- is -75; and, if I of 71 be -4, 4 Of 7 will be 4 times as much. 4 times 7 are 2. Or, by multiplying the denominator of -T by 5, the denominator of 4, it is evident we obtain J of 1T = -75, since the parts into which the number or thing is divided are 5 times as many, and consequently only -1 as large as before. Again, since 5 QUESTIONS. - Art. 137. What is the rule for reducing improper fractions to whole or mixed numbers? Give a reason for the rule. -Art. 138. Hofw do you reduce a compound fraction to a simple one? Give the reason for the operation.

Page 145 SECT. XVIII.] REDUCTION OF COMMON FRACTIONS. 145 of 7 - F of 71 will be 4 times as much; and 4 tires 7 - 2s5 This process will be seen to be precisely like the operation. IEx. 2. Reduce 3 of 4 of A of 6 of -7 to a simple fraction. 5 7 9 11 Ans. 2 OPERATION BY CANCELLATION. 2 Since some of the numerators and > X > x 2 cdenominators to be multiplied together are alike, we may cancel f X $ X VI X 0 < X 1 11 these common factors, according to $ the principles of cancellation. Ur,E. M- Iultiply all the numerators toyether for a new numerator, and all the denominators for a new denominator. NOTE 1. - All whole and mixed numbers in the compound fraction must be reduced to improper fractions, before multiplying the numerators and denominators. NOTE 2. -When there are factors common to both numerator and denominator, they may be cancelled in the operation. EXAMPLES TFOR PRACTICE. 3. What is of of of? Ans. 8-,6 4. What is 7 of 19 of 7? Ans. 51. 5. What is 7 of 9 of 3 of $? Ans. 72. 6. Change - I of ~ of a of 5 of 7 to a simple fraction. A ns, 4 9A g. 7. Required the value of 3 of 4 of 1 of 2 of 5,. Ans. 3. 8. Reduce A of 8 of 9 of T of 3 to a simple fraction. Ans. T3. 9. Reduce - of 14 Of T Of 9 of.4 to a simple fraction. Ans. 2 6 5 5 10. Reduce 15 of - of -.7 to a simple fraction. Anso -. 11. Reduce s of 22 of, of 9A to a whole number. u-a 8 Ans. 3. 12. Reduce 1 of 3 of 4 of 8 of 1 -1 to a simple fraction. Ans. 1 QUEsTIONs. When there are common factors in the numerator and denominator, hoew may the operation be shortened? What is the rule? What must be done with all whole and mixed numbers in the compound fraction? fHow may the operation be shortened by cancelling? ]3

Page 146 146 REDUCTION OF COMMON FIACTIONS. [SECT. XVII1 A COMiMON DENOMINATOR. ART. 139, A common denominator of two or more fractions is a common multiple of their denominators. The least common denominator is the least common multiple. NoTE. - Fractions have a common denominator, when all their denominators are alike. ART. 1l40 To reduce fractions to a common denominator. Ex. 1. Reduce 3, 6, and 7, to a common denominator. Ans. 144 160 168 OPERATION. 3 X 6 X 8 1 4 4 new numerator for = 14 49 5 X 4 X 8 160 "6," " 5 _1g6 7X4X6=l1 6 8'c " 0 7 6 8 4 X 6 X 8 = 1 9 2 common denominator. We first multiply the numerator of 1 by the denominators 6 and 8, and obtain 144 for its numerator. We nest multiply the numerator of 5 by the denominators 4 and 8, and obtain 160 for its numerator; and then we multiply the numerator of j by the denominators 4 and 6, and obtain 168 for its numerator. Finally, we multiply all the denominators together for a commnon denomi?nator, and write it under the several numerators, as in the operation. By this process, since the numerator anld denominator of each fraction are multiplied by the samle numbers, only the -form of the fraction is changed, while the quotient arising from dividing the numerator by the denominator, or the value of the fraction (Art. 133), remlains the same. Therefore, MIultiplying the numerator and denominator of a fraction by the same number does not alter the value of the fraction. RULE. - Multiply each numerator by all the denominators except its own,for the new numerators; and all the denominators toyetherfor a common denominator. NOTE 1. -- COmpound fractions, if any, must first be reduced to simple ones, and whole or mixed numbers to improper fractions. NoTE 2. - Fractions may often be reduced to lower terms, without destroying their common denominator, by dividing all their numer'ators and den-ominators by a common divisor. QUESTIONS. -Art. 139. What is a common denominator of two or more fractions? What is the least common denominator? When have firactions a common denominator? I- Art. 140, How do you find a commnoa denominator of two or more fractions? Give the reason of the operation, What inference is drawn from it? What is the rule for finding a common denominator? How may fi:actions having a common denominator be reduced to lower terims?

Page 147 SFEc,. XVIII.] REDUCTION OF COMMON FRACTIONS. 147 EXAMPLES FOR PRACTICE. 2. Reduce L and 5 to common denominators. Ans. -8, 20 or 92, Io 2 4 YT Y 2' 3. Reduce 7 49 and 1 to a common denominator. Ans. 7a 72 45 4. Reduce -,, and 5, to a common denominator. Ans. 352 231 280 6T-1-5 UT~, 6B-T6'. 5. Reduce 8, 5 and a to a common denominator. Ans, 288 135 216 or 2 5 22 6. Reduce 6, d 7, and,- to a, common denominator. Ans. 160 as84 840 240 20 48 105 30 - -. 9 Gu I 9-6-u wD A6u, ~., or Ae, 120 T27U ART. 141, To reduce fractions to their least common denominator. Ex. 1. educe 2, and 17T, to the least common denomi. nator. OPERATION. 3 3 6 1 2 1 2 common denominator. 2 1 2 4 3 4 X 2 = 8 numerator for 2 --.. 6 2 X 5 = 10 numerator for 2 m 1o=. B 1 2 12 12 X 7 - 7 numerator for -7- = 7 c' 3 X 2 X 2 = 12, the least common denominator. HEaving first obtained a common multiple, or denominator of the given fractions, we take the part of it expressed by each of these fractions separately for their new numerators. Thus, to get a new numerator for 2, we take 23 of 12, the common denominator, by dividing it by 3, and multiplying the quotient 4 by 2. We proceed in this manner with each of the fractions, and write the numerators thus obtained over the common denominator. NOTE. - The change in the terms of the fractions, in reducing them to the least common denominator by this process, depends upon the same principle as explained in the preceding article. RULE. - 1. Find the least (common multiple of the denominators for the least common denominator. 2. Dizvide the least common denominator by the denominator of each of the given fractions, and multiply the quotients by their respective numerators, for the new numerators. NoTE. - Compound fractions must be reduced to simple ones, whole QUESTIONS. - Art. 141. How do you find the least common denominator of two or more fractions? Upon what principle does this process depend? What is the rule for reducing fractions to their least common denominator? What must be done with compound fractions, whole numbers, and mixed numbers?

Page 148 148 ADD1TION OF COMMON FRACTIONS. [SECT. XVTiI. and mixed numbers to improper fractions, and all to their lowest terms before finding the least common denominator EXAMPLES BORio PRACTICE. 2. Reduce, 4, and, to the least common denominator. Ans. 90 96 100 1 05 3. tReduce,4 24 and T2' to the least common denominator, 45 5, I n 14 Ant. I4ss 792 890 360 4. IReduce,,-9, and 7I, to the least common denominator. Ans. 35 36 3 0 e 5. Reduce,3 9 1 and 5, to the least common denominator. Ans. A 2 a8 Il 112 2-' 25' $-9 A 12 6. Reduce I 4 5R T5 and -15 to the least colmmon denomi24 4, 6, 8, 1, na7or. R e duc 12 18 20 45 2- 10 ~nator. Ant.8 ~ 2~24., 24', 24~, 24 24, 24' 7. Reduace 4,, 4, 6, and I, to the least common denominator. Ans. 6. 24 12 9 6 3 S. Reduce. 4, and - -, to the least common denominator. Ans. 3,6 20 1 36, 36, 36' 9. Reduce 74,- 56,- 7, and. 8, to the least common denominator. Ant. 344 244' 308 352 10. Reduce 4, 4, 5, 7, and 9, to the least common denominator. Ans,., 16 20 2V8 36 4' -' 4' 4t 4 ADDITION OF COMMON FRACTIONS. ART. 1 42 ADDITION of Fractions is the process of finding the value of two or more fractions in one sun. ART. 1430 To add fractions that have a common denominator. Ex. 1. Add, 2, 4, 4, and $, together. Ans. 24. OPERATION. These fractions all being 1 2 4 5 6 sevenths, that is, having 7 for'7- + - 7 + 7'+- 8 = 2~. a common denominator, we add their numerators together, and write their sum, 18, over the common denominator, 7. Thus we obtain L7s _ 24, the sum required. Hence, to add fractions having a common denominator, Write their sum over the common denominator, and reduce the fr'action, if necessary. QUESTIONS. - Art. 142. What is addition of fractions? - Art. 143. How wre fractions having a common denominator added? Give the reason.

Page 149 fScc-. XVIII.] ADDITION OF COMMON FRACTIONS. 149 EXAMPLES FOR PRACTICE. 2. Add 4 5 7 8 9, and 1o, together. Ans. 3 1 2 3. Add AI -o 8 - and {T, together. TT) I1T 3. Add 43, 3 8 9 and I, together. Ans. 2v.1 4. Add 23, 4 and 1s, together. Ans. 2.1 5. Add 17 iS 37 and 1l9, together. Ans. 21 —. 6. Add - 9_ Iz and 19, together. Ans. 1 1 2 7. Add 16T 1 35 and 1T9T together. Ans. 1 -6z. YT7 T, TTTT- TT 17T ART. 144, To add fractions that have not a common denominator. Ex. 1. What is the sum of a, a, and 97i? Ans. 1-19 OPERATION. 2 6 8 122 2 4 common denominator. 3 4 6 6 4 X 5 =20 2 1 4 2 8 3 X 3 9 new numerators 2 2 2 i 7 =L 4 ) ______ 12 2X7-14) 1 2.1 Sum of numerators, 4 3 2 X 3 X 2 >< 2 24. Com. denominator, 2 4 24 Having found the common denominator and new numerators, as in Art. 141, we add the numerators together, and write their sum over the common denominator, and reduce the fraction. RULE. - Reduce the given fractions to a common denominator. Add the numerators, and write their sum over the common denominator. NOTE 1. - Mixed numbers must be reduced to improper fractions, and compound fractions to simple fractions, and each fraction to its lowest terms, before attempting to find their common denominator. NOTE 2. - In adding mixed numbers, when deemed most convenient, the fractional parts may be added separately, and their sum added to the amount of the whole numbers. EXAMPLES FOR PRACTICE. 2. What is the sum of, and's? Ans. 2-}-. 3. What is the sum of 9s, ~A, and A? Ans. 1T5z 4. What is the sum of 12 and ~? Ans. 1l4-. 5. What is the sum of a, A, 3, and T12? Ans. 22o 6. Add t-, 8 }, and 2, together. Ans. 1 359 7. Add ~], 5,' and 7, together.. Ans. 1123.'', -4 9 6T-U 8. Add —, 4 9 74, and s8 1 together. Ans. 269 QUESTIONS. - Art. 144. What is the rule for adding fractions not having a common denominator? How may mixed numbers be conveniently added r? 13 23

Page 150 ]50 BUBTRACTION OF COLMMON FRACTIONiY. [EcTr. XVIn. 9. A~dd -,1 3 45 6, and 7, together. ns. 579 Ans. 65x4. 12. Add 1 of s to 1 of Ans. 1. 1 2. Add 1 o f 7 to I I of I Ans. 1Igte i3. Add 23 of a to 6 of 6 oAns, -19 15. Add 1of ~ of 2 to of of. Ans. 2n. 16. Add 3o to 414. Ans. 813 17. Add 4o to 5of Ans. 104. 18. Add 17a to 18-k. Ans. 36A. ARTS. 450 To add any two fractions whose numerators are a unit. Ex. 1. Add - to -. Ans. 92 OPERATMON. We first find the Sum of the denominators, 4 + 5= 9 product of the denomProduct of the denominator, 4 5 20 inators, which is 20, and then their sum, which is 9, and write the former for the denominator of the required fraction, and the latter for the numerator. The reason of this operation will be seen, when we consider that the process reduces the fractions to a common denominator, and then adds their numerators. HIence, to add two fractions whose numerators are a unit, simply W/Vrite the sum of the fiven denominators over their product. ExAMPLES FOR PRACTICE. 2. dAdd 4 to e T, to a, itoa, 5to, I to A 4. Add 2 to 4oto a, to -6 t o, 5. Add I to -1, t to 1 1 to i, t t o ~ L to, 4. Add 1 to 1 to Tl,2t I to I, #to A, to -, - to 6~ 7. Add I to 1AtO t tO u totO~ 6. Add ~to 1, 1 to &, 0 to, to, to to Iol SUBTRACTION OF COMMON FRACTIONS. ART. 146. SUBTRACTION of Fractions is the process of findIng the difference between two fractions. QUESTIoNS. - Art. 145. How can you add two fractions when the numerators are a unit? What is the reason for this? - Art. 146. What is subtraction of fractions?

Page 151 SECT. XVIII.) SUBTRACTION OF COMMON FRACTIONS. 151 ART. 1470 To subtract fractions that have a common denominator. Ex. 1. From 7 take 2 An,'. OPERATION. The fractions both being ninth7s, having 9 for a common denominator, we subtract the less numerator firom the greater, and write the difierence, 5, over the common clenomlinator, 9. Thus we have - as the difference reqtuired. Hence, to subtract fractions having a common denominator, Write the difference of their numerators over the common denomtinators, and reduce thefraction, if necessary. EXAMPPLES FOR PRACTICE. 2. From l take 2 Ans. 5 3. From I9I take -. Ans. T. 4. From 2- take 37 Ans. j. 5. From 1 7 take y9L. Ans. l.1 6. From M6%28 take 150. Ans. 2a 9 7. From 7a take 5 Ans.. 8. From i9~7 take j,. Ans. 45 aRT. 48. To subtract fractions that have not a common denominator. Ex. 1. From I take -17T Ans. 1iOPERATION. 4 1 6 12 4 8 common denominator. 4 3 16 3 3 = 39 new numerators. 12 4X 7=-28 4 X 4 X 3 48. 41 1 difference of numerators. 4 8 common denominator. Having found the common denominator and new numerators as in Art. 141, we subtract the less numerator from the greater, and place the difference over the common denominator. RULE. - Reduce thefractions to a common denominator, then zurite the difference of tho numerators over the common denominator. NoTE. -If the minuend or subtrahend, or both, are compound fractions, they must be reduced to simple ones. QUESTIONS. - Art. 147. How do you subtract fractions having a common denominator? - Art. 148. What is the rule for subtracting fractions not having a common denominator? If the minuend or subtrahend is a compound fractiln, what must be done?

Page 152 152,SUBTRACTION OF COMMON FRACTIONS. [SECT. XVIIL EXAMPLES FOR PRACTICE. 2. From -7A take 4. Ans. 2256. 3. From 19 take 11 Ans. 2. 27U i' 4. From 7 take -9.~ Ans. 43o. 5. From 11 take rf.. Ans. 1.9 6. From 3l take. Ans. 3 7. From 1- take AT. Ans. Of7f.f 8. From 111 take A1 Ans. 94.9 9. From 1 take 1 vAns. ~U TU T~~B' rT 10. From 2 of 9 take I of 2. Ans. 17-3 I1. From A of T9 take T of 44, Ans. T3FU. 12. From 3 of 124 take 2 of 9Tr 7 Ans. 4i. ART. IX499 To subtract a proper fraction or a mixed number from a whole number. Ex. 1. From 16 take 24. Ans. 133. OPERATION. Since we have no firaction fi'om which to subFrom 1 6 tract the J:, we must add 1, equal to 4, to the Take 2 minuend, and say 4 from 4 leaves 4.- We Rem. 1 38 write the 3 below the line, and carry 1 to the 2 in the subtrahend, and subtract as in subtrac. tion of simple numbers. The same result will be obtained, if we Subtract the numerator from the denominator of the fraction, and under the remainder write the denominator, and carry one to the subtrahend to be subtracted from the minuend. NoTE. - When the subtrahend is a mixed number, we may, if we choose, reduce it to an improper fraction, and change the whole number in the minuend to a fraction having the same denominator, and then proceed as in Art, 148. EXAMPLES FOR PRACTICE. 2. 3. 4. 5. 6. From 1 2 1 9 1 3 14 1 7 Take 44 3T 9T1 8 61 Ans. 74 1 5 310 54 1 0 7. From 23 take 13-. Ans. 92. 8. From 47 take 4a. Ans. 46W7. 9. From 139 take 7544. Ans. 6344. QUESTIONS. - Art. 149. How do you subtract a proper fraction or mixed number from a whole number? Give the reason for this rule.

Page 153 SECT. XVIIL. ] SUBTR, ACTION OF COMMON FRACTIONS. 153 ART. 150 f To subtract a mixed number from a mixeO number. Ex. 1. Frorm 92 take 38. Ans. 5z. FIRST OPERA TION. We first reduce the fractional parts to a From 92 = 9 ~ Tk 3 _ 31 common denominator by multiplying thle Take - 3 321 J __ ______ terms of the fraction 2 by 5, the denominator Remn. 452 of the other, thus:X 5 =10; and then the 3 X 7 21 terms of the fraction - by 7, the dlenomiator of th3e first, lthus cuX7al 2 Now, since we cannot take a2I from -o, we add 1 equal to -- to the 5o in the minuend, and obtain W%.Ve next subtract 23 fiom s5 and write the remainder, 2x, below the line, and carry 1 to the 3 in the subtrahend, and subtract as in simple numbers. SECOND OPERATION. rSECOND OP92TION2 In this operation, we reduce From 9 2 65 325 7Take 3 2 the mixed numbers to imaTake 3 — is = 5 u5 a 5- proper fractions, and these Rem.':]-9 -a 5 fractions to a common denominator, as in the first operation. We then subtract the less firaction from the greater, and, reducing the remainder to a mixed number, obtain 52.4, as before. Hence, in performing like examples, we may Reduce the fractional parts, if necessary, to a commnon denominator, and subtract the fractional part of the subtrahlend fiomn tlhat of the minuend, as in Art. 147; remembering to increase the fractional part of the mninuend, when otherw'zise it zwould be less than that of the subtrcahend, before subtracting, by as manycy factional units cas it tcakes to mcake a unit of the fraction (Art. 131), and carry 1 to the whole nunmber of the subtrahend before subtractiny it from the whole number of the nminuend. Or, Reduce the nzixed numnbers to improper fractions, then to a conmmzon denominzator, and subtract the less fraction firom the greater. EXAMPLES FOR PRACTICE. 2. 3. 4. 5. 6. From 97 71 83 9L 10 Take 512 37 44 3} 10 1 Ans. 323 317 322 5a 53 QUESTIONS. -Art. 150. How do you reduce the firactions of mixed numubers to a common denominator? How does it appear that this process Ireducles them to a common denominator? How do you then proceed? What other method of subtracting mixed numbers? How may all like examples be per formed?

Page 154 154 SUBTRACTION OF CO.MMION FRACTIONS. [SECT. XVII1 7. 8. 9. 10. 11. From 1 23 1 6 -3I 1 93 9 71 8 711~ Take 9} 56 1 56 1 8 3 1 93 Ans. 213 1 03 36 7 8443 6 7412. From 194 take 7-3 Ans. 1159 13. From 154 take 84. Aus. 625. 14. From 9-1- take 3T1. Ans, 5?23_. 15. From 71A- take 137. Ans. 5729 16. From 6111 take 3314 Ans. 27-.z17. From a hogshead of wine there leaked out 12-3 gallons; how much remained? Ans. 50) gallons. 18. From $10, $21 were given to Benjamin, $31 to Lydia $12 to Emily, and the remainder to Betsey; what did she receive? Ans. $34. ART. go To subtract one fraction from another, when both fractions have a unit for a numerator. Ex. 1. WVhat is the difference between I and? Ans. 2. OPERATION. Difference of the denominators, 7 - 3 - 4 Product of the denominators, 7 X 3 = 21 We first find the product of the denominators, which is 21, and then their difference, which is 4, and write the former for the denominator of the required fraction, and the latter for the numerator. By this process the fractions are reduced to a common denominator, and their difference found. Hence, to subtract such fractions, we may simply WVrite the difference of the denominators over their product. EXAMPLES FOR PRACTICE. 2. Take 4 from I, I from I, I from 4, 4 from 4. 3. Take 4 from 4, 4 from 4 4, from -4. 4. Take 4 from ~, 4 from 1, T1 from a, 1 from 1. 5. Take fi'om I, I from 4, from 4 1 from.6. Take from, 1 from 1 from -A 1 from 1 6. Takee i~o, 1 1P from 1, ~ — from 1 1~ fr'om ~. It 5 12 4 34 7. Take 4 friom 4,1 4 from 4, 4 from I ~. from 4. QUEsTIONs. — Art. 151. How do you subtract one fraction from another when both fractions have a unit for a numerator? What is the reason for this process?

Page 155 hbEeT. XVIII.] MULTIPLICATION OF COIMMON FRACTIONS. 155 MULTIPLICATION OF COMMON FRACTIONS. ART. i520 IMULTIPLICATION of Fractions is the process of taking one number as many times as there are units in another, when one or both of the numbers are fractions. ART. 1530 To multiply a fraction by a whole number. Ex. 1. Multiply 7 by 4. Ans. 32. FIRST OPERATION. In the first operation we multiply the - X 4' =S Z = 32 numerator of the fraction by the whole number, and obtain 31 for the answer. It is evident that the fraction 7 is multiplied by multiplying its numerator by 4, since the parts taken are 4 times as many as before, while the parts into which the number or thing is divided remain the same. Therefore, Multiplying the numerator of a fraction by any number multiplies thefraction by that number. SECOND OPERATION. In the second operation we divide the 7- XA< 4 = = 32- denominator of the fraction by the whole number, and obtain 31 for the answer, as before. It is evident, also, that the fraction z is multiplied by dividing its denominator by 4, since the parts into which the number or thing is divided are only i as many, and consequently 4 times as large, as before, while the parts taken remain the same. Therefore, Dividing the denominator of a fraction by any number multiplies thefraction by that number. RULE. -Jl Multiply the numerator of the fraction by the whole number. Or, Divide the denominator of the fraction by the whole number, when it can be done without a remainder. EXAxMPLES FOR PRACTICE. 2. Multiply ~ by 9. Ans. 63. 3. Multiply T-8 by 5. Ans. 22. 4. Multiply 16 by 3. Ans. 17. 5. Multiply 1 by 85. Ans. 49. QUESTIONS. - Art. 152. What is multiplication of fractions?- Art. 153. How is a fraction multiplied, by the first operation? Give the reason of the operation. What inference is drawn from it? How is a fraction multiplied, by the second operation? What is the reason of the operation? What inference is drawn from it? What is the rule for multiplying a fraction by a whole number?

Page 156 156 MUTJLTIPLICATION OF COMMON FRACTIONS. [SECT. XVIII 6. IMultiply!- by 83. Ans. 76- 7. Multiply $ 5 by 189. s. Ans. 1661 8. aiultiply L-j 3 by 365. Ans. 352- 6 o 9. Mlultiply s7 by 48. Ans. 431o. I10. If a man receive. of a dollar for one day's labor, what will he receive for 21 days' labor? Ans. 67-T. 11. What cost 561b. of chalk at 3 of a cent per lb.? Ans. $0.42. 12. What cost 3961b. of copperas at -x of a cent per lb.? Ans. $3.24. 13. What cost 79 bushels of salt at 7 of a dollar per bushel? Ans. t;69. ART. 541. To multiply a whole number by a fraction. Ex. 1. Multiply 15 by 3. Ans. 9. FIRST OPERATION. In the first operation we divide the 5 ) 1 5 whole numnber by the denominator of the 3 X 3-9 fraction, and obtain i of it. We then multiply this quotient by 3, the numerator of the fraction, and thus obtain 3 of it, which is 9. SECOND oPERATION. In the second operation we multiply the 1 5 whole number by the numerator of the frae3 tion, and divide the product by the denomni4 5 + 5= 9 nator, and obtain 9 for the answer, as before. Therefore, _Multiplyiny by a fraction is taking the part of the multiplicand denoted by the multiplier. RULE. - Divide the whole number by the denominator of the fraction, when it can be done without a remainder, and multiply the quotient by the numerator. Or, Ml;ultiply the whole number by the numerator of the fraction, and divide the product by the denomzinator. EXAMPLES FOR PRACTICE. 2. Multiply 36 b)y. Ans. 28,. 3. MIultiply 144 by I}. Ans. 88. 4. Miultiply 375 by -~-. Ans. 325. 5. Multiply 2277 by s0. Ans. 1610. 6. Multiply 376 by I. Ans. 243-5. QUESTIONS.- Art. 1514. Iow do you multiply a whole number by a fraction, according to the first operation? How by the second? What inference is drawn from the operation? What is the rule for multiplying a whole number by a fraction?

Page 157 SeCT. XVIII.] MULTIPLICATION OF COMMON FRACTIONS. 157 7. Multiply 471. by T1G Ans. 8a. 8. JMuitiply 871 by r.n Ans. 23 2. 9. Multiply 867 by Tr~R Ans. 6 -'5. ART. 155o To multiply a whole and mixed number together. Ex. 1. Multiply 17 by 64. Ans. 1143. OPERATION. 1 7'We first multiply 17 by 6, the whole 64 number of the multiplier, and then by the 1 0 2 fractional part, 4, which is simply taking aof 17 124 4 of it, and add the two products. 1144 Ex. 2. Multiply 73 by 4. -Ans. 305. OPERATION. We first multiply 3 in the multiplicand by 5 4, the multiplier; thus, 4 times 3 are 12 4 5 equal to 22, which is in effect taking 3 of the } of 4 - 2' multiplier, 4. We then multiply the whole 2 8 number by 4, and add the two products. 3 0 2 Hence, in performing like examples, Multiply the fr'actional part and the whole number separately, and add the products. EXAMPLES FOR PRACTICE. 3. Multiply 93 by 5. Ans. 46. 4. Multiply 123 by 7. Ans. 884. 5. Multiply 9 by 81"~ Ans. 804. 6. Multiply 10 by 71. Ans. 714. 7. Multiply 116 by S. Ans. 94$. 8. What cost 7-l11b. of beef at 5 cents per pound? Ans. $0.37-TL 9. What cost 23 —7-bbl. of flour at $6 per barrel? Ans. $1414. 10. What cost 8-3yd. of cloth at $5 per yard? Ans. $417. 11. What cost 9 barrels of vinegar at $64 per barrel? Ans. $574. QU.STIONS. - Art. 155. What is the rule for multiplying a whole and mixed number together? Does it make any difference which is taken for the multiplier? 14

Page 158 158 MULTIPLICATION OF COMMON FRACTIONS. [SECT. XVII 12. What cost 12 cords of wood at $6.371 per cord? Ans. $76.50. 13. What cost llcwt. of sugar at $93 per cwt.? Ans. $103-. 14. Wrhat cost 4- bushels of rye at $1.75 per bushel? Ans. $7.65$. 15. What cost 7 tons of hay at 117 per ton? Ans. $83s1. 16. What cost 9 doz. of adzes at $10, per doz.? Ans. $958. 17. What cost 5 tons of timber at $3L per ton? Ans. $158. 18. What cost 15cwt. of rice at 87.62~ per cwt.? Ans. $114.37. 19. What cost 40 tons of coal at $8.37- per ton? Ans. $335. ART. 1,15, To multiply a fraction by a fraction. Ex. 1. Multiply 7 by. Ansa. Ans. OPERATION BY CANCELLATION 7 3 7 OPERATION. - - = 3 9 4 X 12 3 To multiply - by 3 is to take 4 of the multiplicand, 7 (Art 154). Now, to obtain 4 of 7, we simply multiply the numerators together for a new numerator, and the denominators together for a new denominator (Art. 138). Therefore, Multiplying one fraction by another is the same as reducing compound fractions to simple ones. RULE. - lultiply the numerators together for a newl numerator, and the denominators together for a new denominator. NOTE. - We can cancel all the common factors in the numerators, and denominators, and then multiply the remaining factors together as before. EXAMPLES FOR PRACTICE. 2. Multiply 7 by 1T' Ans. T171 3. Multiply - by -. Ans, 4. QUESTIONS. - Art. 156. What is the first rule for multiplying one fraction by another? How does it appear that this operation multiplies the fraction of the multiplicand? What is the inference drawn from it? What is the note?

Page 159 SFCT. XVIII.] MULTIPLICATION OF COMMON FRACTIONS. 159 4. Multiply R8 by I. Ans.. 5. M3ultiply ~s by 1. Ans. I. lMultiply 15by G'7 Ans.' 7. IMultiply 1 by A. Ans. R-. 8. MIultiply 6 by 2. Ans. I. 9. What cost 7 of a bushel of corn at ~ of a dollar per bushel? Ans. 7 of a dollar. 10 If a man travels A of a mile in an hour, how far would he travel in of an hour? Ans.' of a mile. 32 4 11. If a bushel of corn will buy 7 of a bushel of salt, how much salt might be bought for 3 of a bushel of corn? Ans. 2 of a bushel. 12. If: of - of a dollar buy one bushel of corn, what will - 3 8 VI~l y 9~ UO LVNCILIV VLI rW rl of 9T of a bushel cost? Ans. 7 of a dollar. 13. If of 4 of -a- of an acre of land cost one dollar, how much may be bought with 2 of $18? Ans. 16 acres. AnT. 157, To multiply a mixed number by a mixed number, it is only necessary to reduce them to improper fractions, and then proceed as in the foregoing rule. Ex. 1. Multiply 4- by 62. Ans. 30-. OPERATION. 3..23 62 23 ~0 92 30 - =30~ $ 3 3 3 EXAMPLES roR PRACTICE. 2. Multiply 7j by 8.3 Ans. 60 3 3. Multiply 41 by 9}. Ans. 45 3 e~. LI~CI~rlY 19 4lrV4 I 4. Multiply 112 by 8-. Ans. 99. TJ 55 -5 5. Multiply 121 by 11g. Ans. 147-. 6. What cost 7 1 cords of wood at $5! per cord? Ans. $412 1 7. What cost 73yd. of cloth at $3- per yard? Ans. 25g. 8. What cost 63 gallons of molasses at 23~ cents per gallon? Ans. $1.521. 9. If a man travel 34 miles in one hour, how far will he travel in 97 hours? Ans. 3'-' QUESTION. - Art. 157. How do you multiply a mixed number by a mixed number?

Page 160 IbO DIVISION OF COMMON FRACTIONS. [SECT. XVIII. 10. What cost 361k 1 acres of land at 8253 per acre? Ans. $9167~-. 11. How many square rods of land in a garden, which is 97-5 rods long, and 4973 rods wide? Ans. 4810W- rods. DIVISION OF COMMON FRACTIONS. ART. 158. DIVISION of Fractions is the process of dividing when the divisor or dividend, or both, are fractions. ART. 059, To divide a fraction by a whole number. Ex. 1. Divide. by 4. Ans. -2. FIRnT OPERATION. In this operation we divide the numerator of the fraction by 4, and write the quotient, 2, over the de4 2 nominator. 9' 9 It is evident this process divides the fiaction by 4, since the number and size of the parts into which the whole number is divided remain the same, while only 1 of the number of parts is expressed by the fraction,. Therefore, Dividing the numerator of a fraction by any numnber divides the fraction by that number. Ex. 2. Divide 7 by 9. Ans. W. SECOND OPERATION. We multiply the denominator of the fraction by the divisor, 9, and write the product under the nu5 9 _5 merator. 7 63 It is evident this process divides the fraction, since multiplying the denominator by 9 makes the number of parts into which the whole number is divided 9 times as many as before, and consequently each part can have but -g of its former value. Now, if each part has but - of its former value, while only the same number of parts is expressed by the fraction, it is plain the fiaction has been divided by 9. Therefore, Multiplying the denominator of a fraction by any number divides the firaction by that number. RULE. - Divide the numerator of the fraction by the whole number, when it can be done without a remainder, and write the quotient over the denominator. Or, Multiply the denominator of the fraction by the whole number, and write the product under the numerator. QUEsTIONS. - Art. 158. W~hat is division of common fractions? - Art. 159. Hlow is the fraction divided by the first operation? What inference may be drawn from this operation? IIow is a fraction divided by the second operation? What inference is drawn from this operation? WThat is the rule?

Page 161 SECT. XVIII.] DIVISION OF COMMON FRACTIONS. 161 EXAMPLES FOR PRACTICE. 3. Divide 6- by 3. Ans. 12A. 4. Divide i by 6. Ans. -. 5. Divide -17 by 12. Ans. arm-. T6. Divide 44 by 8. Ans. r4. 6. Divide 4 by 8. Ans. -. ~7 Divide 27 by 9. Ans. 131. 8. Divide Ad by 15. Ans. 5%. 9. Divide 45-8 by 75 Ans.. 10. Divide 7. by 12. Ans. p TU.W 11. John Jones owns 4 of a share in a railroad valued at $117; this he bequeaths to his five children. What part of a share will each receive? Ans. 4. 12. Divide a by 15. Ans. 4 3. 13. Divide T6 by 28. Ans. 1s. 14. James Page's estate is valued at $10,000, and he has given 2 of it to the Seamen's Society; A of the remainder he gave to his good minister; and the remainder he divided equally among his 4 sons and 3 daughters. What sum will each of his children receive? Ans. $680 ".4 ART, 16go To divide a whole number by a fraction. Ex. 1. How many times will 13 contain #? Ans. 30~.. OPEnrATm~c. For convenience, we inE3 1 RT3 ><Of_ 9 1 vert the terms of the divi3 I - - X = - - 3 0. sor, and then multiply the 7 3 3 whole number by the original denominator, and divide the product by the numerator. The reason of this operation is evident, since 13 will contain 4! as many times as there are sevenths in 13, equal 91 sevenths. Now, if 13 contain 1 seventh 91 times, it will contain 4 as many times as 91 will contain 3, equal to 30k. RULE. — Multiply the whole number by the denominator of the.fraction, and divide the product by the numerator. EXAMPLES FOR PRACTICE. 2. Divide 18 by 47. Ans. 204. 3. Divide 27 by li. Ans. 29 T. 4. Divide 23 by.. Ans. 92. QUESTIONS. - Art. 160. What is the rule for dividing a whole number by a fraction? Give the reason for the rule. 14*

Page 162 162 DIVISION OF COMMON FRACTIONS. [SECT. XVIII 5. Divide 5 by j. Ans. 25. 6. Divide 12 by 3. Ans. 16. 7. Divide 16 by ~. Ans. 32. 8. Divide 100 by }17 Ans. A 1s. 11. 9. I have 50 square yards of cloth; how many yards, 3 of a. yard wide, will be sufficient to line it? Ans. 83~ yards. 10. A. Poor can walk 3 —7 miles in 60 minutes; Benjamin can walk g- as fast as Poor. How long will it take Benjamin to walk the same distance? Ans. 73~ minutes. ART. 1ER, To divide a mixed number by a whole number. Ex. 1. Divide 17" by 6. Ans. 2 4. OPERATION. Htaving divided the whole num6)174 ber as in simple division, we have 2, 53 -; 4 3 6-; 43 a remainder of 5{, which we reS~- -, 8 ~X6- 48 duce to an improper fraction, and 2_4 4 = 24 3 divide it by the divisor, as in 41 8'I 8 Art. 159. Annexing this fraction to the quotient 2, we obtain 24 - for the answer. Hence, to divide a mixed numlller by a whole number, Divide the integral part of the mixed number; and the remainder, reduced if necessary to a simple fraction, divide as in Art. 159. EXAMPLES FOR PRACTICE. 2. Divide 17-3 by 7. Ans. 2~. 3. Divide 18, by 8. Ans. 2 1. 4. Divide 27~- by 9. Ans. 3-1ry~. 5. Divide 311j by 11. Ans. 2A-19 6. Divide 784 by 12. Ans. 6,7. 7. Divide 1589- by 4. Ans. 47 1. 8. Divide 107Tl by 3. Ans. 35 2. 9. Divide $143 among 7 men. Ans. $2-;-. 10. Divide $106g among 8 boys. Ans.,13.2 11. What is the value of 2d of a dollar? Ans. $0.34'. 12. Divide $107-7 among 4 boys and 3 girls, and give the girls twice as much as the boys. Ans. Boy's share, $104:2; Girl's share, 212 — 13. If $14 will purchase 17 of a ton of copperas, what quan. tity will 81 purchase? Ans. 1 1-;cwt. QUESTION.,-Art. 161. How do you divide a mixed number by a whole number?

Page 163 SECT. XVIII.] DIVISION O.F COMMON FRACTIONS. 163 ART., 62 To divide a whole number by a mixed number, Ex. 1. Divide 25 by 4-a. Ans. 5 1OPERATION. We first reduce the divisor and dividend to fifths, 425 and then divide as in whole numbers, 5 5 The reason why the answer is a mixed number, 23)125(50 and not in fifths, is because the divisor and dividend )15 3 were both multiplied by the same number, 5, and therefore their relation to each other is the same as 10 before, and the quotient will not be altered. Hence. Reduce the divisor and dividend to the same parts as are denoted by the denominator of the fraction in the divisor, and then divide as in whole num, bers. EXAMPLES FOR PRACTICE. 2. Divide 36 by 97. Ans. 3 5 3. Divide 97 by 13 I. Ans. 616a. 12, u~vi~nv vo rJJ bVT%~ 167' 4. Divide 113 by 21. Ans. 5 -1541 5. Divide 342 by 14 -A47. Ans. 23-9-. 6. There is a board 19 feet in length, which I wish to saw into pieces 23 feet long; what will be the number of pieces, and how many feet will remain? Ans. 74i pieces. ARTr. t330 To divide a fraction by a fraction. Ex. 1. Divide 7 by A. Ans. 14. OPERATION. In this operation, we invert = 4-7 X 4 _ 6= 13 * the terms of the divisor, and then proceed as in Art. 156. The reason of this process will be seen, when we consider that the divisor, 4, is an expression denoting that 4 is to be divided by 9. Now, regarding 4 as a whole number, we divide the fraction 7 by it, by 7 7 multiplying the denominator; thus, X4= But the divisor, 4, is 9 times too great, since it was to be divided by 9, as seen in the original fraction; therefore the quotient, 7-, is 9 times too small, and must be multiplied by 9; thus, 72X 9_63=1 —i. By this operation, we have multiplied the denominator of the dividend by the numerator of the divisor, and the numerator of the dividend by the denominator of the divisor. Hence the QUESTIONS. — Art. 162. How do you divide a whole by a mixed number? How does it appear that this process does not alter the quotient? -Art. 163. How do you divide a fraction by a fi:raction? Give the reason why this process divides the fraction of the dividend.

Page 164 164 DIVISION OF COMMON FRACTIONS. [SECT. XVII1 RULE. - Invert the divisor, and then proceed as in multiplication oj fractions. NOTE 1. - Factors common to numerator and denominator should be cancelled. NOTE 2. - When the divisor and dividend have a common denominator their denominators cancel each other, and the division may be performed by simply dividing the numerator of the dividend by that of the divisor. EXAMPLES FOR PRACTICE. 2. Divide 7 by % Ans. 1~. 3. Divide 7 by I, Ans. 31. 8.. 2 4. Divide by z. Ans. 52 15 12' 5 5. Divide 2 by 3. Ans. 29. 6. Divide 9-, by 1o Ans. 6a-3. 7. Divide X by -1l. Ans. 425 8. Divide -93 by 3,. Ans. 6. 9. Divide!9 by 7-,. Ans. 27. 10. Divide 2 of 7 by 4 of 2. Ans. 188. 11. Divide 4 of 6 of of by 2 of of. Ans. 9T 12. Divide 3 of 7 of 4 by 2 of 7 of 2. Ans. 39. ART. 1i4. To divide a mixed number by a mixed number, it is only necessary to reduce them to improper fraktions, and pro oeed as in the foregoing rule. (Art. 163.) Ex. 1. Divide 7 by 3. Ans. 22OPERATION. 74 - -9-; 3T _ Ae }- X i, = 2'- = 2. EXAMPLES FOR PRACTICE. 2. Divide 73 -by 4'e Ans. 12. 3. Divide 3' by 71. Ans. As. 4. Divide 11i by 537 Ans. 2-aJ-,j. 5. Divide 4, b)y 17. Ans. 2 —552. 6. Divide 1i63 by 141. Ans. 82. 7. Divide 81' by 91 Ans. An. 8132 8. Divide 3 of 5'- of 7 by - of 38. Ans. 11l. QUESTIONS. - What is the rule for dividing one fraction by another? How may fractions be divided when they have a common denominator? Does this process differ in principle from the other? - Art. 164. How do you divide a mixed number by a mixed number?

Page 165 ScCT, XVII1.J COMPLEX FRACTIONS. 165 COMPLEX FRACTIONS. ART. E50o To reduce complex to simple fractions. Ex. 1. Reduce ~ to a simple fraction. Ans. d81. OPERATION. Since the numerator of a fraction is the a 1 X a = 8 dividend, and the denominator the divisor 3 9f 3 27- (Art. 132), it will be seen by this operation that we simply divide the numerator,', by the denominator, -, as in division offractions. (Art. 163.) 8 Ex. 2. Reduce to a simple fi-action. Ans. 1. OPERATION. MWe reduce the numerator, 8 __1 _ 8 X - 2 _ — 6= AI 8, and the denominator, 41- to 4 9 Xz - improper fractions, and then proceed as in Ex. 1. Ex. 3. Reduce I72 to a simple fraction. Ans. 12. 2 3 OPERATION. VWe here reduce the = - 3 X6 14 12 denominator, 2 of 2, to i of- 2 2 1 - a simple fraction, and then proceed as before. From the preceding illustrations we deduce the following RULE. - Reduce the terms of the complex fraction, if necessary, to theform of a simnple fractioLn. Then divide the numerator of the complex fraction by its denominator. EXAMPLES FO1t PRACTICE. 4. Reduce 12 to a whole number. Ans. 28. 3 5. Reduce -4 to a simple fraction. Ans. Ws. 47 6. Reduce 8 to a simple fraction. Ans. Is. QUESTIONS. — Art. 165. What is the rule for reducing complex to simple fractions? How does this process differ from division of fractions?

Page 166 166 COMPLEX FRACTIONSi. [SEcT. XVII.o 7. Reduce to a simple fraction. Ans. 9o5 S. Change 7-6 to a simple fraction. Ans. 0. 7a4 9. Change 24 to a mixed number. Ans. 21.7 10. Reduce 12 to a simple fraction. Ans. 1.5 11. If 7 is the denominator of the following fraction 14 what is its value when reduced to a simple fraction? - 8 Ans. 7J4-. 12. If 4 is the numerator of the following fraction, what is its value when reduced to a simple fraction? Ans. -. AIT. l1l&o Complex fiactions, after being reduced to simple ones, may be added, subtracted, multiplied, and divided, according to the respective rules for simple fractions. EXAMPLES FOR PRACTICE. 1. Acd ~ and 41 together. Ans. 773 7 2. Add 44 and _ together. Ans. 25f-g. 3. Fromll tale Ans. 2481 9 4k From Af ta 93e Ans. 82' $ ~ 8 by 4 of A~ns. Y 5. Multiply of 62 by 1of Ans T1 3' 6' 6. Mlultiply b5 by A, Ans. 11 p5-x. 7. Divide 7 of 124 by -3m of 83. lAns. 1031 11 QnErSTIoN. - Art. IGG. How do you add, subt.ract, multiply, and divide complex firactions?

Page 167 :EeCT. XVIII.J LEAST COMMON MULTIPLE OF FRACTIONS. 167 GREATEST COMMAION DIVISOR OF FRACTIONS. ART. 6$7, To find the greatest common divisor of two or. more fractions. Ex. 1. What is the greatest common divisor of a, a2 and -I? OPERATION. 4 2 12 -20 30 36 9) 3S 1 T - 41' 4 1' 4T5 Greatest common divisor of the numerators - 2 greatest common divisor Least common denominator of the fractions = 4 5 required. Having reduced the fractions to equivalent fractions with the least common denominator (Art. 141), we find the greatest common divisor of the numerators 20, 30, and 36, to be 2. (Art. 124.) Now, since the 20, 30, and 36, are forty-fifths, their greatest common divisor is not 2, a whole number, but so manyforty-fifths. Therefore we write the 2 over the common denominator 45, and have 2 as the answer. RtULE. -Reduce the fractions, if necessary, to the least cozmon Ilenominator. Thlen find the greatest common divisor of the numerttors, which, written over the least common denominator, will give the o'satest common divisor req/uired. EXAMPLES FOR PRACTICE. 2. What is the greatest common divisor of, 6 and 1-? Ans. 2d3.'What is the greatest common divisor of 1}, 7m8S and? Ans. 42yo -4. What is the greatest common divisor of 15, 21, 4, and 51? Ans. I 5. There is a three-sided lot, of which one side is 166)3ft., another side 1561ft., and the third side 208~ft. What must be the length of the longest rails that can be used in fencing it, allowing the end of each rail to lap by the other'ft., and all the panels to be of equal length? Ans. 101-ft. LEAST COMMON MULTIPLE OF FRACTIONS. ART. 168, To find the least common multiple of fractions. Ex. 1. What is the least common multiple of -2, 1, and 5? Ans. 2 10-.QUESTIONS. -Art. 167. What is the rule for finding the greatest common divisor of fractions? Why, in the operation, was the divisor 2 written over the denominator 45'?

Page 168 163 LEAST COMM1ION MULTIPLE OF FRACTIONS. [eCT. XVII1. OPERATION. 2 11 55 - 1 3 1I TT9 2X 4 ~ 6'Z Y5 mon multiGreatest common divisor of the denominator = 2 pie required. Having reduced the fractions to their lowest terms, we find the least common multiple of the numerators, 1, 3, and 21, to be 21. (Art. 128.) Now, since the 1, 3, and 21, are,'from the nature of a fractions, dividends of which their respective denominators, 6, 2, and 4, are the divisors (Art. 132), the least common multiple of the fractions is notl 21, a whole number, but so many fractional parts of the greatest common divisor of the denominators. This common divisor we find to be 2, which, written as the denominator of the 21, gives 2-1 = 10 as the least number that can be exactly divided by the given fractions. RULE. - Reduce the fractions, if necessary, to their lowest terms. Then find the least common multiple of the numerators, which, written over the greatest common divisor of the denominators, will give the least common multiple required. EXAMPLES FOR PRACTICE. 2. What is the least common multiple of Io,, and --? Ans. 4-. 3, What is the least number that can be exactly divided by r, 2~, 5, 6~, and ~l? Ans. 95. 4. What is the smallest sum of money for which I could purchase a number of bushels of oats, at $ 5 a bushel; a number of bushels of corn, at $-A a bushel; a number of bushels of rye, at $1 a bushel; or a number of bushels of wheat, at $2 a bushel; and how many bushels of each could I purchase for that sum? Ans. $22,; 72 bushels of oats; 36 bushels of corn; 15 bushels of rye; 10 bushels of wheat. 5. There is an island 10 miles in circuit, around which A can travel in - of a day, and B in 7 of a day. Supposing them each to start together from the same point to travel around it in the same direction, how long must they travel before comling together again at the place of departure, and how many miles will each have travelled? Ans. 5} days; A 70 miles; B 60 miles. QUESTIONS. -Art. 168. What is the rule for finding the least common multiple of fractions?'Why is not the least common multiple of the nuDmerators the least common multiple of the fractions?

Page 169 OECT. XVIII.] MISCELLANEOUS EXERCISES. 169 MISCELLANEOUS EXERCISES IN FRACTIONS. 1. What are the contents of a field 76-i rods in length, and 183 rods in breadth? Ans. 8A. 3R. 301p. 2. What are the contents of 10 boxes which are 7 feet long, 1a feet wide, and 1 feet in height? Ans. 16917 cubic feet. 3. From -iT of an acre of land there were sold 20 poles and 200 square feet. What quantity remained? Ans. 22075ft. 4. What cost j of an acre at $1.75 per square rod? Ans. $236.92x 4 5. What cost -39 of a ton at $15s per cwt.? Ans. $49.7313 6. What is the continued product of the following numbers: 142, 113, 54, and 104? Ans. 9184. 7. From - of a cwt. of sugar there was sold $ of it; what is the value of the remainder at $0.124 per pound? Ans. $3.1843. 8. What cost 197 barrels of flour at $7- per barrel? Ans. $1432. 9. Bought a piece of land that was 47,-ar rods in length, and 29-, in breadth; and from this land there were sold to Abijah Atwood 5 square rods, and to Hazen Webster a piece that was 5 rods square; how much remains unsold? Ans. 13668-3- square rods. 10O From a quarter of beef weighing 17531b. I gave John Snow 3 of it; 2 of the remainder I sold to John Cloon. What is the value of the remainder at 84 cents per pound? Ans. $2.04 13 11. Alexander Green bought of John Fortune a box of sugar containing 4751b. for $30. He sold A of it at 8 cents per pound, and 2 of the remainder at 10 cents per pound. WVhat is the value of what still remains at 121 cents per pound, and what does Green make on his bargain? Ans. Value of what remains, $13.199. Green's bargain, $16.979. 12. What cost -1 of an acre at $144 per acre? Ans. $2. 13. Multiply of by of by of 1 of 1-9 Ans. -l _9 I 1 1 4 19 25 1 0. 14. What are the contents of a board 114 inches long, and 41 inches wide? Ans. 494 square inches. 15

Page 170 I 7' 0 FRACTIONS OF COMPOUND NUMBERS. [Snce. XVII 15. Mary Brown had $17.87~; half of this sum was given to the missionary society, and i of the remainder she gave to the Bible society; what sum has she left? Ans. $3.567. 16. What number shall be taken from 12k, and the remainder multiplied by 104, that the product shall be 50? Ans. 8 13s 17. What number must be multiplied by 73, that the product may be 20? Ans. 2a-. 18. What are the contents of a box 8-2 feet long, 3,- feet wide, and 2 1{ feet high? Ans. 68l71#- feet. 19. On 2 of my field I plant corn; on 2 of the remainder I sow wheat; potatoes are planted on 3 of what still remains; and I have left two small pieces, one of which is 3 rods square, and the other contains 3 square rods. How large is my field? Ans. IA. OR. 29p. REDUCTION OF FRACTIONS OF COMPOUND NUMBERS. ART. 16 To reduce a fraction of a higher denomination to a fraction of a lower. Ex. 1. Reduce ~16D of a pound to the fraction of a farthing. Ans. 4fir. OPERATION. 1 X 20 20 20 X 12 240 240 X4_ 960 far. 4-far 2160 2160 2160 2160 2160 2160 -9 OPERAz TION BY CANCELLATION. Since 20s. make a pound, 1 X 7~ X 70 X 4 there must be 20 times as -- -far. many parts of a shilling as 9~{~0 parts of a pound; we therefore multiply -1-o by 20, and obtain 2o s.and obta; and since 12d. make a shilling, there will be 12 times as many parts of a penny as parts of a shilling; hence we multiply 2-o by 12,and obtain iA2o —d. Again; since 4far. make a penny, there will be 4 times as many parts of a farthing as parts of a penny; we therefore multiply jP%0% by 4, and obtain B96Ifar.afar., Ans. RULE. - Multiply the given fraction by the same numbers that woula be employed in reduction of whole numbers to the lower denomination required. QUrMsTIONs. -Art. 167. What is the rule for reducing a. fraction of a higher denomination to the fraction of a lower? Will you explain the opera. tions? Does this process differ in principle from reduction of whole compound numbers?

Page 171 SICT. XVIIlI. FRACTIONS OF COMPOUND NUMBERS. 171 EXAMPLES FOR PRIACTICE. 2. Reduce Tif of a pound to the fraction of a farthing. Ans. 2, 3. What part of a penny is -of a shilling? Ans. 4. What part of a grain is 7-u of a pound Troy? Ans. 2 5. What part of an ounce is - of a cwt.? Ans., 25 6. Reduce T1- of a furlong to the fraction of a foot. Ans. a2 7. What part of a square foot is g of an acre? Ans. ~. 8. What part of a second is ~ g of a day? Ans. R-T 9. What part of a peck is - of a bushel? Ans. 6. 10. What part of a pound is z I of a cwt.? Ans. ~. ART. 170@o To reduce a fraction of a lower denomination to a fraction of a higher. Ex. 1. Reduce 4 of a farthing to the fraction of a pound. Ans. 2y6. OPERATION. 4 4 4 4 4 4 1 >< )4= 3-6; 36 X12= 432; 432 X20 60 2160 OPERATION BY CANCELLATION. Since 4far. make a penny, tj 1 ~i X 12 X there will be d as many pence as B9 XA4 X 12 X 20 = 2160 farthings; therefore we divide the 4 by 4, and obtain -4d. And, since 12d. make a shilling, there will be jl2 as many shillings as pence; hence we divide 4 by 12, and obtain 4 2s. Again, since 20s. make a pound, there will be 2- as many pounds as shillings; therefore we divide;~2 by 20, and obtain 4 for the answer. RULE. - Dicide the given fraction by the same numbers that would be employed in reduction of whole numbers to the higher denomination?'erircd. EXAMPLES FOR PrACTICE. 2. Roduce 7 of a grain Troy to the fraction of a pound. Ans. I QUivSTiONS. — Art. 170. Do you multiply or divide to reduce a fraction of a lower dienominuation to the fraction of a higher? Whaet is the rule?

Page 172 172 FRACTIONS OF COMPOUND NUMBERS. [SECT. XVIII 3. What part of an ounce is -,- of a scruple? Ans. 1. 4. What part of a ton is 4 of an ounce? Ans. 5. What part of a mile is 8 of a rod? Arns..-5-a 6. What part of an acre is R of, a square foot? Ans. D 7. WVhat part of a day is 2 of a second? Ans.. 8. What part of 3 acres is 4 of a square foot? Ans. z 9. What part of 3hhd. is 4 of a quart? Ans. 1T3 10. What part of 6 of a solid foot is a cube whose sides are each ~ of a yard square? Ans. i. ART. 171, To find the value of a fraction in whole numbers of a lower denomination. Ex. 1. What is the value of Ad of 1~.? Ans. 7s. 9d. 1'far. OPERATION. 7 2 O 18)1 4 0(7s. 126 - -'4The reason of this operation will be seen, 1 4o if we analyze the question according to Art 12 169. Thus, T X 20 1_8_-Q-.- 71{s., 18)16 8 ( 9d. and 11 x 12 — Ld 9%'d.; and 1 6 2 x 4:62 1 6 2 16s X4 _~ - - 1~ far. Ans. 7s. 9cl. 1lfar. 6 1 8 ) 24 ('1far. 18 6 1 RULE. - Multiply the numerator of the given fraction by the number required to reduce it to the next lower denomination, and divide the product by the denominator. Then, if there is a remainder, proceed as before, until it is reducea to the denomination required. QUESTION. - Art. 171. What is the rule for finding the value of a fraction in whole numbers of a lower denomination?

Page 173 SECT. XVIII.] FRACTIONS OF COMPOUND NUMBERS. 173 EXAMPLES FOR PRACTICE. 2. What is the value of 7 of a cwt.? Ans. 3qr. 21b. l2oz. 79dr, 3. What is the value of I of a yard? Ans. 3qr. O0na 4. What is the value of 3 of an acre? Ans. IR. 28p. 155ft. 82jin. 5. What is the value of 29 of a mile? Ans. 1 fur. 31rd. Ift. 10in. 6. What is the value of 3 of an ell English? Ans. lqr. -na15. 7. What is the value of a of a hogshead of wine 7 Ans. 18gal. 8. What is the value of 7- of a year? Ans. 232da. 10h. 21m, 49-j1sec. ART. 172, To reduce a simple or compound number to the fractional part of any other simple or compound number of the same kind, Ex. 1. What part of.1. is 3s. 6d, 22far. Ans. A~. OPERATION. In performing this operation, 3s. 6d. 22far. = 512 we reduce the 3s. 6d. 22farl. to L. = 2880 c = t. thirds of far., the lowest denomination in the question, for the numerator of the required fraction, and 1~. to the same denomination for the denominator. We then reduce this fraction to its lowest terms, and obtain ~-x~,. for the answer. R-ULE. - Reduce the given numbers to the lowest denomination men tioned in either of them. Then, write the number which is to becomr the fractional part for the numerator, and the other number for the denominator, of the required fraction. EXAMPLES FOR PRACTICE. 2. Reduce 4s. 8d. to the fraction of 1~. Ans. J7,. 3. What part of a ton is 4cwt. 3qr. 121b.? Ans. -487,,. 4. What part of 2m. 3fur. 20rd. is 2fur. 30rd.? Ans. L,. 5. What part of 2A. 2R. 32p. is 31. 24p.? Ans. ~. 6. What part of a hogshead of wine is 18gal. 2qt.? Ans. T37 QUESTION. Art. 172. What is the rule for reducing a simple or co7ipound number to the fractional part of any other simple or compound number of the same kind? I 5"3

Page 174 17Z4 FRACTIONS OF COMPOUND NUMBERS. [SECT. XVIIL 7. What part of 30 days are 8 days 17h. 20m.? Ans. 5 7. 8. From a piece of cloth containing 13yd. Oqr. 2na. there were taken 5yd. 2qr. 2na, What part of the whole piece was taken? Ans. -. 9. What part of 3 yards square are 3 square yards? Ans. ~o ADDITION OF FRACTIONS OF COMPOUND NUMBERS. ART.!73. To add fractions of compound numbers. Ex. 1. Add 6 of a pound to > of a shilling. Ans. 17s. Ild. OA4Lfar. FIRST OPERATION.'We find the value of each fraes. d. far. Value of 63. S 1. 1 2r6 tion separately, and add the two Value of ~.i = 9 l values together, according to the Value of 7s. 9 1~ 3 rule for adding compound num17 11 0 4 bers. (Art. 101.) SECOND OPERATION. We first reduce 7 7 the fraction of a X2- o shilling to the + _~ 7,1.=2 - 17s. lid. 0 4 far fiaction of a +9@.=. A *-.12;60 @- 8~ * ad an pound, then add the two fractions together and find the value of their sum. (Art 171.) EXAMPLES FOR PRACTICE. 2. Add 14- of a pound to 5 of a shilling. Ans. 7S. lid. 32far. 3. Add together -1o of a ton, 7 of a ton, and 4 of a cwt. Ans. IT. 14cwt. lqr. 56I. 4. Add together 2 of a yard, l of a yard, -4T of a quarter. Ans. lyd. 2qr. 2na. O'in. 5. Add together A-T of a mile, 4 of a mile, -3T of a furlong, and 7T of a yard. Ans. 6fur. 29rd. 3yd. 1ft. 0lin. 6. Add together 9 of an acre, 4 of a rood, and 5 of a square rod. Ans. 1A. OR. 3p. 169ft. 1026in. 7. Sold 4 house-lots; the first7 of an acre, the second r of an acre, the third 2 of an acre, and the fourth 3 of an acre; what was the quantity of land in the four lots? Ans. 3R. 38p. 45':zsft. QUESTIONS. - Art. 173. What is the first method of adding fractions of compound numbers? What is the second?

Page 175 SECT. XVIII.] FRACTIONS OF COMPOUND NUMBERS. 175 SUBTRACTION OF FRACTIONS OF COMPOUND NUMBERS. ART. 1740 To subtract fractional parts of compound numbers. Ex. 1. From 6 of a pound take -4T of a pound. Ans. 9s. 10d. 11Tfar. FIRST OPERATION. We find the value of each v 1 d. far. fraction separately, and subValue of 6~. -— 1 7 1 26 Vaue of T. = 1T 2 tract one from the other, acValue of 41y.= 7 3 1 l cording to the rule for sub9 1 0 -1 9 tracting compound numbers. 90 (Art 102.) SEC'OND OPERATION. We first subtract the less fraction from $ -T- ~-~' - = the greater, and then find the value of 9s. od. 1.9far. their difference. (Art. 171.) EXAMPLES FOR PRACTICE. 2, From ~ of a ton take r6 of a cwt. Ans. l1ewt. Oqr. 76 7lb. 3. From > of a mile take 73 of a furlong. Ans. 5fur. 33rd. 5ft. 6in. 4. From 1o of an acre take 2 of a rood. Ans. 3R. 16p. 154ft. 5. From a hogshead of molasses containing 100 gallons, i of it leaked out; 2 of the remainder I kept for my family; what quantity remained for sale? Ans. 24gal. Oqt. I1~pt. 6. The distance from Boston to Worcester is about 41 miles. A sets out from Worcester, and travels -T of this distance towards Boston; B then starts from Boston to meet A, and, having travelled 4 of the remaining distance, it is required to find the distance between A and B. Ans. 12m. 6fur. 9rd. 5ft. 9-in. 7. A agrees to labor for B 365 days; but he was absent on account of sickness + part of the time; he was also obliged to be employed in his own business -T of the remaining time; required the time lost. Ans. 137da. llh. 13m. 1462sec. 8. From 11 acres, 33 poles, l01I feet of land, I sold 5 to A, ~ of the remainder to B, and fourhouse-lots, each 144 feet square, to C; what is the value of the remainder, at 8~ cents per square foot? Ans. $3937.89 7. QUESTIONS. - Art. 174. What is the first method of subtracting fractions of compound numbers? The second?

Page 176 l 76 QUESTIONS TO BE PERFORMED BY ANALYSIS. [SECT. XVII1. QUESTIONS TO BE PERFORMED BY ANALYSIS. 1. If one yard of cloth cost $4.40, what will - of a yard cost? ILLUSTRATION. - If 1 yard cost $4.40, - of a yard will cost - of $4.40, equal to $0.88; and! will cost 4 times $0.88, equal to $3.52, Ans. 2. If a barrel of flour cost $7.80, what will A- of a barrel cost? Ans. $2.34. 3. If a load of hay cost $17.84, what will X of a load cost? Ans. $15.61. 4. If $786.63 are paid for a cargo of wheat, what is the cost of 1{ of the cargo? Arns. $665.61. 5. What is I! of $87.50? Ans. $80.205. 6. What is j of 17~. 18s. 9d.? Ans. 13o. 9s. 03do 7. What is 4 of 3T. 16cwt. 3qr. 231b.? Ans. 2T. 3cwt. 3qr. 231b. 8. What is of 27A. 3R. 33p.? Ans. 12A. 1IR. 28p. 9. If $3.52 are paid for 4 of a yard of cloth, what is the price of 1 yard? Ans. $4.40. iLLUSTRAION. - If I Of a yard cost $3.52, I will cost a of $3.52, equal to $0.88; and A, or a whole yard, will cost 5 times $0.88, equal to $4.40, Ans. 10. If -% of a barrel of flour cost $2.34, what will be the cost of a whole barrel? Ans. $7.80. 11. When $15.57k are paid for 7 of a ton of hay, what will 1 ton cost? Ans. $17.80. 12. When Ad1 of a cargo of flour cost $665.50, what sum will pay for the whole cargo? Ans. $786.50. i-3. If $73.60- are paid for f- of a ton of potash, what sum must be paid for a ton? Ans. $80.30. 14. Bought 3 of a bale of broadcloth for 13~. 9s. 0-d.; what would have been the cost of the whole bale? Ans. 17~. 18s. 9d. 15. If ~-5 of an acre produce 18cwt. Oqr. 121b. of hay, what quantity will a whole acre produce? Ans. 77cwt. Oqr. Ilb. 16. Bought ~ of a lot of land containing 12A. 1R. 30p.; what were the contents of the whole lot? Ans. 27A. 3R. 39-p. 17. It'- of a ton of potash cost $80.20-6, what is the value of a. ton 7 Ans. 887.50.

Page 177 SECT. XVIII.] QUESTIONS TO BE PERFORMED BY ANALYSIS. 177 18. If -- of a cwt. of sugar cost $5.40, what is the value of 7 of a cwt.? ILLUSTRATION.- If a of a cwt. cost $5.40, 4 will cost I- of 85.40, equal to $1.80; and 4, or a cwt., will cost 4 times $1.80, equal to $7.20. Now, if lcwt. cost $7.20, A of a cwt. will cost - of $7.20, equal to $0.80; and 4 will cost 7 times $0.80, equal to $5.60, Ans. 19. If 7- of a pound of ipecacuanha cost $2.52, what is the value of 4 of a pound? Ans. $1.76. 20. When $80 are paid for 4 of an acre of land, what cost 7 of an acre? Ans. 893.334. 21. If -2a of a carding-mill are worth $631.89, what are - nf it worth? Ans. $401.20. 22. If 4 of a ship and cargo are valued at $141.52, what are 2 of them worth? Ans. $30.50. 23. If the value of 8 of a farm containing 178-4 acres is $1728, what is the price of 4 of the remainder? Ans. $2304. 24. E. Carter's garden is 17-T1 rods long, and 11P9~ rods wide. He disposes of 4 of it for $82.80; what is the value of a of the remainder? Ans. $41.40. 25. When 26~. 12s. 6d. are paid for - of a bale of cloth, what sum should be paid for 7 of the remainder? Ans. 18~. 12s. 9d. 26. If 7cwt.: of sugar cost $28.14, what will 95cwt. cost? ILLUSTRATION. -If 7CWt. cost $28.14, lcwt. will cost I of $28.14, equal to 84.02. In 9$cwt. there are t59cwt.; and if lcwt. cost $4.02; 4cwt. will cost 4 of $4.02, equal to $0.67, and -6- will cost 59 times $0.67, equal to $39.53, Ans. 27. If three tons of hay cost $49, what will 7-T- tons cost? Ans. $120.273. 28. Gave $78.8,0 for 11 tons of coal; what should I give for 34 tons? Ans. $24.674. 7 29. Paid 37~. 18s. 10d. for 3 bales of velvet; what was the cost of 53 bales? Ans. 67~. 19s. 6 d. 30. Gave $40 for 5 yards of broadcloth; what was the price of 194 yards? Ans. $156.574. 31. Paid $360 for 20 barrels of beer; what must be given for 43$- barrels? Ans. $789. 32. If 7 bushels of rye cost $8.75, what cost 18-71 bushels? Ans. $23.291A.

Page 178 178 QUESTIONS TO BE PERF1ORMED BY ANALYSIS. [SECT. XVII. 33. Paid $19.80 for 3 yards of broadcloth; what sum must oe given for 114 yards? Ans. $76. 37. 34. If 9-cwt. of sugar cost $39.53, what must be paid for 7cwt,? ILLUSTRATION. In 9-Gcwt. there are 5-9cwt. If -59-cwt. cost 6 6 6 $39.53, -cwt. will cost 1 of $39.53, equal to $0.67; and 6, or lcwt., will cost 6 times $0.67, equal to $4.02; and 7cwt. will cost 7 times $4.02, equal to $28.14, Ans. 35. When $18- are paid for 3cwt. of sugar, how much may be purchased for $1? How much for $78? Ans. 12%-~-Ocwt. 36. If 33 tons of potash cost $276.18, what will be the value of I ton? Of 75 tons? Ans. $6041.43 3. 37. If 7-1w acres of land cost $875, what will one acre cost? What will 75 acres cost? Ans. $8912.03~7.9 38. If 43 tons of coal cost $70, what will 1 ton cost? What will 86 tons cost? Ans. $1376. 39. For 273 acres of land there were paid $375; what cost L acre? What cost 69 acres? Ans $932.43, -. 40. If 4- tons of hay cost 80.50, what costs I ton? What cost 15 tons? Ans. 8262.50. 41. If 7-4cwt. of sugar cost $62.37, what will lcwt. cost? -What cost 19cwt.? Ans. $160.93. 42. If 7T yards of cloth cost $13.95, what will be the value of 11A yards? ILLUSTRATIoN. - In 73 yards there are -31 of a yard. If 34 of a yard cost $13.95, 1 will cost 3-L of $13.95, equal to $0.45; and 4, or I yard will cost 4 times $0.45, equal to $1.80. In 11l yards there are 3a of a yard. If 1 yard cost $1.80, 1 of a yard will cost V of $1.80 equal to $0.20, and -1_3 will cost 103 times;0.20, equal to $2.060, Ans. 43. When $668.50 are paid for 17'T- acres, what would be the value of 894 acres? Ans. $3457.30. 44. If $1738 are given for ~193 tons of iron, what will be the cost of' 37-1- tons? Ans. 33288. 45. Paid $11 for 1128 feet of boards; how nmany could I have purchased fbr $119-? - Ans. 11480 feet. 46. For 3 tons of potash I received 116cwt. of sugar; required the quantity of sugar that may be received for 11a tons of potash. Ans. 376cwt. 47. For 11a tons of potash I received 376cwt. of sugar

Page 179 6ECT. XVIII.] MISCELLANEOUS QUESTIONS BY ANALYSIS. 179 required the quantity of sugar that should be received for 38 tons. Ans. 116cwt. 48. When $S are paid for 1' yards of broadcloth, how much must be given for 33 yards? Ans. $49. 49. Gave $414 for 20A7x acres of land; what shall be given for 11! acres? Ans. $236. MISCELLANEOUS QUESTIONS BY ANALYSIS. 1. Sold a small farm for $896.50; what was received for ~ — of it? For 7 of it? For 1 of it? Ans. S15. 2. Gave $17A-r fbr 3 barrels of flour; what cost 1 barrel?'What 37 barrels? Ans. $213.0,3.. 3. Sold a house for $3687; what sum was received for $ of it? Ans. $3226.12~. 4. Bought 17z tons of hay for $1878; what is the cost of $ of a ton? Ans. ~7.61 2-577. 5. Bought a hogshead of molasses for $138; what cost I of it? What cost -7 What cost i1? Ans. $30.522 6. When $31-7 are paid for 100 gallons of molasses, what cost 4 of a gallon? Ans. $0.212r. 7. When 12 cents are paid for -A of a gallon of molasses, what will 48 -7 gallons cost? Ans. $16.01$. 8. If. of a barrel of flour cost $32, what will 63 barrels cost Ans. $4811. 9. When $236 are paid for 114} acres, what will be paid for 20T7, acres? Ans. $414. 10. Paid in Liverpool 974~. for 3 bales of cloth; how many bales should be received for 10732~.? Ans. 33 bales. 11. If 63 barrels of flour cost $48, 1, what will 4- of a barrel cost? Ans. $3.287o. 12. If 3. pounds of coffee cost 34 cents, what sum must be paid for 741 pounds? Ans. $6.90 9 13. If 27 tons of hay cost $63, what will be the cost of 16$4 tons? Ans. $381 -. 14. If a piece of land 3 rods square cost $17 41, what will be the cost of 4 square rods? Ans. $771. 15. Paid $314 for 2-cwt. of iron; required the sum to be paid for 689A4cwt. Ans. $76806. 16. For 62, cords of wood J. Holt paid $63; what sum must be paid for 18 cords? Ans. $170.10. 17. Gave $243T-1 for 96 barrels of tar; what quantity could be purchased for $1000? Ans 394ft 2 barrels.

Page 180 180 MISCELLANEOUS QUESTIONS BY ANALYSIS. [SECT. X ITII. 18. Paid $7888.30 for 83 9- acres of wild land; what sum did I pay for each acre, and what would be the cost of 7 acres? Ans. $660.80. 19. Gave 132k. 12s. for 75 tons of starch; what cost 127 tons? Ans. 224~. 5s. 20. For 172 days' work I paid $25.44; what should be paid for 89l days' labor? Ans. $128.64. 21. Sold 7-17T bushels of apples for $7.28; what should I receive for 19j 1 bushels? Ans. $19.12. 22. Paid $4355.52 for 496 pieces of carpeting; what did 37 pieces cost? Ans. $3294.72. 23. If a of 3 of the cost of the Capitol at Washington was $300,000, what was the whole cost? Ans. $2,000,000. 24. Purchased 7 6 thousand of boards for $135.80; what must be paid for 194 thousand? Ans. $359.45. 25. My wood-pile contains 6 cords and 76 cubic feet. If I dispose of 3 of it, what is the value of the remainder at 44 cents per cubic foot? Ans. $23.14 a. 26. I have a field 30 rods square, and having sold 18 square rods to S. Brown, and 82 square rods to J. Smith, what part of the field remained unsold? Ans. 9. 27. Bought 7T. 12cwt. 3qr. 181b. of iron, and having sold 3T. 18cwt. lqr. 201b., what is the value of - of the remainder at 53 cents per lb.? Ans. $242.591. 28. Bought 37 tons of iron at $68.50 per ton, for 4 of which I paid in coffee at $8.50 per cwt., and for the remainder I paid cash. Required the amount of cash paid, and also the value of -the coffee. Ans. Cash, $633.621; Value of the coffee, $1900.87~. 29. A man, having received a legacy of $7896, spent 3 of it in speculations, and the remainder he put in the savings bank, where it continued 15 years. It was then found that the sum deposited had doubled. Required the sum in the bank. Ans. $3948. 30. Bought a piece of broadcloth for $88, and sold k of it to J. Smith, and -1 of the remainder to O. Lake; what is the value of the part unsold? Ans. $37.49 19 31. A gentleman gave - of his estate to his wife, 3 of the remainder to his oldest son, and 3 of what then remained to his daughter, who received $750; required the whole estate. Ans. $12,000. 32, From an acre of land I sold two house-lots, each 100 feet square; what is the value of the remainder, at 8 cents per square foot? Ans. $1884.80.

Page 181 SECT. XIX.] DECIMAL PRACTIONS. 181 ~ XIX. DECIMAL FRACTIONS. ART. 175, A DECIMAL FRACTION is a fraction whose denominator is ten, or the product of a number of tens. Decimal fractions are commonly expressed by writing the numerator only, with a point (.) before it, called the decimal voint or separatrix; thus,.-0 is written.9. 99 " 99 96.99. 999 9 iJ9a "c.999. By examining the foregoing fractions, it will be seen that 9-~1 =.9 can occupy only one place while it remains a proper fraction; 9 -9 --.99, only two places; and 99- 9=.999, only three places; for, if their numerators are increased by l --.1, =ul- —.01, = rI-u =.001, respectively, each fraction becomes a unit or whole number. Hence, The first figure or place of any decimal on the right of the point is tenths, the second hundredths, the third thousandths, 4.c. NOTE. - When a decimal place has no significant figure it must be filled with a cipher. ART. 176, The denominator of -9.9 is 1 with one cipher annexed; the denominator of -9 -.99 is 1 with two ciphers annexed; the denominator of -9-O9 =.999 is 1 with three ciphers annexed. Hence, The denominator of a decimal fraction is I with as many ciphers annexed as the numerator has places. AIT. 1.77 Decimal fractions originate from dividing the unit, first, into 10 equal parts, and then each of these parts into 10 other equal parts, and so on indefinitely. Thus, 1 -- 10 =.-; 1 10 r~ I1V TU T-9 To T~i01; T01-U 10 -- TG'U U'001. Hence, The unit in decimal fractions is divided into 10, 100, 1000,'c., equal parts. QuEsTIONs. - Art. 175. What is a decimal fraction? How are decimal fractions commonly expressed? What is the first figure or place of any decimal? The second? Thethird? &c. Why? What must be done when a decimal place has no significant figure to fill it?-Art. 176. What is the denominator of a decimal fraction?- Art. 177. How do decimal fractions originate? 16

Page 182 182 NUMERATION OF DECIMAL FRACTIONS. [SECT. XIX. ART. 178. If ciphers are placed on the left hand of decimal figures, between them and the decimal point, those figules change their places, each cipher removing them one place to the right; thus, 3. 3, but.03 —- 3, and.003 =- 3 Hence, Every cipher placed on the left of decimal fyures, between them and t/he decimal point, decreases t/he value represented by them the same as dividing by ten. ART. 179, If ciphers are placed on the right hand of decimal figures, their places are not changedl; thus,. 3 — =, and.30 = T-3 --.3. Hence, Ciphers placed on the right hand of decimals do not alter the value represented by them. NUMERATION OF DECIMAL FRACTIONS. ART. 180. The relation of decimals to whole numbers and to each other, and also the names of their different orders and places, may be learned from the following TABLE. 7 B 5 4 3 2 1 * 2 3 4 5 6 7 8 9 cA A 0 a a v24 E-4 E- Ca.8 E v 7 6 5 4 12 4 56 7 89 c3S CS C S Ca c S S a S S a C Ca P4 64 6 64 4 64 64 6 6 64 64 64 64 00 0 0 0 0 I- ro.o / ~ ~. ~ cc ~4 ca' - W Occ Whole Numbers. Decimals. QUESTIONS.- Art. 178. What effect have ciphers placed at the left hand of decimals? Why? - Art. 179. What effect if placed at the right hand? Why? - Art. 180. What may be learned from the table?

Page 183 'SECT. XIX.] NOTATION OF DECIMAL FRACTIONS. 183 The preceding table consists of a whole number and decimal, which, taken together, are called a mixed number. The part on the left of the separatrix is the whole number, and that on the right the decimal. The decimal part is numerated from the left to the right, and its value is expressed in words thus' Two hundred thirty-four millions five hundred sixty-seven thousand eight hundred ninety-three billionths. And the. mixed number thus: Seven millions six hundred fifty-four thousand three hundred twenty-one, and two hundred thirty-four millions five hundred sixty-seven thousand eight hundred ninety-three billionths. From the table and explanations we have the following rule for reading decimals: RULE.- Read the decimal as though it were a whole number, adding the name of the right-hand order. The pupil may write in words, or read orally, the following numbers: 1..5 5..3001 9..72859 2..42 6..0984 10. 12.02003 3..01 7..00013 11. 121.000386 4..908 8..82007 12. 2.3058217 NOTATION OF DECIMIAL FRACTIONS. ART. 181. By examining the decimal table we perceive that tenths occupy the first place, hundredths the second, &e., and that each figure takes its value by its distance from the place of units; therefore, to write decimals, we have the following RULE. - Write the decimal as though it were a whole number, supplying with ciphers such places as have no significant figures. The pupil may write in figures the following numbers: 1. Three hundred seven, and twenty-five hundredths. 2. Forty-seven,and seven tenths. QUESTIONS. - Of what does it consist? What is the number called, when taken together? What is the part on the left hand of the separatrix? The part on the right? What is the value of the decimal? What is the value of the mixed number? What is the rule for reading decimals? —Art. 181. Upon, what does the value of a decimal figure depend? What is the rule for writing decimals?

Page 184 184 ADDITION OF DECIMALS. [SECT. XIX 3. Eighteen, and five hundredths. 4. Twenty-nine,and three thousandths 5. Forty-nine ten thousandths. 6. Eight, and eight millionths. 7. Seventy-five,and nine tenths. 8. Two thousand, and two thousandths. 9. Eighteen, and eighteen thousandths. 10. Five hundred five, and one thousand six millionths. 11. Three hundred, and forty-two ten millionths. 12. Twenty-five hundred,and thirty-seven billionths. ART. 182~, It will be seen that decimals increase from right to left, and decrease from left to right, by the same law as do simple whole numbers; hence they may be added, subtracted, multiplied, and divided, in the same manner. ADDITION OF DECIMALS. ART. 183. Ex. 1. Add together 5.018, 171.16, 88.133, 1113.6,.00456, and 14.178. Ans. 1392.09356. OPERATION. 5.0 1 8 1 7 1.1 6 We write the numbers so that figures of 8 8.1 3 3 the same decimal place shall stand in the same 1 1 306 column, and then, beginning at the right.0 0 4 5 6 hand, add them as whole numbers, and place the decimal point in the result directly under 1 4.1 7 8 those above. 1 392.09356 RULE. - Write the numbers so that figures of the same decimal place shall stand in the same column. Add as in whole numbers, and point off, in the sum,from the right hand as many places for decimals as equal the greatest number of decimal places in any of the numbers added. Proof. - The proof is the same as in addition of simple numbers. EXAMPLES FOR PRACTICE. 2. Add together 171.61111, 16.7101,.00007, 71.0006, and 1.167895. Ans. 260.489775. 3. Add together.16711, 1.766, 76111.1, 167.1,.000007, and 1476.1. Ans. 77756.233117, QUESTIONS.-Art. 182. How do decimals increase and decrease? How may they be added, subtracted, multiplied, and divided? - Art. 183. How are decimals arranged for addition? What is the rule for addition of decimals? What is the proof?

Page 185 SECT. XIX.] SUBTRACTION OF DECIMALS. 185 4. Add together 151.01, 611111.01, 16.5, 6.7, 46.1, and.67896. Ans. 611331.99896. 5. Add fifty-six thousand, andi fourteen thousandths; nineteen, and nineteen hundredths; fifty-seven, Aand forty-eight ten thousandths; twenty-three thousand five, and four tenths; and fourteen millionths. Ans. 79081.608814. 6. What is the sum of forty-nine, and one hundred and five ten thousandths; eighty-nine, and one hundred seven thousandths; one hundred twenty-seven millionths; forty-eight ten thousandths? Ans. 138.122427. 7. What is the sum of three, and eighteen ten thousandths; one thousand five, and twenty-three thousand forty-three millionths; eighty-seven, and one hundred seven thousandths; forty-nine ten thousandths; forty-seven thousand, and three hundred nine hundred thousandths? Ans. 48095.139833. SUBTRACTION OF DECIMALS. ART. 184, Ex. 1. From 74.806 take 49.054. Ans. 25.752, OPERATION. Having written the less number under the greater, so 7 4.8 0 6 that figures of the same decimal place stand in the same 4 9.0 5 4 column, we subtract as in whole numbers, and place the decimal point in the result, as in addition of decimals. 2 5.7 5 2 RULE. - Write the less number under the greater, so that figures of the same decimal place shall stand in the same column. Subtract as in whole numbers, and point off the remainder as in addition of decimals. Proof. -The proof is the same as in subtraction of simple numbers. EXAMPLES FORP PRACTICE. 2. 3. 4. 5. 11.0 7 8 4 7.1 17 4 6.13 8 7.10 7 9.81 8.78195 7.89 15 1.1 1986 1.2 6 8 3 8.3 3 5 0 5 3 8.2 3 85 8 5.9 8 7 1 6. From 81.35 take 11.678956. Ans. 69.671044. 7.- From 1 take.876543. Ans..12345'?. 8. From 100 take 99.111176. Ans..888824. 9. From 87.1 take 5.6789. Ans. 81.4211. QUESTIONS. —Art. 184. What is the rule for subtraction of decimals? What is the proof? 16*

Page 186 186 MULTIPLICATION OF DECIMALS. [SECT. XIX 10. From 100 take.001. Ans. 99.999. 1. From seventy-three, take seventy-three thousandths. Ans. 72.927. 12. From three hundred sixty-five take forty-seven ten thousandths. Ans. 364.9953. 13. From three hundred fifty-seven thousand take twentyeight, and four thousand nine ten millionths. Ans. 356971.9995991. 14. From.875 take.4. Ans..475 15. From.3125 take.125. Ans..1875 16. From.95 take.44. Ans..51. 17. From 3.7 take 1.8. Ans. 1.9. 18. From 8.125 take 2.6875. Ans. 5.4375. 19. From 9.375 take 1.5. Ans. 7.875. 20. From.666 take.041. Ans..625. MULTIPLICATION OF DECIMALS. ART. 185. Ex. 1. Multiply 18.72 by 7.1. Ans. 132.912. OPERATION. We multiply as in whole numbers, and point off 1 8.7 2 on the right of the product as many figures for deci7.1 mals as there are decimal figures in the multiplicand and multiplier counted together. 1 8 7 2 The reason for pointing off decimals in the product 1 3 1 0 4 as above will be seen, if we convert the multiplicand and multiplier into common fractions, and multiply 1 3 2.9 I 2 18 7 2 2_ 78 them together. Thus, 18.72=- 18- = 1J-7S2; and 7.1 7 =. Then - X 1 - 132-&ou =- 132.912, Ans., the same as in the operation. Ex. 2. Multiply 5.12 by.012. OPERATION. Since the number of figures in the product 5.1 2 is not equal to the number of decimals in the.0 1 2 multiplicand and multiplier, we supply the deficiency by placing a cipher on the left hand. 1 0 2 4 The reason of this process will appear, if we 5 1 2 perform the question thus: 5.12 - 51'2 =.0 6 1 4 4 Ans. Tui and.012 = 1T2T. Then X 1 2 T6544Ts-=.06144, Ans., the same as before. Hence we deduce the following QUESTIONS. - Art. 185. In multiplication of decimals how do you point off the product? Will you give the reason for it? When the number of figures in the product is not equal to the number of decimals in the multiplicand and multiplier, what must be done?

Page 187 SECT. XIX.] MULTIPLICATION OF DECIMALS. 187 RULE. —zlT ltiply as in whole numbers, and point off as many.figures /br deci bcals, in t/he product, as there are decimals in the mzultiplicand and 2lmultiplier. J.f there be not so mazy figures in the porlduct as there are decimal places ins the multiplica1d anzd mtdltiplier, slq))ply the deficiency by pr1efixing ciphers. NOTE. - When a decimal number is to be multiplied by 10, 100, 1000, &c., remove the decirmal point as many places to the right as there are ciphers in the multiplier; and if there be not figures enough in the number, annex ciphers. Thus, 1.25 X 10 = 12.5; and 1.7 X 100 = 170. Proof. - The proof is the same as in multiplication of simple numbers. EXAMPLES FOR PRACTICE. 3. Multiply 18.07 by.007. Ans..12649. 4. Multiply 18.46 by 1.007. Ans. 18.58922. 5. Multiply.00076 by.0015. Ans..00000114. 6. Multiply 11.37 by 100. Ans. 1137. 7. Multiply 47.01 by.047. Ans. 2.20947. 8. Multiply.0701 by.0067. Ans..00046967. 9. Multiply 47 by.47. Ans. 22.09. 10. Multiply eighty-seven thousandths by fifteen millionths. Ans..000001305. 11. Multiply one hundred seven thousand, and fifteen ten thousandths by one hundred seven ter thousandths. Ans. 1144.90001605. 12. Multiply ninety-seven ten thousandths by four hundred, and sixty-seven hundredths. Ans. 3.886499. 13. Multiply ninety-six thousandths by ninety-six hundred thousandths. Ans..00009216. 14. Multiply one million by one millionth. Anis. 1. 15. Multiply one hundred by fourteen ten thousandths. Ans..14. 16. Multiply one hundred one thousandths by ten thousand one hundred one hundred thousandths. Ans..01020201. 17, Multiply one thousand fifty, and seven ten thousandths by three hundred five hundred thousandths. Ans. 3.202502135. 18. 3Multiply two million by seven tenths. Ans. 1400000. QIJESTIONS. - What is the rule for multiplication of decimals? What is the proof? How do you multiply a decimal by 10, 100, 1000, &c.?

Page 188 188 DIVISION OF DECIMALS. [SECT. XIX. 19. TIhultiply four hundred, and four thousandths by thirty, and three htundredths. Arns 12012.12012. 20. What cost 461b. of tea at $1.125 per pound? Ans. 851.75. 21. What cost 17.125 tons of hay at $18.875 per ton? Ans. $323.234375. 22. What cost 181b. of sugar at $0.125 per pound? Ans. $2.25. 23. What cost 375.25bu. of salt at $0.62 per bushel? Ans. $232.655. DIVISION OF DECIMALS. ART. 8I[o Ex. 1. Divide 45.625 by 12.5. Ans. 3.65. OPERATION. We divide as in whole numbers, and 1 2.5 ) 4 5.6 2 5 (3.6 5 since the divisor and quotient are the 3 7 5 two factors, which, being multiplied together, produce the dividend, we 8 1 2 point off two decimal figures in the 7 5 0 quotient, to make the number in the two factors equal to the product or divi62 5 dend. 6 2 5 The reason for pointing off will also be seen by performing the question with the decimals in the form of common fractions. Thus, t5.625 -- 45 ss 5 -- 41_520525 and 12.5 =12-5t 125T-25 Then IT6u~25 - 12 5 oyU0__ 5iyrj, -.U9 - 5 X 10 = 46250 5 = 3-= = 3.65 Ans., as before. Ex. 2. Divide.175 by 2.5. Ans..07. OPERATION. We divide as in whole numbers, and since 2.5 ).1 7 5 (.0 7 we have but one figure in the quotient, we 1 7 5 place a cipher before it, which removes it to the place of hundredths, and thus makes the decimal places in the divisor and quotient equal to those of the dividend. The reason for prefixing the cipher will appear more obvious by solving the question with the decimals in the form of common firactions. Thus,.175 _ T175 and 2.5-2 = 5 25 Then lUO 7 5 — + T — YU TU 1 9 T ahTT TI' - -f >< X o Y-7s - 150 - T =.07, Ans., as before. Hence the following QUESTIONS. - Art. 186. In division of decimals how do you point off the quotient? What is the reason for it? If the decimal places of the divisor and quotient are not equal to the dividend, what must be done?

Page 189 SECT. XIX.] DIVISION OF DECIMALS. 189 RULE. - Divide as in whole numbers, and point off as many decimals in the quotient as the decimals in the dividend exceed those of the divisor; but if there are not as nmany, supply the deficiency by prefixing ciphers. NOTE 1. - When the decimal places in the divisor exceed those in the dividend, make them equll. by annexing ciphers to the dividend, and the quotient will be a whole number. NoTE 2. — When there is a remainder after dividing the dividend, ciphers may be arnnexed, and the division continued, the ciphers thus annexed being regarded as deoimals of the dividend; to indicate in atny case that the division does not terIuinate, the sign plus (+-) can be used. NOTE 3. - When a decimal number is to be divided by 10, 100, 1000, &c., remove the decimal point as many places to the left as there are ciphers in the divisor, and if there be not figures enough in the number, prefix ciphers. Thus 1.25 10.125; and 1.7 - 100 =.017. Proof. -The proof is the same as in division of simple nlumbers. EXAiIPiES FOR PRACTICE, 3. Divide 183.375 by 489. Ans..375. 4. Divide 67.8632 by 32.8. Ans. 2.069. 5. Divide 67.56785 by.035. Ans. 1930.51. 6. Divide.567891 by 8.2. Ans..069255. 7. Divide.1728 by 10. Ans..01728. 8. Divide 13.50192 by 1.38. Ans. 9.784. 9. Divide 783.5 by 6.25. Ans. 125.36. 10. Divide 983 by 6.6. Ans. 148.939 -- 11. Divide 172.8 by 1.2. Ans. 12. Divide 1728 by.12. Ans. 13. Divide.1728 by.12. Ans. 14. Divide 1.728 by 12. Ans. 15. Divide 17.28 by 1.2. Ans. 16. Divide 1728 by.0012. Ans. 17. Divide.001728 by 12. Ans. 18. Divide 116.31 by 1000. Ans..11631. 190 Divide one hundred forty-seven, and eight hundred twenty-eight thousandths by nine, and seven tentLs. Ans. 15.24. 20. Divide seventy-five, and sixteen hundredths by five, and forty-two thousand eight hundred one hundred thousandths. Ans. 13.846+o QUESTIONS. - What is the rule for division of decimals? What is note 1? Note 2? Note 3? What is the proof?

Page 190 190 REDUCTION OF DECIMALS. [SECT. XIX 21. Divide six hundred seventy-eight thousand seven hundred sixty-seven millionths, by three hundred twenty-eight thousandths Ans. 2.069-t+. REDUCTION OF DECIMALS. ART. 187. To reduce a common fraction to a decimal. Ex. 1. Reduce - to a decimal. Ans..625. OPERATION. Since we cannot divide the 8 ) 5.0 ( 6 tenths. numerator, 5, by 8, we reduce it 4 8 to tenths by annexing a cipher, and then dividing, we obtain 6 ) 2 0 (2 hundredth. tenth tenths and a remainder of 2 1 6 tenths. Reducing this remain8) 4 0 ( 5 thousandths. der to hundredths by annexing 4 0 a cipher, and dividing, we obAns..625. tain 2 hundredths and a remainOr thus: 8) 5.0 0 o der of 4 hundredths, which being reduced to thousandths by.6 2 5 annexing a cipher, and then dividing again, gives a quotient of 5 thousandths. The sum of the several quotients,.625, is the answer. To prove that 625 is equal to X, we change it to the form of a common fraction, by writing its denominator (Art. 176), and reduce it to its lowest terms. Thus, -ho = 8, Ans. Hence the following RULE. - Annex ciphers to the numerator, and divide by the denominator. Point off in the quotient as many decimal places, as there have been ciphers annexed. EXAMPLES FOR PRACTICE. 2. Reduce - to a decimal. Ans..75. 3. Reduce 3 to a decimal. Ans..875. 4. What decimal fraction is equal to 7,? Ans..4375. 5. Reduce -7 to a decimal. Ans..235294+. 6. Reduce A- to a decimal. Ans..363636+. 7. Reduce -h1- to a decimal. Ans..416666+. NOTE. - In reducing a common fraction to a decimal, when the denominator contains other prime factors than 2 and 5, there cannot be an exact division of the numerator; but, on continuing the division, some figure or figures of the quotient will be continually repeated. A decimal, of which there is a continual repetition of the same figure or figures, is called an igfinite or circulating decimal. The figures that repeat are called repetends. When the repetend is preQUESTIONS. - Art 187. How do you reduce a common fraction to a decimal? How can you prove the answer correct? What is the rule for reducing a common fraction to a decimal?

Page 191 bECT. XIX.] REDUCTION OF DECIMALS. 191 ceded by another decimal, the whole is called a mixed repetencl, and the part not repeating is called the finite part. To mark a repetend a dot (.) is placed over the first and last of the repeating figures. Thus, the answer to example sixth,.36, is a repetend; and the answer to example seventh,.416, is a mixed repetend, of which the figure 6 is the repetend, and the figures 41 the finite part. To change an infinite decimal to an equivalent common fraction, we write the repetend for the numerator, and as many nines as the repetend has figures for the denominator. Thus,.3d _ _9 = --; and the mixed 41 7 5 repetend,.416 = 100- = -- 1 2 A decimal other than a repetend is changed to the form of a common fraction, simply by writing the denominator under the given numerator. (Art. 176.) Thus,.75 -= - - 4; 005 = T — =s. 8. Reduce.875 to a common fraction. 9. Change.4375 to the form of a common fraction. 10. Change.72 to a common fraction. Ans. -11 11. Change.35 to a common fraction.. Ans. tk. 12. What common fraction is equivalent to.23562? Ans. s2332 7 13. Change.093 to an equivalent common fraction. Ans. 7 ART. 188. To reduce a compound number to a decimal of a higher denomination. Ex. 1. Reduce 8s. 6d. 3far. to the decimal of a pound. Ans..428125. OPERATION. We commence with the 3far., and first re4 3.0 0 duce them to hundredths, by annexing two ciphers; and then, to reduce these to the deci1 2 6.7 5 0 0 mal of a penny, we divide by 4, since there will be i as many hundredths of a penny as 2 0 8.5 6 2 5 0 0 of a farthing, and obtain.75d. Annexing this decimal to the 6d., we divide by 12, since.4 2 8 1 2 5 there will be -2 as many shillings as pence; and then the 8s. and this quotient by 20, since there will be Nyx as many pounds as shillings, and obtain.428125~. for the answer HIence the following RULE. - Divide the lowest denomination, annexing ciphers if necessary, by that number which will reduce it to one of the next higher denomination. Then divide as before, and so continue dividing till the decirmal is of the denomination required. QUESTIONS. - What is an infinite decimal? What is a repetendcl? What is a mixed repetend? How is an infinite decinlal changed to the form of a common fraction? -Art. 188. What is the rule for reducing a compound number to a decimal of a higher denomination 7

Page 192 192 REDUCTION OF DECIMALS. [SECT. XIX NOTE. - A compound number may also be reduced to a decimal by first reducing it to a common fraction (Art. 170), and then this fraction to a decimal. (Art. 187.) Thus, 2s. 6d. s= =-3-~ = o.125L. EXAMPLES FOR PRACTICE. 2. Reduce 15s. 6d. to the fraction of a pound. Ans..775. 3. Reduce 5cwt. 2qr. 141b. to the decimal of a ton. Ans..282. 4. Reduce 3qr. 211b. to the decimal of a cwt. Ans..96. 5. Reduce 6fur. 8rd. to the decimal of a mile. Ans..775. 6. Reduce 3R. 19p. 167ft. 72in. to the decimal of an acre, Ans..872595+. ART. 189e To find the value of a decimal in whole numbers of a lower denomination. Ex. 1. What is the value of.9875 of a pound? Ans. 19s. 9d. OPERATION. There will be 20 times as many ten thousandths of a.9 8 7 5 shilling as of a pound; therefore, we multiply the dec2 0 imal,.9875, by 20, and reduce the improper fraction to a mixed number by pointing off four figures on the i 9.7 5 0 0 right, which is dividing by its denominator, 10000. 1 2 The figures on the left of the point are shillings, and.90 0 0 0 those on the right decimals of a shilling. This decimal of a shilling we multiply by 12, and, pointing off as before, obtain 9d., which, added to the 19s., gives 19s. 9d. for the answer. RuLE. - Afultip7y the decimal by that number which will reduce it to the lower denomination, and point off as in multiplication of decimals. Then, multiply the decimal part of the product, and point off as before. So continue till the decimal is reduced to the denominations rc/uired. lThe several whole numbers of the successive products will be the answer.:EXAMPLES FOR PRACTICE. 2. What is the value of.628125 of a pound? Ans. 12s. 6Ed. 3. What is the value of.778125 of a ton? Ans. 15cwt. 2qr. 61b. 4oz. 4. WThat is the value of.75 of an ell English? Ans. 3qr. 3na. 5. What is the value of.965625 of a mile? Ans. 7fur. 29rd. QUESTION. - Art. 189. What is the rule for finding the value of a decimal in whole numbers of a lower denomination?

Page 193 SECT. XIX.] MISCELLANEOUS EXERCISES. 193 6. What is the value of.94375 of an acre? Ans. 3R. 31p. 7. What is the value of.815625 of a pound Troy? Ans. 9oz. 15pwt. 18gr. 8. What is the value of.5555 of a pound apothecaries' weight? Alns. 6; 53 3OD 19 7gr MISCELLANEOUS EXERCISES IN DECIMALS. 1. WVhat is the value of 15cwt. 3qr. 141b. of coffee at $9.50 per cwt.? Ans. $150.953-. 2. Wthat cost 17T. 18cwt. 1qr. 71b. of potash at $53.80 per'ton? Ans. $963.88+. 3. What cost 37A. 3R. 16p. of land at $75.16 per acre? Ans. $2844.80+. 4. What cost 15yd. 3qr. 2na. of cloth at $3.75 per yard? Ans. $59.53+-. 5. What cost 15$ cords of wood at $4.621 per cord? Ans. $71.10+-. 6. What cost the construction of 171m. 6fur. 36rd. of railroad at $3765.60 per mile? Ans. $67263.03+. 7. What cost 27hhd. 21gal. of temperance wine at $15.371 per hogshead? Ans. $420.24+-. 8. What are the contents of a pile of wood, lSft. 9in. long, 4ft. 6in. wide, and 7ft. 3in. high? Ans. 611ft. 1242in. 9. What are the contents of a board 12ft. 6in. long, and 2ft. 9in. wide? Ans. 34ft. 54in. 10. Bought a cask of vinegar containing 25gal. 3qt. Ipt. at $0.371 per gallon; what was the amount? Ans. $9.70-+-. 11. Bought a farm containing 144A. 3R. 30p. at $97.621 per acre; what was the cost of the farm? Ans. $14149.52+-. 12. Sold Joseph Pearson 3T. 18cwt. 211b. of salt hay at $9.37$ per ton. He having paid me $20.25, what remains due? Ans. $16.41+.'13. If i of a cord of wood cost $5.50, what cost one cord? What cost 73 cords? Ans. $48.71+. 14. If 4$ yards of cloth cost -12A, what cost 17 3 yards? Ans. $46.18+. 15. The ship Constantine cost $35000; I of it was sold to Captain Sampson for $9000; ~ of the remainder to T. Lamb for 89200, and the balance to another person at a profit of $500; what was gained in the sale of the whole ship? Ans. $1200. 17

Page 194 194 PERCENTAGE. [SECT. XX ~ XX. PERCENTAGE. ART-. 190. PERCENTAGE and per cent. are terms derived from Latin words, per centumrn, which signify by the hundred. Percentage, therefore, is any rate or sum on a hundred, or it is any number of hundredths. Thus, if an article is bought for $100, and sold for $105, the gain is 5 per cent., because $5 are 10 of $100, or of the original cost. Again, if an article is bought for $25, and sold for $30, the gain is 20 per cent., because $5 are 5 -- of $25, or of the original cost. Since per cent. is any number of hundredths, it is a decimal written in the same manner as hundredths in decimal fractions. Thus, 5 per cent., 25 per cent., &c., are written.05,.25, respectively. (Art. 175.) When the per cent. is more than 100 it is an improper frac tion, and if expressed decimally it becomes a mixed number; thus, 103 per cent., equal to I,o3 is written 1.03. If the per cent. is less than 1, or a part of one hundredth, to be expressed decimally, it must be written at the right of hundredths in the place of thousandths, &c. Thus, ~ of I per cent., 4 of I per cent., 12& per cent., are written.005,.0075,.122, respectively. EXAMPLES. Write decimally 2 per cent.; 3 per cent.; 5 per cent.; 6 per cent.; 7 per cent.; 8 per cent.; 10 per cent.; 12 per cent., 15 per cent.; 25 per cent.; 30 per cent.; 40 per cent.; 50 per cent.; 60 per cent.; 75 per cent.; 100 per cent.; 105 per cent., 115 per cent.; 6~ per cent.; 83 per cent.; 20, per cent.; I of 1 per cent.; 7 of 1 per cent.; ~ of I per cent.; - of 1 per cent.; - of 1 per cent. ART. W9L. To find the percentage on any sum or quantity. Ex. 1. Bought a house for $625, and sold it at 6 per cent. advance; what did I gain by the sale? Ans. $37.50. QUESTIONS. -Art. 190. From what are the terms percentage and per cent. derived, and what do they signify? How, then, will you define percentage? How will you illustrate it? iow is per cent. written when less than 100? How, when more than 100? If the per cent. is a fraction, or contains a fraction, what is the fraction, and, if expressed decimally, what place must it occupy 7

Page 195 SECT. XX.] PERCENTAGE. 195 OPERATION. Since 6 per cent. is A- =.06 Sum, 6$ 2 5 of the original cost, we multiply Rate per cent.,.0 6 $625 by the decimal expression Per cent., or gain, A 7.5 0.063 and point off as in multiplication of decimal fractions. RULE. - Multiply the given qua'ntity or number by the'rate per cent., considered as a decimal, and point of the product as in mulliplication of decimalJfractions. (Art. 185.) NOTE. -If the per cent. contains a fraction that cannot be expressed decimally, or, if thus expressed, would require several figures, it is more convenient to multiply by it as a mixed number. (Art. 155.) EXAMPLES FOR PRACTICE. 2. What is 2 per cent. of $325? Ans. $6.50. 3. What is 5 per cent. of $789? Ans. $39.45. 4. What is 6 per cent. of $856.49? Ans. $51.38,9. 5. What is 7~ per cent. of 765 tons? Ans. 57.375 tons. 6. What is 94 per cent. of $5000? Ans. $490. 7. What is 7 per cent. of $1728? Ans. $15.12. 8. What is 41 per cent. of 587 yards of cloth? Ans. 26.415 yards. 9. I lost 10 per' cent. of $975; how much have I remaining? Ans. $877.50. 10. Sent to Liverpool 5000 bushels of wheat, which cost me $1.25 per bushel; but 25 per cent. of the wheat was thrown overboard in a storm, and the remainder was sold at $2 per bushel; what was gained on the wheat? Ans. $1250. 11. T. Page received a legacy of $8000); he gave 19 per cent. of it to his wife, 37 per cent. of the remainder to his sons, and $2000 to his daughters; what sumn had he remaining? Ans. $2082.40. 12. My tailor informs me it will take 10 square yards of cloth to make me a full suit of clothes. The cloth I am about to purchase is 13 yards wide, and on sponging it will shrink 5 per cent, in width and 5 per cent. in length. How many yards of the above cloth must I purchase for my'" new suit "? Ans. 6yd. 1qr. 1,7,~_-na. 13. A man having $10000, lost 15 per cent. of it in speculation; what sum had he remaining? Ans. $8500. QUESTIONS. - Art. 191. Will you explain the operation for finding the percentage on any sum or quantity? Give the reason for the process. What is the rule?

Page 196 196 SIMPLE INTEREST. [SECT. XXI ~ XXI. SIMPLE INTEREST. ART. X, 2 INTEREST is the compensation which the borrower of money makes to the lender. The rate per cent. is the sum paid for the use of $100, 100 cents, 100~., &c., for one year. The pr-?cipal is the sum lent, on which interest is computed. The amoi2nt! is the interest and principal added together. Leyal interest is the rate per cent. established by law. Uszury is a higher rate per cent. than is allowed by law. The legal rate per cent. varies in the different States and in different countries. In Maine, New Hampshire, Vermont, Massachusetts, Rhode Island, Connecticut, New Jersey, Pennsylvania, Delaware, Mlaryland, Virginia, North Carolina, Tennessee, Kentucky, Ohio, Indiana, Illinois, Iowa, M1issouri, Arkansas, Mississippi, Florida, District of Columbia, and on debts or judgments in favor of the United States, it is 6 per cent. In New York, Michigan, Wisconsin, iinnesota, Georgia, and South Carolina, it is 7 per cent. In Alabama and Texas, it is 8 per cent. In California, it is 10 per cent. In Louisiana, it is 5 per cent. In Canada, Nova Scotia, and Ireland, it is 6 per cent. In England and France, it is 5 per cent. NOTE. - The legal rate, as above, in some of the States, is only that which the law allows, when no particular rate is mentioned. By special agreement between parties, in Ohio, Indiana, Michigan, Illinois, Iowa, and Arkansas, interest can be taken as high as 10 per cent.; in Florida and Louisiana, as high as 8 per cent.; in Texas and Wisconsin, as high as 12 per cent.; and in California, any per cent. In New Jersey, by a special law, 7 per cent. may be taken in the city of Paterson, and in the counties of Essex and Hudson. ART. I93o To find the interest of $1 at 6 per cent. for any given time. Since the interest of $1 is 6 cents, or T- 6 of the principal, for 1 year, or 12 months, for 1 month it will be TI of 6 cents, or I a cent, eq!ial to 5 mills, or W of the principal; and for 2 QUESTIONS. - Art. 192. What is interest? What is rate per cent.? What is the principal? What is the amount? What is legal interest? What is usury? What is the legal rate per cent. in the different States? In Canada, Nova Scotia, and Ireland? In England and France?

Page 197 SECT. XXI.] SIMPLE INTEREST. 197 months, twice 5 mills, or 1 cent, or T- of the principal. Now, since the interest for 1 month, or 30 days, is 5 mills, the interest for 6 days, or 4 of 30 days, will be one mill, or r1ui of the principal. And as 1 day, 2 days, &c., are I, 2, &c., of 6 days, the interest for any number of days less than 6 will be as many sixths of a mill, or six thousandths of the principal, as there are\ days. Also, since the interest for 2 months is 1 cent, or IT, of the principal, for 100 times 2 months, or 200 months, or 16 years 8 mo., it will be 100 cents, or equal the whole principal;. and in the same proportion for any other length of time. Hence, the INTEREST OF $1, AT 6 PER CENT., FOR 1 yr., or 12 mo., is 6 cents, or $0.06, equal -1 r of the prin. - of a yr., or 2 mo., is 1 cent, or $0.01, equal r L of the prin. 9r of a yr., or I mo., is 5 mills, or $0.005. equal, of the prin. of a mo., or 6 da., is 1 mill, or $0.001, equal T 1 of the prin. a of a mo., or 1 da., is I of a mill, or $0.000, equal FFF* prin. ALSO, 16 yr. 8 mo., or 200 mo., is 100 cts., or $1.00, equal the whole prin. 8 yr. 4 mo., or 100 mo., is 50 cts., or $0.50, equal A of the prin. 5 yr. 62 mo., or 664 mo., is 334 cts., or $0.333~, equal 4 of prin. 4 yr. 2 mo., or 50 mo., is 25 cts., or $0.25, equal I of prin. 3 yr. 4 mo., or 40 mo., is 20 cts., or $0.20, equal - of prin. 2 yr. 94 mo., or 334 mo., is 164 cts., or $0.1664, equal ~ of prin. 2 yr. 1 mo., or 25 mo., is 12. cts., or $0.125, equal I of prin. 1 yr. 8 mo., or 20 mo., is 10 cts., or $0.10, equal'-jv of prin. 1 yr. G4 mo., or 162 mo., is 8~ cts., or $0.0834, equal T of prin. -f of a yr., or 10 mo., is 5 cts., or $0.05, equal 9yw of prin. 9 of a yr., or 6 mo., is 3. cts., or $0.0334, equal a of prin.?6 Of a yr.; or 5 mo., is 24 ets., or $0.025, equal w'w of prin. 4 of a yr., or 4 mo., is 2 cts., or $0.02, equal -I of prin. Ex. 1. What is the interest of $1 for 2yr. 7mo. 20da.? Ans. $0.158~. FIRST OPERATION. The interest for 2 years will be Interest for 2y..1 2 twice as much as for 1 year, equal " " 7mo. =.0 3 5 12 cents; and since the interest for " " 20dca. =.O 0 31 2 months is 1 cent, for 7 months it will'e 34 cents. And as the inAns. $ 0.1 5 81 terest for 6 days is 1 mill, for 20 days it will be 34 mills. Adding the several sums together, we have $0.158~ for the answer. QUESTION. - Art. 193. How do you explain the operation? 17*

Page 198 198 SIMPLE INTEREST. [SECT. XXJ SECOND OPERATION. The time, 2y Principal, $ 1.0 0 7mo. 20da., is equal to 2y. L of the prin.,.1 2 5 Int. for 2yr. lmo. Il to o. 3 of the prin.,.0 3 31- Int. for 6mo. 20da. 20da. Now, since the in$ 0.1 5 8~ Int. for 2y. 7mo. 20da. terest on any sum, at 6 per cent., in 200 months equals the principal, for 2y. Imo., or A of 200 months, it will equal A of the principal. We, therefore, take A of the principal, $1.00, equal 12 cents and 5 mills, the interest for 2y. imo. The balance of time, 6mo. 20da., or 6.mo., being &y of 200 months, we take ~- of the principal, equal 3 cents and 3~ mills. as the interest for the 6mo. 20da. We add together the interest for the parts of the whole, and obtain, as by first operation, $0.158t- as the whole interest. RULE 1. - Reckon 6 cents for every YEAR, 1 cent for every Two MONTIIS, 5 mills for the odd month, 1 mill for every 6 days; and for any number of days less than six, as many sixths of a mill as there are days. Or, Reduce the years and months to months, and call half the number of months cents, and one sixth the number of days mills. Or, PRULE 2. - Take such fractional part or parts of the principal as the number expressing the time is of 200 months. EXAMPLES FOR PRACTICE. 2. What is the interest of $1 for ly. 4mo. 6da.? Ans..$0.081. 3. What is the interest of $1 for ly. 9mo. 12da.? Ans. $0.107. 4. What is the interest of $1 for 3y. Smo. 19da.? Ans. $0.223-6. 5. What is the interest of $1 for 2y. Imo. 20da.? Ans. $0.128:~o 6. What is the interest of $1 for 7y. 15da.? Ans. $0.4221. 7. What is the interest of $1 for 3mo. 28d.? Ans. $0.0192. 8. What is the interest of $1 for 4y. 2mo. 5da.? Ans. $0.250}. 9. What is the interest of $1 for 410mo. 3da.? Ans. $0.020. QUESTIONS. - How do you explain the second operation? E'What is the first rule? What is the second rule?

Page 199 1sACT. XXI.] SIMPLE INTEREST. 199 ART. 194. To find the interest on any sum of money at 6 per cent. for any given time. Ex. 1. What is the interest of $926 for 3y. Ilmo. 15da.? What is the amount? Ans. Interest, $219.925; Amount, $1145.925. OPERATION. Principal, $ 9 2 6 Interest of $1,.2 3 71'We find the interest of $1 for the 6 4 8 2 given time to be $0.2374. (Art. 193.) 2 7 7 8 Now, since the interest of $1 is $0.2374, 1 8 5 2 the interest of $926 will be 926 times 4 6 3 as much; therefore we multiply them together. To find the amount, we add Int. $ 2 1 9.9 2 5 the principal to the interest. Hence Prin. 9 2 6 the following Amt. $114 5.9 2 5 RULE. - Find the interest of S 1 for 1tie given time; th7en multijply the principal by tile number denoting this interest, andc point off as in multipl[ication of decimal fractions. (Art. 185.) To find the amount, adcl the princi pal to the interest. NOTE. - If the interest of $1 contains a common fraction, the fraction may be reduced to a decimal, if preferred. The interest may also be multiplied by the number denoting the principal, when it is more convenient. EXAMPLES FOR PRACTICE. 2. What is the interest of $197 for I year? Ans. $11.82. 3. What is the interest of $1728 for 3 years? Ans. $311.04. 4. W1,That is the interest of $69 for 2 years? Ans. $8.28. 5. What is the interest of $1728 for 1 year, 6 months? Ans. $155.52. 6. What is the interest of $16.87 for 1 year, 8 months? Ans. $1.687. 7. Required the interest of $118.15 for 2 years, 6 months. Ans. $17.722. 8. Required the interest of $97.16 for 1 year, 5 months. Ans. $8.258. 9. Required the interest of $789.87 for 1 year, 11 months. Ans. $90.833. QUESTIONS. — Art. 194. Explain the operation for finding the interest on any sum of money at 6 per cent. for any given time. What is the rule? How do you find the amount?

Page 200 '200 SIMPLE INTEREST. [SECT. XXI. 10. Required the amount of $978.18 for 2 years, 3 months. Ans. $1110.234. 11. Required the amount of $87.96 for 1 month. Ans. $88.399. 12. Required the amount of 881.81 for 8 years, 4 months. Ans. $122.715. 13. Required the amount of $0.87 for 7 years, 3 months. Ans. $1.248. 14. What is the interest of $1.71 for 2 years, 2 days? Ans. $0.205. 15. Required the interest of $100 for 8 years, 4 months, 1 day. Ans. $50.016. 16. Required the interest of $3.05 for 2 months, and 2 days. Ans. $0.031. 17. What is the interest of $761.75 for 1 year, 2 months, 18 days? Ans. $55.607. 18. What is the interest of 81728.19 for 1 year, 5 months 10 days? Ans. $149.776. 19. What is the interest of $88.96 for 1 year, 4 months, 6 days? Ans. $7.205. 20. What is the interest of $107.50 for 1 month, 29 days? Ans. $1.057. ART. 195. To find the interest of any sum of money at any rate per cent. for any given time. Ex. 1. What is the interest of $26.25 for 2 years, 4 months, at 7 per cent.? Ans. $4.2875. OPERATION. We first find the interest Principal, $ 2 6.2 5 on the given sum at 6 per Interest of $1 at 6 per cent.,.1 4 cent., and then add to this 1.0 50..0 interest the fractional part of itself, denoted by the 2 6 2 5 excess of the rate above 6 Interest at 6 per cent., $ 3.6 7 5 0 per cent. The excess is A of interest at 6 per cent.,.6 1 2 5 1 per cent.; therefore we add 6 of the interest at 6 Interest at 7 per cent., $ 4.2 8 7 5 per cent. to itself, for the answer. If the rate per cent.had been less than 6, we should have subtracted the fractional part. QUESTION. - Art. 195. Explain the operation for finding the interest on any sum of money at any rate per cent.

Page 201 SECT. XXI.] SIMPLE INTEREST. 201 RULE. - Find the interest of the ziven sum at 6 per cent., and then add to this interest, or subtract from it, such afractional part of itself as the yiven rate is 5yreater or less than 6 per cent. NOTE 1. —If the rate per cent. is 12 per cent., the interest at 6 per cent. nmust be doub)led. NOTEr 2.- If the interest is for years only, it may be found by multiply ing the principal by the interest of I1 for the given time and rate. EXAM3PLES FOR PRACTICE. 1. WVhat is the interest of $144 for one year at 7 per cent.? Ans. $10.08. 2. WVhat is the interest of $850 for 1 year, 7 months, 18 days, at 7 per cent.? Ans. $97.18. 3. What is the interest of $865.75 for 3 years, 9 months, 24 days, at 7 per cent.? Ans. $231.299. 4. What is the interest of $960.18 for 1 year, 2 months, at 7 per cent.? Ans. $78.414. 5. What is the interest of $1728.19 for 3 years, 8 months, 10 days, at 7 per cent.? Ans. $446.929. 6. WXVhat is the interest of $17.90 for 8 months, 4 days, at 7 per cent.? Ans. $0.849. 7. What is the interest of $1165.50 for 5 years, 3 months, 9 days, at 7 per cent.? Ans. $430.36. 8. What is the interest of $1237.90 for 1 year, 7 months, 3 days, at 7 per cent.? Ans. $137.922. 9. What is the interest of $156.80 for 3 years and 3 days, at 3 per cent.? Ans. $14.151. 10. What is the interest of $579.75 for 1 year, 2 months, 2 days, at 5 per cent.? Ans. $33.979. 11. What is the interest of $7671.09 for 2 years, 8 months, 5 days, at 8 per cent.? Ans. $1645.02. 12. What is the interest of $943.11 for 1 month, 29 days, at 9 per cent.? Ans. $1 3.91. 13. What is the interest of $975.06 for 2 years, 7 months, 9 days, at 8{I per cent.? Ans. $829.) 82. 14. What is the amount of $1000 for 3 yearls, 3 mronths, 29 days, at 51 per cent.? Ans. $118 3.1 [. 15. Wh hat is the interest of $765 for 2 years, 9 monfth.s, at I per cent.? Ans. $21.037. 16. What is the interest of $979.15 for 3 years, 2 months, 4 days, at 121 per cent.? Ans. $388.94. QUESTIONS. - What is the rule? What is note first? Note second?

Page 202 202 SIMPLE INTEREST. [SECT. XX1 ART. 196,. Second method of finding the interest of any sum of money, at any rate per cent., for any time. Ex. 1. What is the interest of $26.25 for 3 years, 5 months and 15 days, at 8 per cent.? Ans. $7.262. OPERATION-. IHaving found the inter Principal, $2 6.2 5 est for 1 year and then for Rate per cent.,.0 8 3 years, the interest for 5 Interest for 1 year, months is obtained by first Interest or year taking ~ of 1 year's interest, for 4 months, and then Interest for 3 years, 6.3 0 0 0 4 of this last interest, for Interest for 4mo., 1 of ly.,.7 0 0 0 1 month. rt ^ 1 ^ o A 1 w F n And since 15 days are Interest for imo., - of 4mo.,.175 0 nd since 15 days are Interest or l I mo A of 1 month, we take A Interest for 15da., 1 of lmo.,.0 8 7 5 of 1 month's interest for Int. for 3y. 5mo. 15da., $7.2 6 2 5 the interest of 15 days, and add the several sums together for the answer. RULE. - First find the interest for one year by multiplying the principal by the rate per cent.; and for two or more years multiply this product by the number of years. Find the interest for months by taking the most convenient fractional part or parts of ONE year's interest. Find the interest for days by taking the most convenient fractionao part or parts of ONE month's interest. NOTE. - Many practical men prefer this method of casting interest to any other, but in most questions it is not so expeditious as the preceding. The pupil may be required to solve the questions in interest by both methods. EXAMPLES FOR PRACTICE. 2. What is the interest of $1775 for 7 years? Ans. $745.50. 3. What is the interest of $987 for 3 years, 6 months? Ans. $207.27. 4. Required the interest of $69.17 for 4 years, 9 months. Ans. $19.713. 5. Required the interest of $96.87 for 10 years, 7 months, 15 days. Ans. $61.754. QUESTIONS. - Art. 196. Explain the operation for finding the interest of any sum of money, at any rate per cent., for any time. What is the rule?

Page 203 SECT. XXI.] SIMPLE INTEREST. 203 6. Required the interest of $1.95 for fifteen years, 11 months, 20 days. Ans. $1.868. 7. Required the interest of $1789 for 20 years, 1 month, 25 days. Ans. $2163.199. 8. Required the interest of $666.66 for 6 years, 10 months, 13 days. Ans. $274.775. 9. What is the amount of $98.50 for 5 years, 8 months? Ans. $131.99. 10. What is the amount of $168.13 for 8 years, 5 months, 3 days? Ans. $253.119. 11. What is the amount of $75.75 for 4 years, 2 months, 27 days? Ans. $95.028. 12. Required the amount of $675.50 for 30 years, 3 months, 23 days. Ans. $1904.121. ART. 1I97, To find the interest on pounds, shillings, pence, and farthings, at any rate per cent., for any time. Ex. 1. What is the interest of 25~. 2s. 6d. for 2 years, 6 months, at 6 per cent.? Ans. 3~. 15s. 5d. 2far. OPERATION. We reduce the 2s. 6d. to the 25~. 2s. 6d.- 2 5.1 2 5 ~. decimal of a pound (Art. 188) Interest of 1~..1 5 and, annexing it to the pounds 1 2 5 6 2 5 multiply this principal by the interest of 1~. for the given time. 2 5 1 2 5 The product is the answer in 3.7 6 8 7 5 ~.- pounds and the decimal of a 4d. 2far. pound, which must be reduced to 3~. 15s. 4d. 2far. shillings, pence, and farthings. (Art. 189.) RULE. - Reduce the shillings, pence, andfarthings, to the decimal oJ a pound, and annex it to the pounds; then proceed as in United States money, and reduce the decimal in the result to a compound number. EXAMPLES FOR PRACTICE. 2. What is the interest of 26~. lOs. for 2 years, 4 months, at 5 per cent.? Ans. 3~. is. 10d. 3. What is the interest of 42~. 18s. for 1 year, 9 months, 25 days, at 6 per cent.? Ans. 4~. 13s. 73d. 4. What is the interest of 94~. 12s. 6d. for 4 years, 6 months 7 days, at 8 per cent.? Ans. 34~. 4s. 23d. QUESTIONS. - Art. 197. How do you find the interest on pounds, shillings, pence, and farthings? Repeat the rule.

Page 204 204 SIMPLE INTEREST. [SECT. XXI MISCELLANEOUS EXERCISES IN INTEREST. 1. What is the interest of $172.50 from Sept. )25, 1850, to July 9, 1852? Ans. $18.515. 2. What is the interest of $169.75 from Dec. 10, 1848, to May 5, 1851? Ans. $24,472. 3. What is the interest of $17.18 from July 29, 1847, to Sept. 1, 1851? Ans. $4.214. 4. What is the interest of $67.07 from April 7, 1849, to Dec. 11, 1851? Ans. $10.775. 5. Required the interest of $117.75 from Jan. 7, 1849, to Dec. 19, 1851. Ans. 820.841. 6. Required the interest of $847.15 from Oct. 9, 1849, to Jan. 11, 1853. Ans. $165.476. 7. Required the interest of $7.18 from March 1, 1851, to Feb. 11, 1852. Ans.. $0.406. 8. What is the interest of $976.18 from iMay 29, 1852, to Nov. 25, 1855? Ans. $204.347. 9. I have John Smith's note for $144, dated July 25, 1849; what is due March 9, 1852? Ans. $166.658. 10. George Cogswell has two notes against J. Doe; the first is for $375.83, and is dated Jan. 19, 1850; the other is for 876.19, dated April 23,.1851; what is the amount of both notes Jan. 1, 1852? Ans. $499.141. 11. What is the interest of $68.19, at 7 per cent., firom June 5, 1850, to June 11, 1851? Anls. $4.852. 12. Required the amount of $79.15 from Feb. 17, 1849, to Dec. 30, 1852, at 71 per cent. Ans. $102.119. 13. What is the amount of $89.96 from June 19, 1850, to Dec. 9, 1851, at 84 per cent.? Ans. $100.88o. 14. A. Atwood has J. Smith's note for $325, dated June 5, 1849; what is due, at 71 per cent., July 4, 1851? Ans. $374.022. 15. J. Ayer has D. How's note for $1728, dated Dec. 29, 1849; what is the amount Oct. 9, 1852, at 9 per cent.? Ans. 82160. 16. What is the interest of $976.18 from Jan. 29, 1851, to July 4, 1852, at 12 per cent.? Ans. $167.577. 17. What is the amount of $175.08 from May 7, 1851, to Sept. 25, 1853, at 7 per cent.? Ans. $204.289. 18. What is the amount of $160'from Dec. 11, 1853, to Sept. 9, 1854, at 7 per cent.? Ans. $168.337.

Page 205 XFICT. XXI.1 PARTIAL PAYMENTS. 205 PARTIAL PAYMENTS. ART. 198. A NOTE, or, as it is often called, a promissory note, or note of hand, is an engagement, in writing, to pay a specified sum, either to a person named in the note, or to his order, or to the bearer. A joint note is one signed by two or more persons, who together are holden for its payment; and a joint and several note is one signed by two or more persons, who separately and together are holden for its payment. A neyotiable note is one so made that it can be sold or transferred from one person to another. The maker or drawer of a note is the person who signs it; the payee, promisee, or holder, the person to whom it is to be paid; an endorser, the person who writes his name upon its back to transfer it, or as guarantee of its payment; and the face of a note, the sum for which it is given. Partial payments are part payments of a note or other obligation; and being receipted for by an entry on the back of the note or obligation, are called endorsemenzts. ART. 199, When notes are paid within one year from the time they become due, it has been the usual custom to compute the interest by the following RULE. - F ind the amount of the principal from the time it became due until the time of payment. Then find the amount of each endorsement from the time it was paid until settlement, and subtract their sum from the amount of the principal. Ex. 1. $1234. Boston, Jan. 1, 1853. For value received, I promise to pay John Smith, or order, on demand, one thousand two hundred thirty-four dollars, with interest. John Y. Jones. Attest, Samuel Emerson. On this note are the following endorsements: March 1, 1853, received ninety-eight dollars. June 7, 1853, received five hundred dollars. Sept. 25, 1853, received two hundred ninety dollars. Dec. 8, 1853, received one hundred dollars. What remains due at the time of payment, Jan. 1, 1.854? Ans. $293.12. QUESTIONS. - Art. 198. What is a, note? What a negociable note? A joint note? Who is the maker of a note? Who the payee? Who the endorser? tWhat are partial payments? - Art. 199. What is the rule for co'rputing the interest when there are partial payments, and all are made within one year? 18

Page 206 2 06 PARTIAL PAYMENTS, OPERATION. Principal, $1234.00 Int. from Jan. 1, 1853, to Jan. 1, 1854 (ly.), 74.04 Amount, - - 1308.04 First payment, March 1, 1853, -- $98.00 Int. fiom March 1, 1853, to Jan. 1, 1854 (10mo.), 4.90 Second payment, June 7, 1853, - 500.00 Int. from June 7, 1853, to Jan. 1, 1854 (6mo. 24da.), 17.00 Third payment, Sept. 25, 1853, - 290.00 Int. firom Sept. 25, 1853, to Jan. 1, 1854 (3mo. 6da.), 4.64 Fourth payment, Dec. 8, 1853, - 100.00 Int. friom Dec. 89 1853, to Jan. 1, 1854 (23da.),.38 Amount of payments to be deducted, o o $1014.92 Balance remains due Jan. 1, 1854 o $293.12 2. $987.75. Trenton, ean. 11, 1852. For value received, we jointly and severally promise to pay fames Dayton, or order, on demand, two months fromn date, nine hundred eiyhty-seven dollars seventy-five cents, withl interest after two months. Johnz T. Johnson. Attest, Isaiah Webster. Samuel AJ'tes. On this note are the following endorsements: May 1, 1852, received three hundred dollars. June 5, 1852, received four hundred dollars. Sept. 25, 1852, received one hundred and fifty dollars. What is due Dec. 13, 1852? Ails. $156.94. 3. $800. Indianapolis, July 4, 1852. For value received, I promise to pay Leonard Jo3hnsn, or order, on demand, eiyht hundred dollars, with interest. Attest, Charles True. Samuel I'everpay. On this note are the following endorsements: Aug. 10, 1852, received one hundred forty-four dollars. Nov. 1, 1852, received ninety dollars. Jan. 1, 1853, received four hundred dollars. March 4, 1853, received one hundred dollars. What remains due June 1, 1853? Ans. $88.02. ART. 20o0 In tke United States court, and in most of the courts of the several States, the following rule is adopted fbr computing the interest on notes and bonds, when partial pay ments have been made. QUESTION. - HOw do you explain the operation?

Page 207 SECTo XXJI. PARTIAL PAYMENTS. 207 Com11te the e interest on the principal~ to thMe timne when the first paynenlt was mnade, which equals, or exceeds, either alone or with preceeding pay1menzts, the interest then due. Add thai znlterest to the principal, and from the amount subtract the payment or payrments thus f.r madle. The remainder willbform a new principal; on which compute the interest, prorcediny as before. NOTE. — This.rule is on the principle, that neither interest nor payment should draw interest. This rule is illustrated in the following question: Ex, 1..365.50. Wilmington, Jan. 1, 1,852. For value received, I promise to pay to John Dow, or order, on denanld. t/tree hundred sixty-five dollars fifty cents, wit/l'nlterest. John Smith. Attest, Samuel Webster, On this note are the following endorsements: June 10, 1852, received fifty dollars. Dec. 8, 1852, received thirty dolllrs. Sept. 25, 1853, received sixty dollars. July 4, 1854, received ninety dollars. Aug. 1, 1855, received ten dollars. Dec. 2, 1 55, received one hundred dollars. What, remains due Jan. 7, 1857? Ans. $92.53. OPERATION. Principal. carrying interest from Jan. 1, 1852, to June 10, 1852 - - 365.50 Interest from Jan. 1, 1852, to June 10, 1852 (5 months, 9 days). - 9.68 Amrount, - 35.18 First paysent, June 10, 1852, 50.00 Balance for new principal, - 325.18 Interest from June 10, 1852, to Dec. 8, 1852 (5 months, 28 days), -. 9.64 Amount. - - 334.82 Second payment, Dec. 8, 1852, - 30.00 la3lLnce for new principal, - - 304.82 Int. for Dec, 8, 1852, to Sept. 25, 1853 (9mo. 17 days), 14.58 Amount - - - - - 319.40 Thtird pymrent, Sept. 25, 1853, - - 60).00 Balance for new principal, - - 259.40 Interest from Sept. 25, 1853, to July 4, 1854 (9 months, 9 das), - - - - 12.06 Amount - - - - 271.46 QuESTsoN. — Art. 200. What is the rule generally adopted by the several States for computing the interest on notes and bonds, when partial pay ments have beren made?

Page 208 208 PARTIAL PAYMENTS. [SECT. XXI. Amount brought up, 271.46 Fourth payment, July 4, 1854, - - - 90.00 Balance for new principal, - - 181.46 Interest from July 4, 185-4, to Augl. 1, 1855 (12 mo. 27 cldays), 11. 70 Interest from Aug. 1, 1855, to Dec. 2, 1855 (4 mo. 1 day), 3.66 Amount, 196.82 Fifth payment, Aug. 1, 1855 (a sum less than the interest), - $10.00 Sixth payment, Dec. 2, 1855 (a sum greater than the interest), 100.00 110.00 Balance for new principal, 86.82 Interest from Dec. 2, 1855, to Jan. 7, 1857 (13 months, 5 days), - -. 5.71 Remains due Jan. -7, 1857, $92.53 2. $1666. Philadelphia, June 5, 1848. For value received, I promise to pay J. B. Lippincott 4- Co., or order, on demand, without defalcation, one thousand six hundred sixty-six dollars, with interest. John J. Shellenberyer. Attest, T. Webster. On this note are the following indorsements: July 4, 1849, received one hundred dollars. Jan. 1, 1850, received ten dollars. July 4, 1850, received fifteen dollars. Jan. 1, 1851, received five hundred dollars. Feb. 7, 1852, received six hundred and fifty-six dollars. What is due Jan. 1, 1853.? Ans. $767.08. 3. $960. Detroit, Oct. 23, 1850. On demand, I promise to pay S. S. St. John, or order, nine hundred sixty dollars, for value received, with interest at seven per cent. John Q. Smith. Attest, H. F. Wilcox. On this note are the following indorsements: Sept. 25, 1851, received one hundred forty dollars. July 7, 1852, received eighty dollars. Dec. 9, 1852, received seventy dollars. Nov. 8, 1853, received one hundred dollars. What is due Oct. 23, 1854? Ans. $807.76. 4. $1000. N\ew York, Jan. 1, 1849. Two months after date I promise to pay to S. Duranzd, or QUESTION - Explain the operation.

Page 209 SzE'. 1I. PARTIAL PAYMENTS. 209 order, ane thousand dollars, for value received, with interest after, at seven per cent,. Paul Samlpson, Jr Attest, MWilliam S. Hall. On this note are the following indorsements: AMatrch 1. 1850, received one hundred dollars. Sept. 25, 1851, received two hunldred dollars. Oct. 9, 1852, received one hundred fiftv dollars. July 4, 1853, received twenty dollars. Oct 9, 1853, received three hundred dollars. What is due Dec. 1, 1854? Ans. $567.49. AnT. 201o The following is the rule established by the Supreme Court of the State of Connecticut. 1. "' Compute the interest to the time of the first payment; if that be one year or more from the time the interest commenced, add it to the principal, and deduct the payment from the sum total. if there be after payments made, compute the interest on the balance due to the next payment, and then deduct the payment as above; and in like Tuanner from one payment to another, till all the payments are absorbed; provided the time between one payment and another be one year or more." 2. " But if any payments be made before one year's interest hath accrued, thenr compute the interest on the principal sum due on the obligation for one year,. add it to the principal, and compute the interest on the sum paid from the time it was paid up to the end of the year; add it to the sum paid, and deduct that sum from the principal and interest added together." 3.'" If any paIyments be made of a less sum than the interest arisen at the time of such payment, no interest is to be computed, but only on the principal sum for any period." Ex. 1. $500. Hartford, July 1, 1854. For value received, I promnise to pay.1 D0e, or order, o?, demand, five hundred dollars, with interest. D. P. Page. On this are the following indorsements: Sept. 1, 1855, received one hundred dollars. April 1, 1856, received one hundred forty-four dollars. Jan. 1, 1857, received ninety dollirs, fifty cents. Dec. 1, 1858, received one hundred sixty-eight dollars, ve cents. What is due Oct. 1, 1859? Ans. $92.40. I [f f year extends beyond the time when the note becomes due,'find the amount of the remaining principal to the time of settlement; find also the amuount of the indorsement or indorsenments, if any, from the time they were paid to tbe time of settlement, and subtract their sum from the amount of the principal. QUESTION. - Art. 201. What is the Connecticut rule for computing interest eo notes and bonds, when partial payments have been made? 1,8*

Page 210 210 PROBLEMS IN INTEREST. [SECT. XX1. PROBLEMS IN INTEREST. ART. 202. A PROBLEM in ← arithmetic → is a question proposed which requires solution. ART. 203. In the preceding questions in interest, five terms or things have been mentioned; namely, the Interest, Amount, Rate per cent., Time, and Principal. The investigation of these involves five problems: I. To find the interest; II. To find the amount; III. To find the rate per cent.; IV. To find the time; V. To find the principal. With one exception, any three of the preceding terms being given, a fourth may be found by the rules deduced from the solution of the problems. But if the rate per cent., time, and amount, are given, an additional rule is necessary to find the principal, which will form a sixth problem; but, from its connection with Discount, its solution will be deferred until that subject is considered. The Problems I. and II. have already been examined (Art. 194), and we now proceed to an examination of those remaining. ART. 204, Problem III. To find the rate per cent., the principal, interest, and time, being given. Ex..1. The interest of $300 for 2 years is $48; what is the rate per cent.? Ans. 8 per cent. OPERATION. We find the interest on the $3 0 0 principal for 2 years at 1 per.0 2 cent., and divide the given inter$6.0 0 )4 8.0 0 ( 8 per cent. est by it. 4 8.0 0 Since the interest of 81 at 1 per cent. for 2 years is 2 cents, the interest of $300 will be 300 times as much, equal to $6. Now, if $6 is 1 per cent., $48 will be as many per cent. as $6 is contained times in $48, which gives 8 per cent. for the answer. RULE. - Divide the given interest by the interest of the given sum at 1 per cent. for the given time, and the quotient will be the rate per cent. required. QUESTIONS. -Art. 202. What is a problem in ← arithmetic → ?- Art. 203. How many terms or things have been given in the preceding questions in interest? Name them. What does an investigation of these terms involve? Name them. How many terms are given in each problem in order to find a, fourth? What two problems have been examined? - Art. 204. What is Problem III.? Explain the operation. What is the rule for finding the rate per cent., the principal, interest, and time, being given?

Page 211 bEOT. XXI.] PROBLEMS IN INTEREST. 211 EXAMPLES FOR PRACTICE. 2. The interest of $250 for 1 year, 3 months, is $28.125; what is the rate per cent.? Ans. 9 per cent. 3. If I pay $8.82 for the use of $72 for 1 year, 9 months, what is the rate per cent.? Ans. 7 per cent. 4. A note of $500, being on interest 2 years, 6 months amounted to $550; what was the rate per cent.? Ans. 4 per cent. 5. The interest for $700 for 1 year, 6 months, is $63; what'is the rate per cent.? Ans. 6 per cent. 6. If I pay $53.781 for the use of 8922 for 1 year, 2 months, what is the rate per cent.? Ans. 5 per cent. ART. 205e Problem IV. To find the time, the principal, interest, and rate per cent., being given. Ex. 1. For how long a time must $300 be on interest at 6 per,ent. to gain $36? Ans. 2 years. OPERATION. We find the interest on the $3 0 0 given principal for 1 year, by.0 6 which we divide the given in$1 8.0 0 ) 3 6.0 0 (2 years. terest. 3 6.0 0 Since the interest of $1 for 1 year is 6 cents, the interest of $300 will be 300 times as much, equal to $18. Now, if it require 1 year for the given, principal to gain $18, it will require as many years to gain $36 as $18 is contained times iu $36, which gives 2 years for the answer. RULE. - Divide the given interest by the interest of the given principal for 1 year, and the quotient will be the time. EXAMPLES FOR PRACTICE. 2. If the interest of $140 at 6 per cent. is $42, for how long a time was it on interest? Ans. 5 years. 3. How long a time must $165 be on interest at six per cent. to gain $14.85? Ans. 1 year, 6 months. 4. How long must $98 be on interest at 8 per cent. to gain $25.48? Ans. 3 years, 3 months. 5. A note of $680 being on interest at 4 per cent. amounted to $727.60; how long was it on interest? Ans. 1 year, 9 months. QUESTTONS. - Art. 205. What is Problem IV. Explain the operation. What is the rule for finding the time, the principal, interest, and rate per cent., being given?

Page 212 2 2 COMLPOUND INTEREST. [SECT. XXIL AaRT. a28,1 Proble3m V. To find the principal, the interest, time, and rate per cent., being given. Ex. 1. What principal at 6 per cent. will gain $36 in 2 years? Ans. $300. OPERATION. We find the interest of $1 for.0 t; int. of $1 flor ly. 2 yearts, by which we divide the'2 given interest..1 2 ) 3 6.0 0 ( $3 0 0 principal. Sinpe it equires 2 years for a prin(cipal of $1 to gain 12 cents, it will require a principal of as many dollars to gain 63G as.12 is contained tinmes in $36. Thus, $36.00 +.12 = 300 ior the anSweLr. UILE. - -Divide the gyiven interest or amount by the. interest or amount of $1 for the given rate and time, and the quotient will be the vprincipal. EXAMIPLES FOR PRACTICE. 2. WNhat principal will gain $24.225 in 4 years, 3 months, at: six per cent.? Ans. $95. 3.'What principal will gain $5.11 in 3 years, 6 months, at 8 per cent.? Ans. 818.25. 4. The interest on a certain note at 9 per cent. in 1 year and 8 months amounted to $42; what was the full amount of the note? Ans. $280o ~ XXIi. - COMPOUND INTEREST. ART. 20'6, COMPOUNrn INTEREST iS interest on the principal and interest together, when the interest is not paid.at the end of the year, or when dule. - The law specifies that the borrower of mnoney shall pay the lender a certain sumn for the use of' $100 for a year. Now, if he does not pay this sum at the end of the year, it is no more than just that he should pay interest for the -use of it as long as he shall keep it in his possession. The com-putation of compound interest is based upon this principle. QUrEsTToxs.- Art. 206. What is Problem V.? EIxplain the operation. What is the rule for finding the principal, the interest, time, and rate per cent., being given? - Art. 207, What is compoundinterest? On what prin ciple is it based?

Page 213 SECT. XXII.] COMPOUND INTEREST. 213 ART. 208~ To find the compound interest of any sum of money at any rate per cent. for any given time. Ex. 1. Wrhat is the compound interest of $500 for 3 years, 7 months, and 12 days, at 6 per cent.? Ans. $117.541. OPERATION. Pr incipel, $5 0 0 Interest of $1 for 1 year,.0 6 Interest for 1st year, 3 0.0 0 500 5 0 0 Amount for ist year, 5 3 0.0 0.06 Interest for 2d year, 3 1.8 0 0 0 5 3 0.0 0 Amount for 2d year, 5 6 1.8 0.0 6 Interest for 3d year, 3 3.7 0 8 0 5 6 1.8 0 Amount for 3d year, 5 9 5.5 0 8 Interest of $1 for 7mo. 12da.,.0 3 7 4168556 1786524 Interest for 7mo. 12da., 2 2.0 33 7 9 6 5 9 5.5 0 8 Amount for 3y, 7mo. 12da., 6 1 7.5 4 1 7 9 6 Principal subtracted, 5 00 Compound interest, $1 1 7.5 4 1 7 9 6 We first multiply the principal by the interest of $1 for 1 year, and add the interest thus found to the principal for the amount, or new principal. We then find the interest on this amount for 1 year, and proceed as before; and so also with the third year. For the months and days we find the interest on the amount for the last year, and, adding it as before, we subtract the original principal from the last amount for the answer. RULE. -TFind the interest of the given sum for one year, and add it to the princepal; then find the amount of this amount for the next year; and so coti7nue, until the tiLe of seltlemnent. If there are rtronths and days in the given time, find the amount for them on the amouo.lt for the last year. Sulitract the principal from the last amount, and the remainder zs the compound interest. QuEsTioNs.- Art. 208. Explain the operation in computing compound interest. What is the rule 1

Page 214 214 COMPOUND INTEREST. [SEcT. XXTL. Nore. -1. If the interest is to be paid semi-annually, quarterly, monthly, or daily, it must be computed for the half-year, quarter-year, month, or day, and added to the principal, and then the interest computed on this, and each succeedinyg amount thus obtnained, up to the time of settlement. 2. When partial payments have been matde on notes at compoulnd in terest, the rule is like that adopted in Art. 199. EXAMPLES FOR PRACTICE. 2. What is the compound interest of $761.75 for 4 years? Ans. i 99.9(t41. 3. WThat is the amount of 867.25 for 3 years, at compound interest? - Ans. $80.095. 4. What is the amount of $78.69 for 5 years, at 7 per cent.? Ans. $110.364. 5. What is the amount of $128 for 3 years, 5 months, and 1 8 days, at compound interest? Ans. $156.717. 6. What is the compound interest of $76.18 for 2 years, 8 months, 9 days? Ans. $12.967. ART. 209. There is a more expeditious method of computing compound interest than the preceding, by means of the following TABLE. Showing the amount of $1, or 21, for any number of years not exceeding 20, at 3, 4, 5, 6, and 7 per cent., compound interest. Years. 3 per cent. 4 per cent. 5 per cent. 6 per cent. 7 per cent. Years. 1 1.030000 1.040000 1.050000 1.060000 1.070000 1 2 1.060o00 1.081600 1.102500 1.123600 1.144900 2 3 1.092727 1.124864 1.157625 1.191016 1.225043 3 4 1.125508 1.169858 1.21.5506 1.262476 1.310796 4 5 1.159274 1.216652 1.276281 1.338225 1.402552 5 6 1.194052 1.265319 1.340095 1.418519 1.500730 6 7 1.229873 1.3156931 1.407100 1.503630 1.605781 7 8 1.266770 1.368569 1.477455 1.59,3848 1.718186 8 9 1.304773 1.423311 1.551328 1.689478 1.838459 9 10 1.3439. 16 1.480244 1.628894 1.790847 1.967151 10 11 1.384233 1.539454 1.710339 1.898298 2.104852 11 12 1.425760 1.601032 1. 795856 2.012196 2.252191 12 13 1.4685,;3 1.665073 1.885649 2.13',2928 2.408345 13 14 1.512589 1.731676 1.979931 2.260(903 2.57534 14 15 1.557967 1.800943 2.078928 2. 3B)6558 2. 750(32 15 16 1.60-1706 1.872981 2.182874 2.5-140351 2.-95216 16 17 1.652847 1.947900 2.292018 2.692772 3 1 58815 1 7 18 1.702433 2.025816 2.406619 2.854339 3.379 9932 1 8 19 1.7535,06 2.106849 2.526950 3.025599.616527 19 20 1.806111 2.191123 2.653297 3.207135 3.869t685 20 QUESTIONS. — If the interest is to be paid semi-annually, quarterly, &e., how is it computed? How, when partial payments have been made?

Page 215 SECT. XXII.] COMPOUNI) INTEitES1. 215 Ex. 1. What is the interest of $240 for 6 years, 4 months, and 6 days, at 6 per cent.? Ans. $107.593. OPERATION. Amount of $1 for 6 years, 1.4 1 8 5 1 9 Principal, 24'0 56740760 2837038 Amount of principal for 6 years, 3 4 0.444560 Interest of $1 for 4mo. 6da.,.0 2.1 34044456 68088912 Interest of amount for 4mo. 6da., 7.14 9 3 3 5 7 6 Amount added, 340.444560 Amount for 6y. 4mo. 6da., 347.593 8 9 5 76 Principal subtracted, 2 4 0 Interest for given time, $1 0 7.5 9 3 8 9 5 7 6 We find the amount of $1 for 6 years in the table, and multiply it by the principal, and obtain the amount for 6 years. We then find tile interest on this amount for the 4 months and 6 days, and add it to the first amount, and from this sum subtract the principal for the answer. hence, to find the compound interest by use of the table, Multiply the amount of $1 for the given rate and time, as found in he tadle, by the principal, and the product will be the amount. Subtract the principal from the amount, and the remainder will be the corn pound interest. If there are months and days, proceed as in the foregoing rule. EXAMPLES FOR PRACTICE. 2. What is the interest of $884 for 7 years, at 4 per cent.? Ans. $279.283. 3. What is the interest of $721 for 9 years, at 5 per cent.? Ans. 8397.507. 4. What is the amount of $960 for 12 years, 6 months, at 3 per cent.? Ans. $1389.26. 5. What is the amount of $25.50 for 20 years, 2 months, and 12 days, at 7 per cent.? Ans. $100.058. 6. What is the amount of $12 for 6 months, the interest to be added each month? Ans. $12.364+. 7. What is the amount of $100 for 6 days, the interest to be added daily? Ans. $100.10004.

Page 216 216 DISCOUNI. [SECT. XXIIL ~ XXIII. DISCOUNT. ART. 210, DISCOUNT is an allowance or deduction made for the payment of money before it is due. The present worth of any sum is the principal, which, being put at interest, will amount to the given sum in the time for which the discount is made. Thus, $100 is the present worth of $106, due one year hence at 6 per cent.; for $100 at 6 per cent. will amount to $106 in this time; and $6 is the discount. NoTE. - Business men, however, often deduct five per cent., or more, from the face of a bill due in six months, or a percentage greater than the legal rate of interest. ART. 2i1 The interest or percentage of any sum cannot properly be taken for the discount; for we see, from the preceding illustration, that the interest for one year is the fractional part of the sum at interest, denoted by the rate per cent. for the numerator, and 100 for the denominator; and the discount for one year is the fractional part of the sum on which discount is to be made, denoted by the rate per cent. for the numerator, and 100 plus the rate per cent. for the denominator. Thus, if the rate per cent. of interest is 6, the interest for one year is fTO of the sum at interest; but if the rate per cent. of discount is 6, the discount for one year is T6- of the sum on which discount is made. ART. 21,. In discount, the rate per cent., time, and the sitm on which the discount is made, are given to find the present worth. These termn correspond precisely to Problem VI. in interest, in which the time, rate per cent., and amount, are given to find the principal. (Art. 203.) ART. 213. To find the present worth and the discount on any sum, at any rate per cent., for any given tinie. Ex. 1. What is the present worth of $25.44, due one year hence, discounting at 6 per cent.? What is the discount? Ans. $24 present worth; $1.44 discount. QUESTIONS. -Art. 210. What is discount? What is the present worth of any sum of money? How illustrated? - Art. 211. Are interest and discount the same? Explain the difference. Which is the greater, the interest or discount on any sum, for a given time? -Art. 212. What terms are given in discount, and what is required? To what do these correspond in interest?

Page 217 SECT, XXIII ] DISCOUNT. 217 OPERATION. _Amount of>!, 1.0 6 ) 2 5.4 4 ( p2 4, present worth. 44 2 $2 5.4 4, sum or amount. 4.2 4 2 4.0 0, present worth. $1.4 4, discount. We find the amount of $1 for the given time, by which we divide the given sum, and obtain the present worth. Then, subtracting the present worth from the given sum, we obtain the discount. Since the present worth of $1.06, due one year hence, at 6 per cent., is $1, it is evident the present worth of $25.44 is as many dollars as $1.06 is contained times in $25.44. $25.44- 1.06 $24. Hence the following RULE.- Find the amount of $1 for the given time and rate; by which divide the given sum, and the quotient will be the present worth. The present worth subtracted from the given sum will give the discount..NOTE. - The discount may be found directly by making the interest of $1 for the given rate and time the numerator of a fraction, and the amount of $1 for the given rate and time the denominator, and then multiply the given sum by this fraction. ExAMPLES von PRACTICE. 2. What is the present worth of $152.64, due 1 year hence? Ans. $144. 3. WVhat is the present worth of $477.71, due 4 years hence? Ans. $385.25. 4. What is the discount of $172.86, due 3 years, 4 months hence? Ans. $28.81. 5. What is the discount of $800, due 3 years, 7 months, and 18 days hence? Ans. $143.186. 6. Samuel Heath has given his note for $375.75, dated Oct. 4, 1852, payable to John Smith, or order, Jan. 1, 1854; what is the real value of the note at the time given? Ans. $349.697. 7. Bought a chaise and harness of Isaac MIorse for $125.75, for which I gave him my note, dated Oct. 5, 1852, to be paid in 6 months; what is the present value of the note, Jan. 1, 1853? Ans. $123.81. QUESTIONS. - Art. 213. Explain the operation for finding the present worth and discount. Give the reason of the operation. What is the rule? What other method is given? 19

Page 218 21.8 COMMISSION AND BROKERAGE. [SECT. XXIV ~ XXIV. C0OMMISSION, BROKERAGE, AND STOCKS. ART. %24. Commission is the percentage paid to an agent factor, or commission merchant, for buying or selling goods, or transacting other business. Brokerage is the percentage paid to a dealer in money and stocks, called a broker, for making exchanges of money, nego. tiating different kinds of bills of credit, or transacting other business. Stocks is a general name given to government funds, state bonds, and to the capital of moneyed incorporations, such as banks, insurance, railroad, manufacturing, and mining companies. Stocks are usually divided into equal shares, the market value of which is often variable. When stocks sell for their original value they are said to be at par; when for more than their original value, above par, or in advance; when for less than their original value, below par, or at a discount. The premium, or advance, and the discount on stocks, are generally computed at a certain per cent. on the original value of the shares. The rate per cent. of commission or brokerage is not regulated by law, but varies in different places, and with the nature of the business transacted. Commission and brokerage are computed in the same manner. ART. 2g1S To find the commission or brokerage on any sum of money. Ex. 1. A commission merchant sells goods to the amount of 8879; what is his commission at 3 per cent.? Ans. $26.37. Since commission is a percentage on the given sum, the eom. mission on $879, at 3 per cent., will be $879 X.03 == $26.37. RULE. Find the percentage on the given sum at the given rate per cent., and the result is the commission or brokerage. (Art. 191.) QUESTIONS. - Art. 214. What is commission? What is brokerage? What is stock? Into what are stocks divided? When are stocks at par? When above par? When below par? How is the premium or discount on stocks computed? How is commission and brokerage computed -? Art. 215. What is the rule?

Page 219 8ECT. XXIV.] COi MISSION AND BROKERAGE. 219 EXAMPLES FOR PRACTICE 2. WVhat is the commission on the sale of a quantity of cotton goods valued at $5678, at 3 per cent.? Ans. 8170.34. 3. A commission merchant sells goods to the amount of $7896, at 2 per cent.; what is his commission? Ans. $157.92. 4. liy agent in Chicago has purchased wheat for me to the amount of 61728; what is his commission, at 1 per cent.? Ans. $25.92. 5. My factor advises me that he has purchased,'on my account, 97 bales of cloth, at $15.50 per bale; what is his commission, at 24 per cent.? Ans. $37.587. 6. My agent at New Orleans informs me that he has disposed of 500 barrels of flour at $6.50 per barrel, 88 barrels of apples at $2.75 per barrel, and 56cwt. of cheese at $10.60 per cwt.; what is his commission, at 33 per cent.? Ans. $153.21. 7. A broker negotiates a bill of exchange of $2500 at I per cent. commission; what is his commission? Ans. $12.50. 8. A broker in New York exchanged $46256 on the Canal Bank, Portland, at 4 of 1 per cent.; what did he receive for his trouble? Ans. $57.82. 9. A broker in Baltimore exchanged $20500 on the State Bank of Indiana, at I of 1 per cent.; what was the amount of his brokerage? Ans. $102.50. AaRT. 2it, To find the commission or brokerage on any sum of money, when it is to be deducted from the given sum, and the balance invested. Ex. 1. A merchant in Cincinnati sends $1500 to a commission merchant in Boston, with instructions to lay it out in goods, after ideducting his' commission of- 2. per cent.; what is his commission? Ans. $36.586. OPERATION. $1500 ~ 1.025 = j1463.414. $1500 - $1463.414 = $36.586. Since the agent is entitled to 21 per cent. of the amount he lays out, it is evident he requires $1.02- to purchase goods to the amount of $1. Hence, he can expend for goods as many dollars as 1.02A is contained times in $1500. On dividing we find he can expend $1463.414, which, being subtracted from $1500, the amount sent him, leaves as his commission $36.586. QUESTION. - How do you find the commission or brokerage when it is to be subtracted from the given sum?

Page 220 220 COMMISSION AND BROKERAGE. LSECT. XXIV RULE. - Divide the given sum by I increased by the per cent. of con.mission, and the quotient will be the sum to be invested. Subtract the sum to be investedfrom the yiven sum, and the remnain der wzll be the commission. EXAMPLES FOR PRACTICE. 2. A town agent has $2000 to invest in bank stock, after deducting his commission of 1 per cent.; what will be his commission, and what the sum invested? Ans. $29.557 commission; $1970.443 sum invested. 3. A shoe-dealer sends $5256 to his agent in Boston, which he wishes him to lay out for shoes, reserving his commission of 3 per cent.; what is his commission? Ans. $153.088. 4. A broker expends $3865.94 for merchandise, after deducting his commission of 4 per cent.; what was his commission, and what sum did he expend? Ans. $148.69 commission; $3717.25 sum expended. 5. I have sent to my agent at Buffalo, N. Y., $10000, which I wish him to expend for flour, after deducting his commission of 31 per cent.; what will be his commission, and also the value of the flour purchased? Ans. $314.76+ commission; $9685.23+- value of flour. Anr. 237X, To find the value of stocks, when at an advance or at a discount. Ex. 1. What is the value of $2150 railroad stock, at 7 per cent. advance? Ans. $2300.50. OPERATION. $2150 X.07 = $150.50; $2150 + $150.50 = $2300.50. RULE. - Find the percentage on the given sum, and add or subtract, according as the stock is at an advance or at a discount. (Art. 191.) EXAMPLES FOR PRACTICE. 2' What must be given for 10 shares in the Boston and l3aine Railroad, at 15 per cent. advance, the shares being $100 each.? Ans. $1150. 3. What must be given for 75 shares in the Lowell Railroad, at 25 per cent. advance, the original shares being $100 each? Ans. $9375. QUESTIONS. - What is the rule? - Art. 217. How do you find the value of etocks, when at an advance or at a discount? What is the rule?

Page 221 6Ecif XRV.] BANKING. 221 4. What is the purchase of $8979 bank stock, at 12 per cent. advance? Ans. $10056.48. 5. What is the purchase of $1789 bank stock, at 9 per cent. below par? Ans. $1627.99. 6. A stocklholder in the Illinois Central Railroad sells his right of purchase on 5 shares of $100 each at 12 per cent. advance; what is the premium? Ans. $60. 7. Wrhat is the value of 20 shares canal stock, at 12j per cent. discount, the original shares being $100 each? Ans. $1750. 8. What is the value of 15 shares in the Livingston County Bank, at 8S per cent. advance, the original shares being $100 each? Ans. $1623.75. 9. Bought 87 shares in a certain corporation, at 12 per cent. below par, and sold the same at 192 per cent. above par; what sum did I gain, the original shares being $175 each? Ans. $4795.87. ~ XXV. BANKING. ART. 218, A BANK is a joint stock company, established fo~ the purpose of receiving deposits, loaning money, dealing in exchange, or issuing bank notes or bills, as a circulating medium, redeemable in specie at its place of business. The capital of a bank is the money paid in by its stockholders, as the basis of business. Banking is the general business commonly transacted at banks. NoTE. - The persons chosen by the stockholders to manage the affairs of the bank are called its board of directors, who select one of their own number as president, and some person as cashier. The president and cashier sign the bills issued, which also are, in some instances, counntersigned by some state officer. The cashier superintends the bank accounts; and another person, called the teller, usually receives and pays out money. A check is an order drawn on the cashier of the bank for money. QUESTIONS. - Art. 218. What is a bank? What is the capital of a, bank? What is banking? Who choose the directors? Who choose the president and cashier? Who sign the bills issued? Who superintends the accounts? Wiho receives and pays out the money? What is a check? 1 9*

Page 222 2_22 BANK DISCOUNT. [SECT. XXIV BAN,, DISCOUNT. ART. Uio90 BANK DISCOUNT iS the simple interest of a note draft, or bill of exchange, deducted from it in advance, or before it becomes due. The interest is computed, not only for the specified time, but also for three days additional, called days of grace. Thus, if a note is given at the bank for 60 days, the interest, which is called the discount, is computed for 63 days; and if the note is paid within this time, the debtor complies with the requirements of the law. The legal rate of discount is usually the same as the legal rate of interest; and the difference between bank discount and true discount is the same as the difference between interest and true discount. A note is said to be discounted at a bank, when it is received as security for the money that is paid for it, after deducting the interest for the time it was given. The sum paid is called the avails or present worth of the note. ART. 220. To find the present worth and the bank discount of any note or sum of money for any rate per cent. and time. Ex. 1. What is the bank discount on $842 for 90 days, at 6 per cent.? What is the present worth? Ans. $13.051 discount; $828.949 present worth. OPERATION. Sum discounted, $ 8 4 2 $ 8 4 2.0 0 0 Interest of $1,.01 5 5 1 3.0 5 1 4 2 1 0 $ 8 2 8.9 4 9 present worth. 4210 842 Bank discount, $1 3.0 5 1 0 We find the interest of $1 for 93 days, by which we multiply the sum discounted, and the product is the discount. We then subtract the discount from the given sum, and obtain the present worth. QuEsTIoNs. -Art. 219. What is bank discount? When is it paid? Is interest computed fcr more than the specified time? What are these three additional days called? How will you illustrate this? What is the legal rate of discount? What is the difference between bank discount and true discount? When is a note said to be discounted at a bank? What is the sum paid for it called? - Art. 220. Explain the operation for finding the bank discount on any sum.

Page 223 SECT. XXV.] BANK DISCOUNT. 223 RULE. — Find the interest on the note, or sum discounted, for the given rate and time, including THREE days of grace, and this interest is the discount. Subtract the discount from the face of the note or sum discounted, ana the remainder is the present worth. EXAMPLES FOR PRACTICE. 2. What is the bank discount on $478 for 60 days? Ans. $5.019. 3. What is the bank discount on $780 for 30 days? Ans. $4.29. 4, What is the bank discount on $1728 for 90 days? Ans. $26.784. 5. How much money should be received on a note of $1000, payable in 4 months, discounting at a bank where the interest is 6 per cent.? Ans. $979.50. 6. W'hat sum must a bank pay for a note of $875.35, payable in 7 months and 15 days, discounting at 7 per cent.? Ans. $836.542. 7. What are the avails of a note of $596.24, payable in 8 months and 9 days, discounted at a bank at 8 per cent.? Ans. $562.85. 8. What is the bank discount of a draft of $1350.50, payable in I year, 4 months, at 5 per cent.? Ans. $90.596. ART. 22l, To find the amount for which a note must be given at a bank, to obtain a specified sum for any given time. Ex. 1.. For what amount must a note be given, payable in 90 days, to obtain $500 from a bank, discounting at 6 per cent.? Ans. $507,872. OPERATION. We subtract the interest of $1.0 0 0 0 $1 for 93 days, at six per cent., Int. of $1 for 93da.,.0 1 5 5 from $1, and divide the given Present worth of $1,.9 8 4 5 sum by the remainder, for the answeor. $500 e+.9845 = $507.872 Since $0.9845 requires $1 principal for the given time, $500 will require as many dollars principal as $0.9845 is contained times in $500; and $500 - $0.9845 = $507.872. Hence the QUESTIONS. - What is the rule? — Art. 221. Explain the operation for finding the amount for which a note must be given at a bank to obtain a specified sum for a, given time.

Page 224 224 INSURANCe. [SECT. XXVI RULE. - Divide the yiven sum by the present worth of $1 for th given time and rate per cent. of bank discount, including THREE day of grace, and the quotient will be the answer. EXAMPLES FOR PRACTICE. 2. For what sum must I give my note at a bank, payable in 4 months, at 6 per cent. discount, to obtain $300? Ans. $306.278. 3. A merchant sold a. quantity of lumber, and received a note payable in 6 months; he had his note discounted at a bank, at 6 per cent., and received $4572.40. What was the amount of his note? Ans. $4716.245. 4. A gentleman wishes to take $1000 from the bank; for what sum must he give his note, payable in 5 months, at 6 per cent. discount? Ans. $1026.167. 5. The avails of a note, discounted at the bank for 8 months, at 71 per cent., were $483.56; what was the face of the note? Ans. $509.345. ~ XXVI. INSURANCE. ART. 222. INSURANCE iS indemnity obtained, by paying a certain sum, against such losses of property or of life as are agreed upon. The party taking the risk is called the insurer or underwriter, and the party protected, the insured. The policy is the written obligation, or contract, entered into between the parties. Premium is the amount of percentage paid on the property insured for one year, or any specified time. As a security against fraud, property is not usually insured for its whole value, nor is the insurer or underwriter bound to indemnify the insured for a loss more than is specified in the policy. QUESTIONS. - What is the rule? - Art. 222. What is insurance? What is the party called that takes the risk? What is the party called that is protected? What is the policy? What is the premium? Is property usually insured to its whole value?

Page 225 SECT. XXVII.] CUSTOM-HOUSE BUSINESS. 225 ART. 223, To find the premium on any amount of property insured. Ex. 1. What is the premium on $485 at 2 per cent.? Ans. $9.70 OPERATION. $485 X.02 = $9.70. RULE. - Find the percentage on the given sum, and the result is the premium. (Art. 191.) EXAMPLES rOR PRACTICE. 2. What is the premium on $868 at 12 per cent.? Ans. $104.16. 3. What is the premium on $1728 at 15 per cent.? Ans. $259.20. 4. A house, valued at $3500, is insured at 1- per cent.; what is the premium? Ans. 861.25. 5. A vessel and cargo, valued at $35000, are insured at 34 per cent.; now, if this vessel should be destroyed, what will be the actual loss to the insurance company? Ans. $33687.50. 6. A cotton factory and its machinery, valued at $75000, are insured at 21 per cent.; what is the yearly premium? and if it should be destroyed, what loss would the insurance company sustain? Ans. $1875 premium; $73125 loss. ~ XXVII. CUSTOM-HOUSE BUSINESS. ART. 221o DUTIES are sums of money required by govern ment to be paid on imported goods. All goods from foreign countries brought into the United States are required to be landed at particular places, called porlts of entry, where are custom-houses, at which the duties or revenue is collected. Duties are either specific or ad valorema. A specifc duty is a certain sum paid on a ton, hundred weight, yard, gallon, &c. QUESTIONS. -Art. 223. What is the rule for finding the preimiumn on aily amount of property insured?-Art. 224. What are duties? Where are duties collected? What is a specific duty?

Page 226 226 CUSTOM-I-HOUSE BUSINESS. VSECT. XXVII. An ad valoreim duty is a certain per cent. paid on the actua cost of the goods in the country fiom which they are imported Draft is an allowance for waste made in the weight of goods. Tare is an allowance made for the weight of the ca-sk, box &c., containing the commodity. Leakage is an allowance for waste made on liquors. Gross weight is the weight of the commodity, together with the cask, box, bag, &c., containing it. Net weight is what remains after all allowances have been made. By the present tariff, all duties are levied on the ad valoremn principle.e It has been decided that no allowances for tare, draft, breakage, &e., are applicable to imports subject to ad valorem duties, except actual tare, or-weight of a cask, or package, and- the actual drainage, leakage, or damage. The collector may cause these to be ascertained, when he has any doubts as to what they are. AnT. 225o To find the ad valorem duty on goods or merchcandise. Ex. 1. At 25 per cent., what is the ad valoriem duty on 165 yards of broadcloth, at $5 per yard? Ans. $206.25. OPERATION. 165 >< 5 825; $ 825 X.25 - 206.25, duty. RULE. - Find the percentaye on the cost of the goods, and the result is the ad valoremn duty. (Art. 191.) EXYXAMPLES FOr PRACTICE. 2. What is the duty on 17281b. of copper sheathing, invoiced at 83200, at 20 per cent. ad valorem? Ans. $640. 3. What is the duty on 22311b. of Russian iron, at 30 per cent. ad valorem; the cost of the iron being 4 cents per lb.? Ans. $26.772, duty. 4. What is the duty on 16911b. of lead, at 20 per cent. ad valorem; the value of the lead being 5 cents per pound? Ans. $16.91, duty. QUESTrIoNS. - What is an ad valorem duty? What is draft? Tare? Gross weight? Net weight? - Art. 225. What is the rule for finding the ad valo orem cluty?

Page 227 SECT. XXVIII.] ASSESSMENT OF TAXES. 227 5. What is the duty on 10 hogsheads of molasses, each hogs. head gauging 150 gallons gross, the actual wants being 5 gallons to each hogshead, and the cost of the molasses 25 cents per gallon; duty 20 per cent. ad valorem? Ans. $72.50, duty. 6. What are the net weight and duty, at 30 per cent. ad valorem, on 13 boxes of sugar, weighing gross 450 pounds each; actual tare 15 per cent., and the cost of the sugar being 8 cents per pound? Ans. 49721lbs., net weight; $119.34, duty. 7. What is the duty on an invoice of woollen goods, which cost in Liverpool 1376~. sterling, at 30 per cent. ad valorem; the pound sterling being $4.84? Ans. $1997.95+. 8. What is the duty on an invoice of goods, which cost in Paris $2340, at 80 per cent. ad valorem? Ans. $1872. 6 XXVIII. ASSESSMENT OF TAXES. ART. 226. A TAX is a sum of money assessed by government for public purposes, on property, and in most states on persons. Taxes may be either direct or indirect. A direct tax is one imposed on the income or property of an individual; an indirect tax is one imposed on the articles for which the income or property is expended. A poll or capitation tax is one without regard to property, on the person of each male citizen, liable by law to assessment. A person so liable is termed a poll. Immovable property, such as lands, houses, &c., is called real estate. All other property, such as money, notes, cattle, furniture, &c., is called personal property. The method of assessing taxes is not precisely the same in all the states, yet the principle is virtually the same. The following is the law regulating taxation in Massachusetts (Revised Statutes, p. 79): QUESTIONS. -Art. 226. What is a tax? What is a direct tax? What an indirect tax? What is real estate? What is personal property? What is a poll or capitation tax? What is a poll? Is the method of assessing taxes the same in all the states?

Page 228 223 ASSESSMENT OF TAXES. [SECT. XXVIIL " The assessors shall assess upon the polls, as nearly as the same can be conveniently done, one sixth part of the whole sum to be raised; provided the whole poll tax assessed in any one year upon any individual for town and county purposes, except highway taxes, shall not exceed one dollar and fifty cents; and the residue of said whole sum to be raised shall be apportioned upon property;" that is, on the real and personal estate of individuals which is taxable. ArT. 22 a To assess a town or other tax. Ex. 1. The tax to be assessed on a certain town is $2200. The real estate'of the town is valued at $60000, and the per. sonal property at $30000. There are 400 polls, each of which is taxed $1.00. What is the tax on $1.00? What is A's tax, whose real estate is valued at $2000, and his personal property at $1200, and who pays for 2 polls? OPERATION. $1.00 X 400 = $400, amount assessed on the polls. $2200 - $400 = $1800, am't to be assessed on the property. $60000 + $30000 -$90000, amount of taxable property. $1800 -$- 890000 = $0.02, tax on $1.00. $2000 X.02 = $40, A's tax on real estate. $1200 X.02 = $24, A's tax on personal property. $1.00 X 2 = $2, A's tax on 2 polls. $40 + $24 + $2 =- 66, amount of A's tax. Hence, in assessing taxes, it is necessary to have an inventory of the taxable property, and, if -a levy on the polls is to be included, there should be also a complete list of taxable polls. Having these, we then Multiply the tax on each poll by the number of taxable polls, and subtract the product from the whole sum to be raised, which gives the sum to be raised on the property. Tize sum to be raised on property divided by the whole taxable property, will give the sum to be paid on each dollar of property taxed. Each man's taxable property multiplied by the sum to be paid on $1, wzth his poll tax added to the product, will give the amount of his tax. EXAMPLES FOR PRACTICE. 2. The town of L. is taxed $3600. The real estate of the town is valued at $560,000, and the personal property at $152,500. There are 600 polls, each of which is taxed $1.25. What is the per cent. or tax on $1.00? and what is B's tax, QUESTIONS. -What is the law regulating taxation in Massachusetts?Art. 227. What is the rule for assessing taxes?

Page 229 SOcT. XXVIII.I ASSESSIVENT OF TAXES. 229 whose real estate is valued at $4100, and his personal property cat $1800, he paying for four polls? Ans. $.004, tax on $1; $28.60, B's tax. 3. What is the tax of a nonresident, having property in the sa.me town, worth $15800? Ans. 863.20. 4. What is D's tax, who pays for 3 polls, and whose real estate is valued at $40000, and his personal property at $23600? Ans. $258.15. ART. 228. The operation of assessing taxes may be facilitated by the use of a table, which can be easily made after having found the tax on $1. Ex. 1. A tax of $3900 is to be assessed on the town of P. The real estate is valued at $840000, and the personal property at $210000; and there are 500 polls, each of which is taxed $1.50. What is the assessment on $1? Ans. $.003. Having found the tax on $1 to be $.003, before proceeding to make the assessment on the inhabitants of the town, we find the tax on $2, $3, &c., and arrange the numbers as in the following TABLE. $1 gives $.003 $20 gives $.06 $300 gives $.90 2'".006 30 ".09 400 A" 1.20 3".009 40 ".12 500 " 1.50 4 ".012 50 ".15 600 "6 1.80 5 6.015 60.18 700 " 2.10 6 ".018 70 6.21 800 "6 2.40 7 ".021 80 s.24 900 " 2.70 8 ".024 90 ".27 1000 6 3.00 9 ".027 100 ".30 2000 " 6.00 10 4.030 200 "6.60 3000 4 9.00 2. What is E's tax, by the above table, whose property, real and personal, is valued at $1860, and who pays 3 polls? Ans. $10.08. OPERATION. Tax on $ 1 0 0 0 is $ 3.0 0 We find the tax on $1000 8 0 0" 2.4 0 in the table, and then on 6 0 ".1 8 $800, and then on $60, G6. 3 p ols" and to these sums add the ~ polls 1 4.5 0'tax on the 3 polls for the Valuation, $1 8 6 0 $ 1 0.0 8, Tax, answer. QUESTIONS. - Art 228. 2low may the operation of assessing taxes be facilitated'! How is the above table formed? 20

Page 230 230 EQUATION OF PAYMENTS. 3, What is F's tax, whose real estate is valued at $6535, ana his personal property at $3175, and who pays for 6 polls? Ans. $38.13. 4. What is Mrs. G's tax, who has property to the amount of $7980? Ans. $23.94. 5. If H pays for 2 polls, and has property to the amount of $4790, what is his tax? Ans. $17.37. 6. M's real estate is valued at $9280, and his personal prop, erty at $3600; what is his tax, if he pays for 4 polls? Ans. $44.64. ~ XXIX. EQUATION OF PAYMENTS. ART. 9, EQUATION OF PAYMENTS is the process of' finding the average or mean time when the payment of several sums, due at different times, may all -be made at one time, without loss either to the debtor or creditor. ART. 23,. To find the average or mean time of payment, when the several sums have the same date. Ex.. John Jones owes Samuel Gray $100; $20 of which is to be paid in 2 months, $40 in 6 months, $30 in,8 months, and $10 in 12 months; what is the average time for the payment of the whole sum? Ans. 6mo. 12da. OPERATION. It is evident that the interest $20 X 2 =!- 4 0 of $20 for two months is the $40 X 6 - 2 4 0 same as the interest of $1 for $ 3 0 X 8 2 4 0 40 months; and of $40 for 6 $1 0 X 12 12 0 mo., the same as of $1 for 240 mo.; and of $30 for 8 mo., the $1 0 0 1 00 ) 6 40 ( 6 mo. same as of $1 for 240 m^.; and 6 0 0 of $10 for 12 mo., the c:me as of $1 for 120 mo. Hen e, the 4 0 interest of all the sums.o the 3 0 times of their payment, is the same as the interest of $1 for 40 10 0 ) 1 2 00 (12 de. + 240o 240 + 120 - =6`0 mo. 1200 Nowit $1 require 640 mo to gain a certain sum, $20 +-[ QUESTIONS. -Art. 229. What is equation of payments? - Art. 230. Why in the example, do we multiply the $20 by 2?

Page 231 SECT. XXIX.] EQUATION OF PAYMENTS. 231 $40 + $30 + $10 = $100 will require only, Td of this time; and 640 mo. + 100 = 6 mo. 12 da., the average or mean time for the payment of the whole. Hence the following iULE. - Multiply each payment by its own time of credit, and divide the sum of the products by the sum of the payments. NOTE 1. - This is the rule usually adopted by merchants, but it is not peifectly correct; for ifI owe a man $200, $100 of which I was to pay down, and the other $100 in two years, the equated time for the payment of both sums would be one year. It is evident that, for deferring the payment of the first $100 for 1 year, I ought to pay the amount of $100 for that time, which is $106; but for the other $100, which I pay a year before it is due, I ought to pay the present worth of $100, which is $94.8833; whereas, by equation of payments, I only pay $200. NOTE 2.- When a payment is to be made down it has no product, but it must be added with the other payments in finding the average time. EXAMPLES FOR PRACTICE. 2. John Smith owes a merchant in Boston $1.000, $250 of which is to be paid in 4 months, $350 in 8 months, and the remainder in 12 months; what is the average time for the payment of the whole sum? Ans. nmo. 18da. 3. A gentleman purchased a house and lot for $1560, 4 of which is to be paid in 3 months, 5 in 6 months, 6 in 8 months, and the remainder in 10 months; what is the average time of payment? Ans. 7 9 1 months. 4. Samuel Church sold a farm for $4000; $1000 of which is to be paid down, $1000 in one year, and the remainder in 2 years; but he afterwards agreed to take a note for the whole amount; for what time must the note be given? Ans. 15 months. 5. A wholesale merchant in Boston sold a bill of merchandise to the amount of $5000 to a retail merchant of Exeter, N. II.; he is to pay 4 of the money down, I of the remainder in 6 months, 2 of what then remains in 9 months, and the rest at the end of the year. If he wishes to pay the whole at once, what will be the average time of payment? Ans. 6mo. 27da. AgT. 23A To find the average or mean time of paymept, when the several sums have different dates. Ex. 1. Purchased of James Brown, at sundry times, and on QUESTIONS. - What is the rule for equation of payments? Is the rule perfectly correct? Explain why it is not. When a payment is to be made down, what is to be done with it?

Page 232 232 EQUATION OF PAkMENTS. [SEeCT. XXIX various terms of credit, as by the statement annexed. WVhen is the,mediumz time of payment? Jan. 1, a bill amounting to $360, on 3 months' credit, Jan. 15, do. do. 186, on 4 months' credit. larch 1, do. do. 450, on 4 months' credit. Mlay 15, do. do. 300, on 3 months' credit. June 20, do. do, 500, on 5 months' credit. Ans. July 25, or in 115 da. OPERATION. Due April 1, $ 3 60 May 15, 186X 44- 8 184 July 1, 450X 91= 40950 Aug. 15, 300X1 36= 40800 ZNov. 20, $500 X 233 = 116500 1 7 9 6 )2 0 6 4 3 4 ( 1 454i days. 1796 2683 1 7 9 6 8874 7184 169 0 We first find the time when each of the bills will become due. Then, since it will shorten the operation and bring the same result, we take the first time when any bill becomes due, instead of its date, for the period from which to compute the average time. Now, since April 1 is the period from which the average time is computed, no time will be reckoned on the first bill, but the time for the payment of the second bill extends 44 days beyond April 1, and we multiply it by 44. (Art. 230.) Proceeding in the same manner with the remaining bills, we find the average time of payment to be 115 days nearly, from April 1, or on the 25th of July. IHence, in like cases, Find the time when each of the sums becomes due. Multiply each sum by the number of days intervening between the date of its hecomingn due and the earliest date on which any sum becomes due. Thlen proceed as in the rule (Art. 230), and the quotient will be the average time required, in days,from the earliest payment. NOTE. - In the work, if there be a fraction of a day less than 1, it ntay be rejected; but if more than 4, it may be reckoned as 1 day. QUESTIONS. - Art. 231. What is the rule for finding the average time, when there are different dates? By what other method can'you obtain nearly the same result?

Page 233 SecT XXIX.] EQUATION OF PAYMENTS. 233 EXAMPLES FOR PRACTICE, 2. I have purchased several parcels of goods, at sundry times, and on various terms of credit, as by the following statement. What is the average time for the payment of the whole? Jan. 1, 1856, a bill amounting to $175.80, on 4 months' cr. 16, a" do. do. 96.46, on 90 days'' Feb. 11, "6 do. do. 78.39, on 3 months' 6 " 23, " do. do. 49.63, on 60 days' " Mar. 19, 6" do. do. 114.92, on 6 months' 6 Ans. May 30, or in 45 da. 3. Sold S. Dana several parcels of goods, at sundry times, and on various terms of credit, as by the following statement: Jan. 7, 1854, a bill amounting to $375.60, on 4 months' cr. April 18, " do. do. 687.25, on 4 months' 46 June 7, 1" do. do. 568.50, on 6 months' " Sept. 25, " do. do. 300.00, on 6 months'" Nov. 5, " do. do. 675.75, on 9 months' " Dec. 1,'; do. do. 100.00, on 3 months' 6 What is the average time for the payment of all the bills? Ans. Dec. 24, or in 231 da. 4. The following is my account against G. M. Holbrook, and I wish to ascertain the average time of payment. Jan. 1, 1857, 97 yards of broadcloth, at $4.50, on 3 mos.' cr. Feb. 10, 6' 7 bales of cotton cloth, " 18.50, on 60 days' AMay 1, 66 9 tons of iron, " 45.00, on 4 mos.' 6 June 15, " 11 hhds. of molasses, " 12.00, on 30 days' " July 5, " 8 doz. shovels, 6 9.00, on 2 mos.'" Sept. 25, "6 4cwt. of sugar, " 6.50, on 1 mo.'s Dec. 1, 66 8 chests of tea, " 15.00, on 90 days' Ans. July 16, or in 106 da. 5. The following is an account of my bills against J. Crowell: Jan. 1, 1854, a bill amounting to $300, on 6 months' credit. June 1, "6 do. do. 500, on 5 months' " Sept. 1, " do. do. 200, on 6 months' " Feb. 1, 1855, do. do. 800, on 8 months' " July 1, 1856, do. do. 400, on 9 months' 6 Dec. 1, " do. do. 900, on 7 months' " May 1, 1857, do. do. 100, on 3 months' 6 What is the average time of payment on the above bills? Ans. March 9, 1856, or in 20 mo. 8 da. 20*

Page 234 234 EQUATION OF PAYMENTS. [SECT. XXIX ART. 232 To find the average or mean time of paying the balance of a debt, when partial payments have been made before the debt is due. Ex. 1. I have purchased goods to the amount of $800, on a credit of 6 months. At the end of 2 months I pay $100, and at the end of another month I pay $200 more. How long, in equity, after the expiration of the 6 months, ought the balance to remain unpaid? Ans. 2 months. OPERATION. The interest on the $ 100 X 4 = 400 $100 for 4 months is $2 00 X 3 = 6 O 0 equal to the interest of $1 for 400 months; and 3 0 0 5 0 0 ) 1 0 0 0 the interest of the $200 8 0 0-$ 3 0 0 = $ 5 0 0 2 mo. for 3 months, to that of $1 for 600 months; and thus the interest on both partial payments, at the expiration of the 6 months, is-equal to the interest of $1 for 400 +- 600, or 1000 months. To equal this credit of interest, the balance of the debt, which we find to be $500, should remain unpaid, after the 6 months, 5-6- of 1000 months, or 2 months. RULE. - Multiply each payment by the time, in months or days, zt was made before it became due, and divide the sum of the products by the balance remaining unpaid. The quotient will be the average time required. EXAMPLES FOR PRACTICE. 2. Sold, March 11, 1855, James Stone goods to the amount of $1850, on a credit of 4 months. I received from him, April 7, $400; May 15, $270; and June 20, $350. WVhen in equity should I receive the balance? Ans. Sept. 22, 1855. 3. Bought, June 12, 1855, of William Jones, goods to the amount of $1200, on a credit of 8 months. I paid him, September 1, $400; November 1, $200; and December 1, $100, When in equity can he require the balance of me? Ans. Aug. 17, 1856. 4. I sold, September 25, 1855, John Eckles 144 barrels of flour, at $12 per barrel, and 370 bushels of wheat, at $3 per bushel, on 6 months' credit. I received of him, September 25, $1000; November 1, $800; and December 21, $600. When ought I to be paid the remainder? Ans. June 14, 1858. QUESTION. - Art. 232. What-fs the rule for finding the average time of paying the balance of a debt, when partial payments have been made?

Page 235 SECT. XXIX.] EQUATION OF PAYMENTS. 235 5. Wilson Seymour bought March 20, 1855, of a merchant in Troy, merchandise to the amount of $2000, on 6 months' credit. I-e pays down $500; May 10, $350, and June 7, $400. When ought he to pay the balance? Ans. May 18, 1856. ART. 233. To equate an account containing items of both debit and credit. Ex. 1. At what time did the balance of the following account become due, allowing that each item drew interest from its date? Dr. Martin Jordan in account with David Hill 4 Co. Cr. 1856. 1856. Jan. 22, To merchandise, $89 00 Jan. 4, By merchandise, $77 00 6s 24, 6" " 76 00 Apr. 16, " C 40100 Feb. 20, " " 25 00 May 14, 6 6 14300 "' 23, 6" " 210100 April 4,9 " I I 189 00 May 21,'6 66 30 00 Ans. February 9, 1856. OPERATION. Debits. Credits. Jan. 22, $89 Jan. 4, $77 C( 24, 76X 2 = 152 April 16, 40X103= 4120 Feb. 20, 25 X 29= 725 May 14, 143X131=18733.' 23, 210 X 32: 6720 - days. April 4, 189 X 73.13797 260 )22853(872 May 21, 30 X 120 = 3600 2080 -.... days. 2080 619 )24994(40o 2053 2476 1820 234 233 Average date of purchase, 40 days Average of credits, 88 days from from Jan. 22, or on March 2. Jan. 4, or on April 1. Difference between March 2 260 and April 1 = 30 dasys, 30 359) 7800 (212 6 1 days. $619 - $260: $359, or balance, 718 620 359 261 22 days back from March 2 = February 9. QUESTION. — Art. 233. How do you equate an account having items of debit and credit?

Page 236 236 EQUATION OF PAYMENTS. [SECT. XXIX On equating each side of the account (Art. 230), we find the dehits $619, became due 40 days from January 22, or on iMarllch 2; and the credits, $260, became Clue 88 days from January 4, or on April 1. If thle account had been settled on oMarch 2, it is evident the credits, $2G0, would have been paid 30 days, or the time from March 2 to April 1, before having become due. This would have been a loss of the use or interest of that sum to the credit side of the account, and a corresponding gain to the debit side. Now, as the settlement is required to be one of equity, we find how long it will take the balance of the account, $359, to gain the same interest that $260 would gain in the 30 days. If it takes $260 to gain a certain interest in 30 days, it would take $1 to gain the samne interest 260 times 30 days, or 7800 days; and $359 to gain the same amount of interest "-E of 7800 days, or 22 days nearly. I-Hence, the balance became due 22 days back of March 2, or on February 9, which is the answer sought. In this example, the time was counted back from the average date of the larger amount, since it became due first; but when that amount becomes due last, the time is counted forward from its average time. RULE. - Find the average time of each side becoming due. Multiply the amount of the smaller side by the number of days between the two average dates, and divide the product by the balance of the account.'The quotient will be the time of the balance becoming due, counted Jfrom the average date of the laryer side, BACK when the amount of that side is due FIRST, but FORWARD when it is clue LAST. NoTE. -Having the average time of a balance becoming due, its CAsH VALUE can be ascertained wvhen the balance is due before the time of setthing the account, by adding to it the interest up to the time of settlement e and when duse after that time, by finding the present worth (Art. 213), fromn the time of settlement to the time of the balance becoming due. EXrAMIPLES FOR PRACTICE. 2. In settling the following account, when did the balance become due, the merchandise items being on 6 months'credit? Dr. Hiram Lewis in account with Joseph Warren. Cr. 1854. 1854. Feb. 16, To merchandise, $375 80 Mar. 20, By cash, $30000 April 8, " " 43218 June 17,' merchandise, 371 50 May 17, " " 320 15 July 4, " cash, 200 00 July 13, " 6 158 12 Sept. 25, "merchandise, 85 20 Ans. March 3, 1855. QUJESTIONS. - What is the rule? How can the cash value of the balance of an account be found?

Page 237 OECT. XXX.] RATIO. 237 3. Edward Doton owes Daniel Stetson, 1855, May 1, for merchandise, $500; May 15, for timber, $400; June 14, for a horse, $300; July 24, for bill of labor, $100. Stetson owes Doton, 1855, March 7, for a pleasure-boat, $400; April 2, for merchandise, $200; May 6, for merchandise, $300; June 13, for a carriage, $120. Allowing all the items to be on 6 months' credit, when will the balance of the account become due? Ans. April 27, 1856. ~ XXX. RATIO. ART. 2340 RATIO is the relation, in respect to magnitude or value, which one quantity or number has to another of the same kind, or the quotient arising from the division of one number by another. Thus, the ratio of 6 to 3 is 2. Of the two numbers necessary to form a ratio, the first is called the antecedent, and the last the consequent. Thus, in the example given, 6 is the antecedent, and 3 the consequent. When there is but one antecedent and one consequent, the ratio is called a simple ratio. The antecedent and consequent are also called the terms of the ratio. ART. 5235~ A ratio may be expressed in two ways. The ratio of 6 to 3 may be expressed by two dots between the terms, thus, 6: 3; or in the form of a fraction, by making the antecedent the numerator and the consequent the denominator, thus, a. The terms of a ratio must be of the same kind, or such as may be'educed to the same denomination, in order that they may have a ratio to each other. Thus, shillings have a ratio to shillings, and shillings to pounds, &c.; but shillings have not a ratio to gallons, nor pounds to days, because they are not commensurable. ART. 236. A ratio may be either direct or inverse. A direct ratio is when the antecedent is divided by the consequent; an inverse ratio is when the consequent is divided by the antecedent. Thus, the direct ratio of 6 to 3 is 6, and the inverse ratio of 6 to 3 is 3, or 1. The direct ratio of one quantity or number to another is found by dividing the nuTmber whose ratio is required, which is QUESTIONS. - Art. 234. What is ratio? How many numbers are necessary to form a ratio? What are the antecedent and consequent called? - Art. 235. What two ways are there of expressing a ratio? - Art. 236. What is a direct ratio? What an inverse ratio?

Page 238 '238 RATIO. [SECT. XXX the antecedent, by the number with which it is compared, which is the consequent. The inverse ratio is found by reversing this process. EXAMPLES FOR PRACTICE. 1. WVhat is the direct ratio of 9 to 3? Ans. 3. Of lS to 6? Of 16 to 4? Of-24 to 12? Of 20 to 5? Of 15 to 3? 2. What is the direct ratio of 7 to 21? Ans. ~. Of 4 to 28? Of 6 to 30? Of 9 to ll? Of 9 to 99? Of 30 to 907 3. WVhat is the direct ratio of 60 to 12? Of 132 to 11? Of 40 to 120? Of 32 to 96? Of 200 to 50? Of 144 to 1728? Of 360 to 60? 4. What is the inverse ratio of 10 to 5? Ans. -. Of 27 to 81? Of 16 to 48? Of 72 to 9? Of 11to 88? 5. What is the direct ratio of 2~. 5s. to 9s.? Ans. 5. Of 9in. to ift. 6in.? ART. 237. A compound ratio consists of two or -more simple ratios, whose corresponding terms are to be multiplied together. Thus, The simple ratio of 8 4 is 2 And " " of 1 2:3 is 4 The compound ratio of 8 X 1 2 4 X 3 is 2 X 4 Or " " of 96:12 is8 When a compound ratio is composed of two equal ratios, it is called a duplicate ratio; when of three, it is called a triplicate ratio, &c. The simple ratio of 4: 2 is 2',' d" of 6:3 is 2 "4 4 6" of 8 4 is 2 The triplicate ratio of 4 X 6 X 8: 2 X 3 X 4 is X 2 X 2 X 2 Or " " of 192:24 is8 If the terms of a ratio are both multiplied or divided by the same nzumber, the ratio is not altered. Thus, the ratio of 8:2 is 4; the ratio 8 X 2:2 X 2 is 4; and the ratio of 82 2 2is 4. QUESTIONS.- Art. 237. What is a compound ratio? mhat a, duplicate ratio? What a triplicate ratio? What is the effect of multiplying or dividing the terms of a ratio?

Page 239 SECT. XXXI.] PROPORTION. 239 ~ XXXI. PROPORTION. ART. 238. PROPORTION is an equality of ratios. Thus the ratios 9: 3 and 12: 4 are equal, and when united form a proportion. Proportion is usually expressed by four dots between the two ratios; thus, the proportion in the preceding example is written 9: 3:: 12: 4, and is read, 9 is to 3 as 12 to 4. The numbers which form a proportion are called proportionals. The first and third are called antecedents, the second and fourth are called consequents; also, the first and last are called extremes, and the remaining two the means. ART. 239, Any four numbers are said to be proportional to each other when the first contains the second as'many times as the third contains the fourth; or when the second contains the first as many times as the fourth contains the third. Thus, 9 has the same ratio to 3 that 12 has to 4, because 9 contains 3 as many times as 12 contains 4. ART. 240. If the antecedents or consequents of a proportion, or both, are divided by the same number, they are still proportionals. Thus, dividing the antecedents of the proportion 4: 8:: 10: 20 by 2, we have 2: 8:: 5: 20; dividing the consequents by 2, we have 4: 4:: 10: 10; and dividing both the consequents and antecedents by 2, we have 2: 4:: 5: 10; each of which is a proportion, since if we divide the second term of each by the first, and the fourth by the third, the two quotients will be equal. The effect is the same when the terms are multiolied by the same number. ART. 24l. The product of the extremes of a proportion is equal to the product of the means. Thus, the proportion 14: 7:: 18: 9 may be expressed fractionally, 14- — 8I. Now, if we reduce these fractions to a common denominator, we have 1-62 — 1=-; but in this operation we multiplied together the two extremes of the proportion, 14 and 9, and the two means, 18 and 7; thus, 14 X 9 - 18 X 7. QUESTIONS. - Art. 238. What is proportion? How is proportion expressed? What are the numbers called that form a proportion? Which are the extremes? Which the means? - Art. 239. When are numbers said to be in proportion to each other? - Art. 240. What is the effect of dividing the antecedents or consequents of a proportion? Of multiplying them? - Art. 241. How does the product of the extremes compare with that of the means?

Page 240 240 SIMPLE PROPORTION. [SECT. XXXL ART. 242. If the extremes and one of the means are given the other mean may be found by dividing the product of the extremes by the given mean. Thus, if the extremes are 3 and 24, and the given mean 6, the other mean is 12; because 24 X 3 =72; and 72. 6 - 12. ART. 243. If the means and one of the extremes are given the other extreme may be found by dividing the product of the means by the given extreme, Thus, if the means are 8 and 16, and the given extreme 4, the other extreme is 32; because 16 X 8 - 128; and 128 - 4 =32. SIMPLE PROPORTION. ART. 244. SIMPLE PROPORTION is an expression of the equality between two simple ratios. NOTE. - Simple Proportion is sometimes called the Rule of Three. ART. 245. Method of stating and solving questions in Simple Proportion. Ex. 1. If 7ib. of sugar cost 56 cents, what will 361b. cost? Ans. $2.88. OPERATION. Since 71b. have the same ratio to Extreme. Mean. Mlean. 361b. as 56 cents, the cost of the 7 lb.: 3 6 lb.: 5 6 cts. former, have to the cost of the lat3 6 ter, we have the first three terms 3 3 6 of a proportion given, namely, one of the extremes and the two means. 1 6 8 Now, to ascertain which of these 7 ) 2 0.1 6 terms are the means, and which the $2.8 8 Extreme. extreme, we arrange them in the order of a proportion, or state the question, by making 56 cents the third term, because it is of the same kind, and has the same proportion to the required answer, or fourth term, as the first has to the second. From the nature of the question, since the answer will be more than 56 cents, or the third term, the second term must be greater than the first; we make 361b. the second term, and 71b. the first, and then proceed as in Art. 246. QuEsTIONs. - Art. 242. If the extremes and one of the means are given, how can the other mean be found? - Art. 243. When the means and one of the extremes are given, how can the other extreme be found? —Art. 244. What is simple proportion? How many terms are given in questions in simple proportion?

Page 241 SECT. XXX1.] SIMPLE PROPORTION. 241 BY ANALYSIS. - If 71b. cost 56 cents, 1 pound will cost I of 56 cents, which is 8 cents. Then, if 1lb. cost 8 cents, 361b. will cost 36 times as much; that is, 36 times 8 cents, which are $2.88, Ans. as before. Ex. 2. If 76 barrels of flour cost $456, what will 12 barrels cost? Ans. $72. QPERATION. We state this question by making bar. bar. $. $456 the third term, because it is of 7 6: 1 2:: 4 5 6 the same kind of the required answer. 1 2 Then, since the answer must be less 7 6 ) 5 4 7 2 ( $7 2 than $456, because 12 barrels will A6) 5 4 7 2 ($7 2 cost less than 76 barrels, we make 12 5 3 2 barrels, the smaller of the other two 1 5 2 terms, the second term, and 76 bar1 5 2 rels the first term, and proceed as before. BY ANALYSIS. - If 76 barrels cost $456, 1 barrel will cost -7-1 of $456, which is $6. Then, if 1 barrel cost $6, 12 barrels will cost 12 times as much; that is, $72, Ans. as before. Ex. 3. If 3 men can dig a well in 20 days, how long will it take 12 men? Ans. 5 days. OPERATION. Since the required answer is days, we men. men. days. make 20 days the third term. And as 12 1 2: 3:: 20 men will dig the well in less time than 3 3 men, the answer must be less than 20 days. 1 2 6 0 Therefore we make 3 men the second term, and 12 men the first, and proceed as in the 5 days. other examples. BY ANALYSIS.- If 3 men dig the well in 20 days, it will take one man 3 times as long, that is, 60 days. Again, we say, If one man dig the well in 60 days, 12 men would dig it in I~ of 60 days, that is, 5 days, Ans. as before. From the preceding examples we deduce the following RULE. - Write the given number that is of the same kind as the required fourth term, or answer, for the third term of the proportion. Of the other two numbers, write the larger for the second term, and the less for the first, when the answer should exceed the third term; but write the less for the second term, and the larger for the first, when the answer should be less than the third term. Multiply the second and third terms together, and divide their product by the first. QuESTIONS. - What is meant by stating the question? Which of the terms given in the example do you make the third? Why? Which the second? Why? Which the first? Why? After the question is stated, how do you obtain the answer? 21

Page 242 242 SIMPLE PROPORTION. [SECT. XXXL NoTE 1.- When the first and second terms consist of different denominations, they must be reduced to the same denomination; and when the third term is a compound number, it must be reduced to the lowest denomination mentioned in it. The answer will be the same denomination as the third term. NOTE 2. - TO shorten the operations, factors common to the dividend and divisor may be cancelled. NoTE 3. - The pupil should perform these questions by analysis as well as by proportion, and- introduce cancellation when it will abbreviate the operation. Ex. 4. If 16 bushels of wheat are worth $24, what are 96 bushels worth? Ans. $144. OPERATION BY CANCELLATION. We first state the question as dibu. bu. $. rected in the rule, and then write the second and third terms above a hori 6 9 6 zontal line, with the sign of multiplication between them, for a divi9 0 X 2 4 dend, and the first term below the -$1 4 4 line for a divisor, and cancel the a 0t common factors. By this method of analysis we first place the $24, which is of the same BY ANALYSIS AND CANCELLATION. kind of the required answer, above 6 a line for a dividend; and then say, $2 4 X 9 0 Since $24 is the price of 16 bushels, = $14 4 1 bushel will cost 1 of $24, and exY 0 press the division by placing the 16 below the line for a divisor. Now, since we have an expression for the price of 1 bushel, we next express the multiplication of it by 96 bushels, the price of which is required, and then cancel as before. EXAMPLES FOR PRACTICE. 5. What cost 9 gallons of molasses, if 63 gallons cost $14.49? Ans. $2.07. 6. What cost 97 acres of land, if 19 acres can be obtained for $337.25? Ans. $1721.75. 7. If a man travel 319 miles in 11 days, how far will he travel in 47 days? Ans. 1363 miles. 8. When $120 are paid for 15 barrels of mackerel, what will be the cost of 79-barrels? Ans. $632. QuESTIoNs. - What is the rule for simple proportion? How should the pupil perform the questions? How do you state the question and arrange the terms for cancellation? What do you cancel? How do you arrange the terms for cancellation by analysis?

Page 243 EUCT. XXXI.] SIMPLE PROPORTION. 243 9. If 9 horses eat a load of hay in 12 days, how many horses would it require to eat the hay in 3 days? Ans. 36 horses. 10. When $5.88 are paid for 7 gallons of oil, what cost 27 gallons? Ans. $22.68. 11. When $10.80 are paid for 91b. of tea, what cost 1471b Ans. $176.40. 12. What cost 27 tons of coal, when 9 tons can be purchased for $85.95? Ans. $257.85. 13. If 15 tons of lead cost $105, what cost 765 tons? Ans. $5355.00. 14. If 16hhd. of molasses cost $320, what cost 176hhd.? Ans. $3520.00. 15. If 15cwt. 3qr. 171b. of sugar cost $124.67, what cost 76cwt. 2qr. 191b.? Ans. $600.56+. 16. If the cars on the Boston and Portland Railroad go one mile in 2 minutes and 8 seconds, how long will they be in pass ing from Haverhill to Boston, the distance being 32 miles? Ans. lh. 8min. 16sec. -17. If a man travels 3m. 7fur. 18rd. in one hour, how far will he travel in 9h. 45min. 19sec.? Ans. 38m. 2fur. 32rd+ 18. A fox is 96 rods before a greyhound, and while the fox is running 15 rods the greyhound will run 21 rods; how far will the dog run before he can catch the fox? Ans. 336 rods. 19. If 5 men can reap a field in 12 hours, how long would it take them if 4 men were added to their number? Ans. 62 hours. 20. Ten men engage to build a house in 63 days, but 3 of their number being taken sick, how long will it take the rest to complete the house? Ans. 90 days. 21. If a 4 cent loaf weighs 5oz. when flour is $5 per barrel, what should it weigh when flour is $7.50 per barrel? Ans. 3~oz. 22. If 7 men can mow a field in 10 days when the days are 14 hours long, how long would it take the same men to mow tbe field when the days are 13 hours long? Ans. 101~ days. 23. If 291b. of butter will purchase 401b. of cheese, how many pounds of butter will buy 791b. of cheese? Ans. 57l 1b. 24. If X of a yard cost g of a dollar, what will {1 of a yard cost? Ans. $0.76-7gf. yd. yd. $. -a:1+A::%5; W X Gil X l= is $0.76iu, Ans.

Page 244 244 SIMPLE PROPORTION. [S:crT. XXXI 25. If 47 yards of cloth cost $27, what will 191 yards cost9 Ans. $11.50. yd. yd. $. $ 23 47: 19~: 2-; 2X X = _ -- 8.11.50, Ans. 26. If for 4-T1 yards of velvet there be received 114 yards of calico, how many yards of velvet will be sufficient to purchase 100 yards of calico? Ans. 39Q89 yards. 27. A certain piece of labor was to have been performed by 1,4 men in 36 days, but a number of them having been sent away, the work was performed in 48 days; required the number of men discharged. Ans. 36 men. 28. James can mow a certain field in 6 days, John can mow it in 8 days; how long will it take John and James both to mow it? Ans. 38 days. 29. A. Atwood can hoe a certain field in 10 days, but with the assistance of his son Jerry he can hoe it in 7 days, and he and his son Jacob can hoe it in 6 days; how long would it take Jerry and Jacob to hoe it together? Ans. 9,- days. 30. Bought a horse for $75; for what must I sell him to gain 10 per cent.? 1.00: 1.10::$75 $82.50, Ans. 31. Bought 40 yards of' cloth at $5.00 per yard; for what must I sell the whole amount to gain 15 per cent.? Ans. $230.00. 32. My chaise cost $175.00, but, having been injured, I am willing to sell it at a loss of 30 per cent.; what should I receive? Ans. $122.50. 33. Bought a cargo of flour on speculation at $5.00 per barrel, and sold it at $6.00 per barrel; what did I gain per cent.? Ans. 20 per cent. 34. Bought a hogshead of molasses for $15.00, but, it not proving so good as I expected, I sell it for $12; what do I lose per cent.? Ans. 20 per cent. 35. Bought a hogshead of molasses for $27.50, at 25 cents per gallon; how much did it contain? Ans. 110 gallons. 36. A certain farm was sold for $1728, it being $15.75 per acre; what was the quantity of land? Ans. 109A. 2R. 3427p. 37. A certain cistern has 3 cocks; the first will empty it in 2 hours, the second in 3 hours, and the third in 4 hours; in what time would they all empty the cistern together? Ans. 55 5 minutes.

Page 245 SECT. XXXI.J COMPOUND PROPORTION. 245 COMPOUND PROPORTION. ART. 246. COMPOUND PROPORTION is an expression of the equality between a compound and a simple ratio. It is employed in performing such questions as require two or more operations in Simple Proportion. ART. 247. Method of stating and solving questions in Compound Proportion, sometimes called the Double Rule of Three. Ex. 1. If $100 will gain $8 in 12 months, what will $600 gain in 10 months? Ans. $40. OPERATION. In' stating Extreme. Mean. this question, $ 1 0 0 $ 6 0 0 Mean. this question, $100 $600 a$. 8we make $8, 1 2 mo. 1 0 mo. $ 8 the gain,which is the same 0O0OX X8- 480 $ 4 0, Extreme. name of the 100 x 12 1200' required answer, the third term. Then, taking $100 and $600, two of the remaining terms of the same kind, we inquire if the answer, depending on these alone must be greater or less than the third term; and since it must be greater, because $600 will gain more than $100 in the same time, We make $600 the second term, and $100 the first. Again, we take the two remaining terms, and make 10 mo. the second term, and 12 mo. the first, since the same sum would gain less in 10 mo. than in 12 mo. We then find the continued products of the second and third terms, and divide it by the continued product of the first terms, for the answer. Hence the following RULE. - Make that number which is of the same kind as the answer required the third term of a proportion. Of the remaining numbers, take any two, that are of the same kind, and consider whether an answer, depending upon these alone, would be greater or less than the third term, and place them as directed in Simple Proportion. Then take any other two, and consider whether an answer, depending only upon them, would be greater or less than the third term, and ar range them accordingly; and so on until all are used. QUESTIONS. — Art. 246. What is compound proportion? For what is it employed?- Art. 247. By what other name is it sometimes called? In stating the question, which of the numbers do you make the third term? Why? What do you do with the remaining terms? How do you know which of the two to take for the second term? Which for the first? After all the terms have been arranged, how do you find the answer? What is the rule for compound proportion? 21*

Page 246 246 COMPOUND PROPORTION. [SECT. XXXI Multiply the product of the second terms by the third, and divide the result by the product of the first terms. The quotient will be the fourth term, or answer. NoTE. - The following questions should be performed not only by the rule, but by an analysis and cancellation. Ex. 2. If $100 will gain $6 in 12 months, what will $800 gain in 8 months? OPERATION BY CANCELLATION. $100: $800 } 12 mo.: 8 mo. $ 6 We state the question according to 8 4 the -rule, and then write the second 8 0 0 X $ X 0 and third terms for a dividend and = $ 3-2 the first terms for a-divisor, and can[I 0 0 X I $ cel the common factors.! BY ANALYSIS AND CANCELLATION. By this method of analysis 4 8 we say, if $6 are the gain of $ X $ mo. X $$0 32 $100 inl 12 mo., in 1 mo. the mo. X $ 0 $30 gain of $100 will be - as much, or -T, and in 8 mo. 8 times as much, or $1-X28. Again, if $100 gain $-q — in 8 mo., $1 will gainTl0H of it, or $26 X 800 and $800 will gain 800 times as much, or $w612 X00, the $12 X 100' 12 X 100 same as in the operation. Cancelling the common factors, we obtain $32 for the answer. EXAMPLES rOR PRACTICE. 3. If $100 gain $6 in 12 months, in how many months will $800 gain $32? Ans. 8 months. 4. If $100 gain $6 in 12 months, how large a sum will it require to gain $32 in 8 months? Ans. $800. 5. If $800 gain $32 in 8 months, what is the per cent.? Ans. 6 per cent. 6. If 15 carpenters can build a bridge in 60 days when the days are 15 hours long, how long will it take 20 men to build the bridge when the days are 10 hours long? Ans. 671 days. QUESTIONS. - How does the author say the questions under this rule should be performed? How are questions stated for cancellation? Which terms are taken for the dividend? Which for the divisor? What are cancelled?

Page 247 SECT. XXXI.] COMPOUND PROPORTION. 247 7. If a regiment of soldiers, consisting of 939 men, can eat 351 bushels of wheat in 3 weeks, how many soldiers will it require to eat 1404 bushels in 2 weeks? Ans. 5634 soldiers. 8. If 8 men spend $64 in 13 weeks, what will 12 men spend in 52 weeks? Ans. $384. 9. If 8 horses consume 42 bushels of grain in 24 days, how many bushels will suffice 32 horses 48 days? Ans. 336 bushels. 10. If 6 men in 16 days of 9 hours each build a wall 20 feet long, 6 feet high, and 4 feet thick, in how many days of 16 hours each will 24 men build a wall 200 feet long, 16 feet high, and 6 feet thick? Ans. 90 days. 11. If a man travel 117 miles in 15 days, employing only 9 hours a day, how far would he go in 20 days, travelling 12 hours a day? Ans. 208 miles. 12. If 12 men in 15 days can build a wall 30 feet long, 6 feet high, and 3 feet thick, when the days are 12 hours long, in what time will 30 men build a wall 300 feet long, 8 feet high, and 6 feet thick, when they work 8 hours a day? Ans. 240 days. 13. If the carriage of 5cwt. 3qr. 150 miles cost $24.58, what -must be paid for the carriage of 7cwt. 2qr. -151b. 32 miles, at the same rate? Ans. $6.97-. -14. A received of B $9 for the use of $600 for 6 months; now B3 wishes to hire of A $1800 until the interest shall amount to the same sum. How long may he keep it? Ans. 2 months. 15. If 15 oxen or 20 cows will eat 3 tons of hay in 8 weeks, how much hay will be sufficient for 15 oxen and 8 cows 12 weeks? Ans. 6 3 tons. 16. If 5 men, by laboring 10 hours a day, can mow a field of 30 acres in 10 days, how long will it require 8 men and 7 boys, provided each boy can do T7 as much as a man, to mow a field containing 54 acres? Ans. 7 3 3 days. 17. If 2 men can build 123 rods of wall in 6- days, how long will it take 18 men to build 247 2 rods? Ans. 14 days. 18. If 248 men, in 5- days of 11 hours each, dig a trench of 7 degrees of hardness, and 232 fMeet long, 32 feet wide, and 2~ feet deep, in how many days of 9 hours each will 24 men dig a trench of 4 degrees of hardness, and 337z feet long, 55 feet wide, and 31 feet deep? Ans. 132 days.

Page 248 248 PROFIT AND LOSS. [SecT. XXXII ~ XXXII. PROFIT AND LOSS. ART. 248. PROFIT and Loss is the process by which merchants and other traders estimate their gain or loss in buying and selling goods. The following questions may be performed either by analysis or by proportion. ART. 249. To find the profit or loss per cent. when the cost and Belling price are given. Ex. 1. If I buy flour at $4 per barrel, and sell it at $5 per barrel, what is the gain per cent.? Ans. 25 per cent. OPERATION. $5 $4 - $1; = 1.00 -- 4 =.25, or 25 per cent. By subtracting the cost from the selling price, we find the gain per barrel to be $1. Now, if the gain is $1 on $4, on $1 it will be I of $1 = i of a dollar,.or 25 per cent. OPERATION BY PROPORTION. $5 - $4 = _$1; $4: $1:: 1.00:.25, that is, 25 per cent. 2. If I buy flour at $5 per barrel, and sell it at $4 per barrel, what is the loss per cent.? Ans. 20 per cent. OPERATION. 5 - $4 $1; = = 1.00 -5.20, or 20 per cent. By subtracting the selling price from the cost, we find the loss per barrel to be $1. Now, if the gain is $1 on $4, on $1 it-will be 4 of $1 -- of a dollar, or 20 per cent. From this analysis and that of the preceding example, it is seen that the operation is equivalent to making the gain or loss the numerator of a fraction, and the cost the denominator, and then reducing this fraction to a decimal; or, in short, to simply dividing the gain or loss by the cost. OPERATION BY PROPORTION. $5 - $4 = $1; $5: $1: 1l00 per cent: 20 per cent. RULE 1.- Divide the gain or loss by the cost, and the quotient will be the gain or loss per cent. Or, RULE 2. - As the cost is to the gain or loss, so is 100 per cent. to thegain or loss per cent. QUESTIONS. — Art. 248. What is profit and loss? What is the first rule for finding the profit or loss in buying or selling goods? What is the second rule?

Page 249 BUCT. XXXII.] PROFIT AND LOSS. 249 EXAMPLES FOR PRACTICE. 3. Bought 40 yards of broadcloth at $5.40 per yard, and I sell a of it at $6 per yard, and the remainder at $7 per yard; what do I gain per cent.? Ans. 152Q per cent. 4. A merchant purchased for cash 50 barrels of flour, at $5 per barrel, and immediately sold the same on 8 months' credit, at $5.98 per barrel; what does he gain per cent.? Ans. 15 per cent. 5. A grocer bought a hogshead of molasses, containing 100 gallons, at 30 cents per gallon; but 30 gallons having leaked out, he disposed of the remainder at 40 cents per gallon. Did he gain or lose, and how much per cent.? Ans. Lost 62 per cent. 6. A gentleman in Rochester, N. Y., purchased 3000 bushels of wheat, at $1.12~ per bushel. He paid 5 cents per bushel for its transportation to N. Y. city, and then sold it at $1.371 per bushel; what did he gain per cent.? Ans. 17,a per cent. 7. J. Morse bought, in Lawrence, a lot of land 7-aT rods square, for $5 per square rod. HIe %old the land at 5 cents per square foot; what did he gain per cent.? Ans. 1721 per cent. ART. 250. To find the selling price when the cost and the gain or loss per cent. are given. Ex. 1. If I buy flour at $4 per barrel, for how much must I sell it per barrel to.gain 25 per cent.? Ans. $5. OPERATION. $4 X.25 - $1.00; then $4 + $1 - $5, Ans. It is evident, if I sell the flour for 25 per cent. gain, I sell it for.25 more than it cost. Therefore, if I add to the cost.25 of the cost, the sum will be the price per barrel for which the flour must be sold; as seen in the operation. OPERATION BY PROPORTION. 1.00 +.25 - 1.25; 1.00: 1.25 $4: $5, Ans. 2. If I buy flour at $5 per barrel, for what must I sell it per barrel to lose 20 per cent.? QUESTION. - Art. 250. Explain how you find the selling price when the cost bud the gain or loss per cent. are given.

Page 250 250 PROFIT AND LOSS. [SScT. XXXII OPERATION. $5 X.20 = $1.00; $5. - $1 = $4, Ans. It is evident, if I sell the flour for 20 per cent. loss, I sell it for.20 less than it cost. Therefore, if I subtract from the cost.20 of the cost, the remainder will be the price per barrel for which the flour must be sold; as seen in the operation. OPERATION BY PROPORTION. 1.00-.20 =.80; 1.00:'.80 —:: $5: $4, Ans. IRULE 1. - Find the percentage on the cost at the given rate per cent., and add it to the cost, or subtract it fromz the same, according as the selling price is to be that of profit or loss. Or, RULE 2. - As 1 is to 1 with the profit per cent. added, or lossper cent. subtracted, expressed decimally, so is the given price to the price required. EXAMPLES FOR PRACTICE. 3. Bought a hogshead of molasses, containing 120 gallons, for 30 cents per gallon, but it not proving so good as was expected, I am willing to lose 10 per cent. on the cost; what shall I receive for it? Ans. $32.40. 4. A grocer bought a hogshead of sugar, weighing net 8cwt. 3qr. 51b., for $88; for what must he sell it per pound to gain 20 per cent.? Ans. 12 cents per pound. 5. J. Simpson bought a farm for $1728; for what must it be sold to gain 12 per cent., provided he is to wait 8 months, without interest, for his pay? Ans. $2012.77+. 6. J. Fox purchased a barrel of vinegar, containing 32 gallons, for $4; but 8 gallons having leaked out, for how much must he sell the remainder per gallon to gain 10 per cent. on the cost? Ans. $0.18A per gallon. 7. Bought a horse for $90, and gave my note to be paid in 6 months, without interest; what must be my cash price to gain 20 per cent. on my bargain? Ans. $104.84+. 8. H. Tilton bought 7cwt. of coffee at $11.50 per cwt., but finding it injured, he is willing to lose 15 per cent.; for how much must he sell the 7cwt.? Ans. $68.42-. ART. 251. To find the cost when the selling price and the gain or loss per cent. are given. Ex. 1. If I sell flour at $5 per barrel, and by so doing make 25 per cent., what was the cost of the flour? Ans. $4 per barrel. QUESTIONS. -Art. 250. What is the first rule for finding at what price goods must be sold to gain or lose a given per cent.? What is the second rule?

Page 251 SECT. XXXIII.I PROFIT AND LOSS. 251 OPERATION. $5.00 +. 1.25 = $4, Ans. Since the gain evidently is 25 per cent. of the cost, the selling price, $5, is equal to the cost increased by 25 per cent. of the cost; or, as it may be expressed, equal to 1.25 of the cost. Hence, the cost must be as many dollars as the $5 contains times 1.25; and, dividing, we obtain $4, the answer required, OPERATION BY PROPORTION. 1.00.-.25 =1.25; 1.25: 1.00-: $5: $4, Ans. 2. If I sell flour at $4 per barrel, and by so doing lose 20 per cent., what was the cost of the flour? Ans. $5 per barrel. OPERATION. $4.00 -.80 = $5, Ans. Since the loss evidently is 20 per cent. of the cost, the selling price, $4, is equal to the cost decreased by 20 per cent. of the cost; or, as it may be expressed, equal to.80 of the cost. Hence, the cost must be as many dollars as the $5 contains times.80; and, dividing, we obtain $5, the answer required. OPERATION BY PROPORTION. 1.00 —.20 =.80;.80: 1.00::$4: $5, Ans. RULE 1. - Divide the selling price by 1 increased by the gain per cent., or by 1 decreased by the loss per cent., expressed decimally, and the quotient will be the cost. Or, RULE 2. - As 1 with the gain per cent. added, or loss per cent. subtracted, expressed decimally, is to 1, so is the selling price to the cost. EXAMPLES FOR PRACTICE. 3. Having used my chaise 16 years, I am willing to sell it for $80; but by so doing I lose 621 per cent.; what was the cost of the chaise? Ans. $213.331. 4. If I sell wood at $7.20 per cord, and gain 20 per cent., what did the wood cost me per cord? Ans. $6 per cord. 5. J. Adams sold 40 cases of shoes for $1600, and gained 18 per cent.; what was the first cost of the shoes? Ans. $1355.93+. QUESTIONS.- Art. 251. What is the first rule for finding the cost, when the selling price and the gain or loss per cent. are given? What is the second rule?

Page 252 252 PROFIT AND LOSS. [SECT. XXXIL 6. Sold 17 barrels of flour at $8 per barrel, for which I received a note payable'in 3 months. This note I had discounted at the Granite Bank, but, on examining my account, I find I have lost 10 per cent. on the flour; what was the cost of it? Ans. $148.76 —. ART. 252. The selling price of goods and the rate per cent. being given, to find what the gain or loss per cent. would be, if sold at another price. Ex. 1. If I sell flour at $5 per barrel, and gain 25 per cent., what should I gain if I were to sell it for $7 per barrel? OPERATION. The solution of this question involves two principles: First, to find the cost of the flour per barrel. (Art. 251.) Thus, $5.00 + 1.25 = $4.00, the cost per barrel. Second, to find the gain per cent. on the cost when sold at $7 per barrel. (Art. 248.) Thus, $7 - $4 = $3; 3.00. 4-.75, or 75 per cent. OPERATION BY PROPORTION. 1.00 +.25 =1.25; $5: $7:: 1.25: 1.75; 1.75 - 1.00 =.75, that is, 75 per cent. RULE 1. —Find the cost (Art. 251), and then the gain or loss per cent. on this cost at the proposed sellinzq price. (Art. 248.) Or, RULE 2. - As the first price is to the proposed price, so is 1 with the gain per cent. qf the first price added, or the loss per cent. of the first price subtracted, to 1 with the gain per cent. of the proposed price added, or with the loss per cent. of the proposed price subtracted. NOTE. -If the result by the last rule exceeds 1.00, the excess is the gain per cent.; but, if it is' less than 1.00, the deficiency is the loss per cent. EXAMPLES FOR PRACTICE. 2. Sold a quantity of oats at 28 cents per bushel, and gained 12 per cent.; what per cent. should I gain or lose, if I were to sell them at 24 cents per bushel? Ans. Lose 4 per cent. 3. S. Rice sold a horse for $37.50, and lost 25 per cent:; what would have been his gain per cent. if he had sold him for $75? Ans. 50 per cent.;4. S. Phelps sold a quantity of wheat for $1728, and took QUESTTONS.- Art. 252. What is the first rule for finding what gain or loss is made by selling goods at another price when the selling price and rate per cent. are given? What is the second rule? If the answer exceeds $100 what is the excess? If it is less than $100, what is the deficiency?

Page 253 SECT. XXXII.] PROFIT AND LOSS. 253 a note payable in 9 months without interest, and made 10 per cent. on his purchase; what would have been his gain per cent. if he had sold it to James Wilson for $2000 cash? Ans. 33- per cent. MISCELLANEOUS EXERCISES IN PROFIT AND LOSS. 1. A horse that cost $84, hhving been injured, was sold for $75.60; what was the loss per cent.? Ans. 10 per cent. 2. Sold a horse for $75.60, and lost 10 per cent. on the cost; but, if I had sold him for $97.44, what per cent. should I have gained on the cost of the horse? Ans. 16 per cent. 3. M. Star sold a horse for $97.44, and gained 16 per cent.; what would have been his loss per cent. if he had sold the horse for $75.60, and what his actual loss? Ans. Loss 10 per cent. $8.40 loss. 4. If I buy cloth at $5 per yard, on 9 months' credit, for what must I sell it per yard for cash to gain 12 per cent.? Ans. $5.35+-. 5. A. Pemberton bought a hogshead of molasses, containing 120 gallons, for $40; but 20 gallons having leaked out, for what must he sell the remainder per gallon to gain 10 per cent. on his purchase? Ans. $0.44. 6. H. Jones sells flour, which cost him $5 per barrel, for $7.50 per barrel; and J. B. Crosby sells coffee for 14 cents per pound, which cost him 10 cents per pound; which makes the greater per cent.? Ans. H. Jones makes 10 per cent. most. 7. J. Gordon bought 160 gallons of molasses, but having sold 40 gallons, at 30 cents per gallon, to a man who proved a bankrupt, and could pay only 30 cents on the dollar, he disposed of the remainder at 35 cents per gallon, and gained 10 per cent. on his purchase; what was the cost of the molasses? Ans. $41.45+. 8. D. Bugbee bought a horse for $75.60, which was 10 pec cent. less than his real value, and sold him for 16 per cent. more than his real value; what did he receive for the horse, and what per cent. did he make on his purchase? Ans. Received $97.44, and made 288 per cent. 9. A merchant bought 70 yards of broadcloth that was 1~ yards wide, for $4.50 per yard, but the cloth having been wet, it shrunk 5 per cent. in length, and 5 in width; for what must the cloth be sold per square yard to gain 12 per cent.? 22 Ans. $3.19-+. 22

Page 254 254 PARTNERSHIP. [SECT. XXXIII ~XXXIII. PARTNERSHIP, OR COMPANY BUSINESS. ART. 253. PARTNERSHIP is the association of two or more persons in business, with an agreement to share the profits and losses in proportion to the amount of capital stock, or the value of the labor and experience of each. The association is called a Firm, or Comnpany; the money or property invested is called the Joint Stock, or Capital; each of the owners is called a Partner, and the profit or gain the Dividend. ART. 254. To find each partner's share of the profit or loss when the stock is employed for the same time. Ex. 1. John Smith and Henry Gray enter into partnership for three years; Smith puts in $4000, and Gray $2000. They gain $570. What is each man's share of the gain? Ans. Smith's gain, $380; Gray's gain, $190. OPERATION. $ 4 0 0 0, Smith's stock, V, Smith's part of the stock. $ 2 0 0 0, Gray's " z 0= o, Gray's part of the stock. $ 6 0 0 0, Whole stock. Then 2 of $ 5 7 0, the whole gain, - $ 3 8 0, is Smith's share of the gain. And of$ 7 0, " " i= $1 9 0, is Gray's share of Proof, $ 5 7 0 the gain. Since the sum of $4000 and $2000, equal to $6000, is the whole stock, it is evident that Smith's part of the stock is B4ur = 23; and that Gray's part is Aoo ~ a. Then, since each man's gain must be in proportion to his stock,. of $570, = $380, is Smith's share of the gain; and ~ of $570, = $190, is Gray's share of the gain. OPERATION BY PROPORTION. $ 6 000': $ 5 70':: $ 4 000'$: $ 3 8 0, Smith's gain. $6000: $570:: $ 2 000: $19 0, Gray's gain. QUESTIONS. — Art. 253. What is partnership? What is the association called? What the property invested? What are the owners called? What the profit or loss?

Page 255 SiECT. XXXIII.j PARTNERSHIP. 255 RULE 1. - Multiply the whole gain or loss by each partner's fractional part of the stock, and the product will be the gain or loss of each. Or, RULE 2. - As the whole stock is to each partner's stock, so is the whole gain or loss to each partner's gain or loss. EXAMPLES FOR PRACTICE. 2. Three merchants, A, B, and C, engaged in trade. A put in $6000, B put in $9000, and C put in $5000. They gain $840. What is each man's share of the gain? Ans. A's gain $252, B's gain $378, C's gain $210. 3. A bankrupt owes Peter Parker $8750, James Dole $3610, and James Gage $7000. His effects, sold at auction, amount to $6875; of this sum $375 are to be deducted for expenses, &c. What will each receive of the dividend? Ans. Parker, $2937.75 1 ~;o Dole, $1212.03 62; Gage, $2350.20802T. 4. A merchant, failing in trade, owes A $500, B $386, 0 $988, and D $126. His effects are sold for $100. What will each man receive? Ans. A receives $25.00, B $19.30, C $49.40, D $6.30. 5. A, B, and C, engaged in trade. A put in $700, B put in $300, and C put in 100 barrels of flour. They gained $90; of which sum C took $30 for his part; what will A and B receive, and what was C's flour valued per barrel? Ans. A receives $42, B $18, C's flour $5 per barrel. ART. 255, To find each partner's share of the profit or loss, when the stock is employed for unequal times. Ex. 1. Josiah Brown and George Dole trade in company Brown put in $600 for 8 months, and Dole put in $400 for 6 months. They gain $60. What is each man's share of the gain? OPERATION. $ 6 0 0 X 8 $ 4 8 0 0, Brown's money for 1 month. 4 =28 oo 2, Brown's part of stock. $ 4 0 0 X 6 - $ 2 4 0 0, Dole's money for 1 month. 24o0o = -, Dole's part of stock. $ 7 2 0 0, Whole stock for 1 month. Then 2 of $60, the whole gain, = $40, is Brown's share of gain. And - of $60, " " " = $20, is Dole's " " QUESTION. -Art. 254. What is the rule for finding the shares of profit or loss when the stock is employed for the same time?

Page 256 256 PARTNERSHIP. LSECT. XXXII1. It is evident that $600 for 8 months is the same as $600 X 8 = $4800 for 1 month, because $4800 would gain as much in 1 month as $600 in 8 months. And for the same reason, $400 for 6 months is the same as $400 X 6 = $2400 for 1 month. The question is, therefore, the same as if Brown had put in $4800 and Dole $2400 for 1 month each. The whole stock would then be $4800 + $2400 $7200, and Brown's share of the gain would be $8 00 = of $60 - $40. Dole's share will be 24 of $60 = 20 OPERATION BY PROPORTION. $4800 $7200: $4800:: $60: $40, B's share. $2400 7 2 00: $24 00:: $ 60: $2 0, D's share. $7200 RULE 1. — Multiply each partner's stock by the time it was in trade, and consider each product a numerator, to be written over the sum of the products, as a common denominator. Then, multiply the whole gain or loss by each of these fractions, and the product will be the gain or loss of each partner. Or, RULE 2. —Multiply each partner's stock by the time it was in trade; then, as the sum of these products is to each product, so is the whole gain or loss to each partner's gain or loss. EXAMPLES FOR PRACTICE. 2. A, B, and C, trade in company. A put in $700 for 5 months; B put in $800 for 6 months; and C put in $500 for 10 months. They gain $399. What is each man's share of the gain? Ans. A's gain $105, B's gain $144, C's gain $150. 3. Leverett Johnson, William Htyde, and William Tyler, formed a connection in business, under ihe firm of Johnson, Hyde & Co. Johnson at first put in- $1000, and at the end of 6 months he put in $500 more. Hyde at first put in $800, and at the end of 4 months he put in $400 more; but, at the end of 10 months, he withdrew $500 from the firm. Tyler at first put in $1200, and at the end of 7 months he put in $300 more, and at the end of 10 months he put in $200. At the end of the year they found their net gain to be $1000. What is each man's share? Ans. Johnson's gain $348.02w31, Hyde's $273.78-5-, Tyler's $378.19-3lT. 4. George Morse hired of William Hale, of Haverhill, his best horse and chaise for a ride to Newburyport, for $3.00, with the privilege of one person's having a seat with him. Having QUESTIONS. — Art. 255. What are the rules for finding the shares of profit or loss when the stock is employed for unequal times? Why do you multiply each man's stock by the time it was in trade?

Page 257 S cT. XXXIII.] PARTNERSHIP. 257 rode 4 miles, he took in John Jones, and carried him to Newburyport, and brought him back to the place from which he took him. What share of the expense should each pay, the distance frbm Haverhill to Newburyport being 15 miles? Ans. Morse pays $1.90, Jones pays $1.10. 5. J. Jones and L. Cotton enter into partnership for one year. January 1, Jones put in $1000, but Cotton did not put in any until the first of April. What did he then put in, to have an equal share with Jones at the end of the year? Ans. $1333.331. 6. S, C, and D, engage in partnership, with a capital of $4700. S's stock was in trade 8 months, and his share of the profits was $96; C's stock was in the firm 6 months, and his share of the gain was $90; D's stock was in the firm 4 months, and his gain was $80. Required the amount of stock which each had in the firm. S's stock $1200. Ans. C's stock $1500. D's stock $2000. 7. A, B, and C, engage in trade. A put in $300 for 7 months, B put in $500 for 8 months, and C put in $200 for 12 months; they gain $85; what share of the gain does each receive? Ans. A $21, B $40, and C $24. 8. A and B engage in trade, with $500. A put in his stock for 5 months, and B put in his for 4 months. A gained $10, and B gained $12; what sum did each put in? Ans. A $200, B $300. 9. A and B trade in company. A put in $3000, and at the end of 6 months put in $2000 more; B put in $6000, and at the end of 8 months took out $3000; they trade one year, and gain $1080; what is each man's share of the gain? Ans. A's share is $480, B's $600. 10. Four men hired a pasture for $50. A put in 5 horses for 4 weeks; B put in 6 horses for 8 weeks; C put in 12 oxen for 5 weeks, calling 3 oxen equal to 2 horses; and D put in 3 horses for 14: weeks. I-low much ought each man to pay? Ans. A: $6.66;, B $16.00, C $13.33~, and D $14.00. 11. A, B, and C contract to build a piece of railroad for $7500. A employs 30 men 50 days; B employs 50.men 36 days; and C employs 48 men and 10 horses 45 days, each horse to be reckoned equal to one man, and he is also to have $112.50 for overseeing the work. How much is each man to receive? Ans. A receives $1875; 1B $2250; C $3375. 22*

Page 258 258 CURRENCIES. [SECT. XXXIV ~ XXXIV. CURRENCIES. ART. 256, The currency of a state or country is its money or circulating medium of trade. In the United States, the gold silver, and copper coins of the country, foreign coins whose value has been fixed by law, and bank notes, redeemable in specie, pass as money. The legal tender, however, in payment of debts, is gold and silver. The intrinsic value of foreign coins is their mint value, or that depending upon the weight and purity of the metal of which they are made; their commercial value is the price they will bring in the market, and their legal value is that fixed by law. The value of foreign coins, as fixed by present laws of the United States, is shown in the following TABLE OF FOREIGN CURRENCIES. Pound Ster. of G. Britain, $4.84 Ounce of Sicily, $2.40 Pound Ster. of Br. Prov.,) Pagoda of India, 1.84 Nova Scotia, N. Bruns., 4.00 Tael of China, 1.48 Newfoundland, and Can., Milrea of Portugal, 1.12 Dollar of Mexico, Peru, 1}00 Milrea of Azores,.831 Chili, and Cen. Amer., * Ducat of Naples,.80 Specie Dollar of Sweden } Rupee of British India,.44A and Norway, Marco Banco of Hamburg,.35 Specie Dol. of Denmark, 1.05 Franc of France and Bel.,.18%6Rix Dollar of Bremen,.78 Livre Tournois of France,.18A Rix Dol., or Thaler, of Leghorn Livre,.16 Prussia and Northern.69 Lira of Lombardy, Vene-.16 States of Germany, tian Kingdom,.16 Ruble, silver, of Russia,.75 Lira of Tuscany,.16 Guilder of Netherlands,.40 Lira of Sardinia,.18 % Florin of Netherlands,.40 Real Plate of Spain,.10 Florin of South of Ger.,.40 Real Vellon of Spain,.05 The legal currency of this countr, previous to 1786, was sterling money, or that of pounds, shillings, and pence. On the adoption of the currency of dollars and cents, there were in circulation colonial notes, or bills of credit, which had depreciated in value. This depreciation being greater in some sections QUESTIoNS. - Art. 256. What is currency? What pass for money in the United States? What is the intrinsic value of foreign coins? What is the commercial value? What is the legal value? -Art. 257. Mention some of the foreign coins whose value has been fixed by law. What was the currency of this country previous to 1786?

Page 259 SECT. XXXIV.] CURRENCIEb. 259 than in others, gave rise to the variation, in the states, as to the number of shillings equivalent to a dollar, as shown in the following TABLE. F New Eng. States, J Virginia, 6s. = 3 ~., called New Eng. currency, Kentucky, (of which 1~. = $3}; Is.= 162cts. [ Tennessee, J New York, $1 in Ohio, = 8sE~., called New York currency, Michigan, of which 1~. = $2.; Is. = 12.ets. North Carolina, [Pennsylvania, iin New Jersey, = 7s. 6d. =I~., called Pennsylvania curDelaware, rency; of which ~..= $23; ls.= 13cts. Maryland, J S Georgia, = = 4s. 8d.:- ~., called Georgia currency; South Carolina, of which 1~. = $42; Is. = 21lcts. F Canada, 1 $1 iNova Scotia, = 5s. = 4i., called Canada currency; of $ in) New Brunswick, (which 1~. = $4; Is. = 20cts. Newfoundland, J =4 _1?6Ass 2-5J ~. called English or Ster$1 in Great Britain, ling money; of which 1~. = $4.84; Is. = 241cts. NOTE. - The old currencies of the states are no longer used in keeping accounts, yet the price of articles is still named by some traders in the old currency of their state. REDUCTION OF CURRENCIES. ART. 257, REDUCTION of Currencies is the process of finding the value of the denominations of one currency in the denominations of another. ART. 258. To reduce pounds, shillings, pence, and farthings, of the different currencies, to United States money. Ex. 1. Reduce 18~. 15s. 6d. New England currency to United States money. Ans. $62.5 8. OPERATION. We first reduce the shillings 18~. 15s. 6d. -- 18.775~. and pence to the decimal of a 18.775 ~. = $62.5"8- pound (Art. 188), and then an nexing it to the pounds, we divide QUESTIoNS. - What gave rise to the variation in the old currency of this.country? Repeat the table. How are the old currencies of the states now used? - Art. 260. What is reduction of currencies?

Page 260 260 CURRENCIES. [SECT. XXXIV. the sum by 3w, because 6s., or a dollar in this currency, is -~ of a pound, and thus obtain the answer in dollars and the decimal of a dollar. Hence the following RULE. - Divide the given sum expressed in pounds and decimals of a pound, by the value of $1 expressed in a fraction of a pound. The quotient will be the value in dollars. EXAMPLES FOR PRACTICE. 2. Change 144~. 7s. 6d. of the old New England currency to United States money. Ans. $481.25. 3. Change 74~. is. 6d. of the old currency of New York to United States money. Ans. $185.18,. 4. Change 129~. of the old currency of Pennsylvania to United States money. Ans. $344. 5. Change 84~. of the old currency of South Carolina to United States money. Ans. $360. 6. Change 144~. 4s. of Canada and Nova Scotia currency to United States money. Ans. $576.80. 7. Change 257~. 8s. 6d. English or sterling money to United States money. Ans. $1245.937. ART. 259. To reduce United States money to pounds, shillings, pence, and farthings, of the different currencies. Ex. 1. Reduce $152.625 to old New England currency. Ans. 45~. 15s. 9d. OPERATION. Since 6s., or a dollar, in this cur$152.625 X, — 45.7875~. rency, is -j3 of-a pound, we mul45.7875~. = 45g. 15s. 9d. tiply the given sum by the fraction if, and reduce the decimal to shillings and pence. (Art. 189.) RULE.- Multiply the given sum expressed in dollars by the value of $1 expressed in a fraction of a pound. The quotient will be the value in pounds. EXAMPLES FOn PRACTICE. 2. Change $481.25 to the old currency of New England. Ans. 144~. 7s. 6d. QUESTIONS. - Art. 260. How do you reduce United States money to pounds, shillings, pence, and farthings, New England currency? Why multiply by -TT~.? How would you reduce United States money to pounds, &c., Ohio currency? How, to Pennsylvania currency? What is the general rule? - Art. 261. What is the rule for reducing United States money to pounds, shillings, pence, and farthings, of the different currencies?

Page 261 SECt. XXXV. EXCHANGE. 261 3. Change $185.18- to the old currency of New York. Ans. 74~. ls. 6d. 4. Change $344 to the old currency of Pennsylvania. Ans. 129~. 5. Change $360 to the old currency of South Carolina. Ans. 84~. 6. Change $576.50 to Canada and Nova Scotia currency. Ans. 144~. 2s. 6d. 7. Change $1245.93,7 to English or sterling money. Ans. 257~. 8s. 6d. ART. 269, To reduce any foreign currency to United States money, and United States money to any foreign currency, when the value of a unit of the foreign currency is known (Art. 256), simply Multiply or divide, as the case may require, by the value of the unit of the given currency expressed in United States money. EXAMPLES FOR PRACTICE. Ex. 1. Reduce 123 rubles, silver, of Russia, to United States money. Ans. $92.25. 2. Reduce $27.90 to francs. Ans. 150 francs. 3. What is the value of 121 thalers of Prussia in United States money? Ans. $83.49. 4. What is the value of $165.20 in florins? Ans. 413 florins. 5. A merchant purchased tea in China to the amount of 216 taels. What did it cost in United States money? Ans. $319.68. 6 How many reals, plate of Spain, are equal to $5137.90? Ans. 51379. ~ XXXV. EXCHANGE. Anr. 261. EXCHANGE, in commerce, is the paying or receiving of money in one place for an equivalent sum. in another, by means of drafts, or bills of exchange. QUESTIONS. —Art. 260. How do you reduce any foreign currency to United States money, and United States money to any foreign currency? Art. 261. What is exchange?

Page 262 262 EXCHANGE. [SECT. XXXV A bill of exchange is a written order, to some person at a dlo tance, to pay a certain sum, at an appointed time, to another person, or to his order. The person who draws a bill is termed the maker, or drawer; the person for whom it is drawn, the buyer, taker, or remitter; the person on whom it is drawn, the drawee, and after he has accepted, the acceptor. The person who endorses it is termed the endorser; and the person in whose legal possession the bill may be at any time is termed the holder, or possessor. Exchange is at par when a certain sum, at the place from which it is remitted, will pay an equal sum at the place to which it is remitted. It is said to be at a premium, or above par, when the balance of trade is against the place from which the bill is remitted; and below par when the balance of trade is in favor of the place from which the bill is remitted. INLAND BILLS. ART. 262, An INLAND BILL of exchange, or draft, is one of which the drawer and drawee are both residents of the same country. ART. 263. To find the value of an inland bill, or draft, Add to the face of the bill, or draft, the amount of premium, or subtractfrom theface of the bill, or draft, the amount of discount. EXAMPLES FOR PRACTICE. Ex. 1. What is the value of the following bill of exchange, or draft, at 11 per cent. discount? Ans. $445.22. $452. Boston, March 6, 1856. At sight, pay to William Dura, or order, four hundred and fifty-two dollars, value received, and charge the same to my account. To LEWIS FONTENAY, EDWIN DANTON. Merchant, New Orleans. 2. A merchant in Chicago purchased a bill on New York for $1164, at 1 per cent. premium; what did he pay? Ans. $1175.64. QUESTIONS. - What is a bill of exchange? Who is the maker, or drawer, of a bill 7 Who is the buyer, taker, or remitter? Who is the drawee? Who is the endorser? Who is the holder, or possessor? When is exchange at par? When at a premium? When at a discount? What is an inland bill, or draft? How do you find the value of an inland bill, or draft?

Page 263 SECT. XXXV.] EXCHANGE. 263, 3. What costs a bill on Burlington, Iowa, for $4000, at 2per cent. discount? Ans. $3900. 4. What costs a bill on Buffalo for $450, at 5 of 1 per cent. discount? Ans. $447.1 83. 5. What costs a draft fTiom the Girard Bank, Philadelphia, on the Bank of Commerce, Boston, for $2517.70, at 8 of 1 per cent. premium? Ans. $2520.84+. FOREIGN BILLS. ART. 264. A FOREIGN BILL of exchange is one of which the drawer and drawee are residents of different countries. Foreign bills are usually drawn in sets; that is, at the same time there are drawn two or more bills of the same tenor and date, each containing a condition that it shall continue payable only while the others remain unpaid. NoTE. - Each bill of a set is remitted in a different manner, in order to guard against loss or delay; and when one of the set has been accepted and paid, the others become worthless. EXCHANGE ON ENGLAND. ART. 265. The exchange value, in the United States, of the pound sterling of Great Britain, is that of its former legal value, or $44 = $4.444, which is considerably below either its intrinsic or commercial value. The commercial value is generally about 9 per cent. more than this exchange, or nominal par value. Thus, nominal par value being = $4.44a To which we add 9 per cent. premium, =.40 The commercial par value will be = $4.844. Therefore, when the nominal exchange between the United States and Great Britain exceeds 9 per cent. premium, it is above true par; when less, it is below true par. ART. 266. To find the value in United States currency of a bill on England. Ex. 1. What should be paid for the following bill at 91 per cent. premium? Ans. $4866.66.23 QUESTIONS. -Art. 264. What is a foreign bill of exchange? flow are foreign bills usually drawn? Why? - Art. 265. What is the exchange value of the pound sterling of Great Britain, in United States money? How does this differ from the commercial or true par value?

Page 264 264 EXCHANGE. [SECT. XXXV Exchange for ~1000. New York, May 16, 1 S56. Thirty days after sight of this first of exchange (second ana third of the same tenor and date unpaid), pay J. W. Hathaway 4. Co., or order, in London, one thousand pounds sterling, value 1 eceived, and place the same to my account. To BATES, BARING & Co., London. RUFUS W. KING. OPERATION. 1~.+.09-~. = 1.095~.; 1.095 X $4- = $4.866~; 1000 X $4.866- $4866.66a. We add to 1~. the premium on 1~., and obtain 1.095~., which, multiplied by $49-, or $4W, the nominal value of a pound, gives $4.8662 as the value of a pound at the given rate of ex. change; and 1000 multiplied by this value of a pound gives $4866.662 as the value of the.bill. If there had been with the pounds shillings, pence, or farthings, they would have been reduced to a decimal of a pound, and as such annexed to the pounds in the operation. RULE. - Multiply the amount of the bill, expressed in pounds ana decimals of a pound, by the value of one pound at the given rate of exchange, and the product will be the value in dollars. EXAMPLES FOR PRACTICE. 2. A merchant in Boston wishes to purchase a bill of 572~. 10s., on Liverpool, the premium being 81 per cent.; what will it cost him in dollars and cents? Ans. $2760.722. 3. If J. C. Sherman, of Chicago, should remit to London 1200~., exchange being at 91 per cent., what will be the cost of the bill in United States money? Ans. $5826.6623. ART. 267, To find the amount of a bill on England, which can be purchased for a given sum of United States currency. Ex. 1. When exchange is at 91 per cent. premium, what will be the amount of a bill on London which I can purchase for $4866.662? Ans. 1000~. OPERATION. 1~. +.09~. = 1.095~.; 1.095 X $_4-_ = $4.866 2; $4866.66 2 - $4.8662 - 1000~. QUESTIONS. -Art. 266. What is the rule for finding the value of a bill on England in United States currency? - Art. 267. What is the rule for finding the amount of a bill on England, which can be purchased for a given sum of United. States currency?

Page 265 SECT. XXXV.] EXCHANGE. 265 We find, as by Art. 266, the value of one pound at the given rate of exchange. The given sum, $4866.662~, we divide by the value of a pound, and obtain 1000~. as the required amount of the bill. RULE. - Divide the given sum by the value of one pound at the given rate of exchange, and the quotient will be the amount in pounds and decimals of a pound. EXAMPLES FOR PRACTICE. 2. J. Reed, of Cincinnati, proposes to make a remittance to Liverpool of $1640, exchange being at 81 per cent. premium; what will be the amount of the bill he can remit for that sum? Ans. 340~. is. lOd. 3. A merchant wishes to remit $500 to England, exchange being at 10 per cent. premium; what will be the amount of the bill he can purchase for that sum? Ans. 102~. 5s. 5d.+. EXCHANGE ON FRANCE. ART. 268, In France accounts are kept in francs and centimes. The centimes are hundredths of a franc. All bills of exchange on France are drawn in francs, and are bought, sold, and quoted, as at a certain number of francs to the dollar. ART. 269, To find the value in United States currency of a bill on France, Divide the amount of the bill by the value of one dollar in francs, and the quotient will be the value in dollars. EXAMPLES FOR PRACTICE. Ex. 1. What must be paid, in United States currency, for a bill on Paris of 2380 francs, exchange being 5.15 francs per dollar? Ans. $462.13+. 2. How many dollars will purchase a bill on Havre of 30000 francs, exchange being 5.17I francs per dollar? Ans. 85797.10+. 3. What is the value of a bill on Paris of 62500 francs, exchange being 5.12 francs per dollar? Ans. $12207.03+. ART. 270. To find the amount of a bill on France, which can be purchased for a given sum of United States currency, QUESTIONS.- Art. 268. How are accounts kept in France? How are all bills of exchange on France drawn? - Art. 269. How do you find the value in United States currency of a bill on France? - Art. 270. How do you find the amount of a bill on France, which can be purchased for a given sum of United States money? 22

Page 266 266 DUODECIMALS. [SECT. XXXV. Multiply the given sum by the value of one dollar in francs and the product will be the amount of the bill in francs. Ex. 1. Alfred Walker, of New York, pays $2500 for a bill on Paris, exchange being 5.12 francs per dollar. What was the amount of the bill in francs? Ans. 12800. 2. When exchange on France is at 5.13 francs per dollar, a bill of how many francs should $700 purchase? Ans. 3591. 3. Morton and Blanchard, of Boston, wish to remit $675 to Paris, exchange being 5.16 francs per dollar; what will be the amount of the bill of exchange they can purchase with the money? Ans. 3483 francs ~ XXXVI. DUODECIMALS. ART. 271. DUODECIMALS are a kind of compound numbers in which the unit, or foot, is divided into 12 equal parts, and each of these parts into 12 other equal parts, and so on indefinitely; thus, 4z, Tsi, &c. Duodecimals decrease from left to right in a twelve-fold ratio; and the different orders, or denominations, are distinguished from each other by accents, called indices, placed at the right of the numerators. Hence the denominators are not expressed. Thus, 1 inch or prime, equal to T12 of a foot, is written 1 in. or 1'. 1 second " A " 1". 1 third " IA " " 1"'. 1 fourth 1" I " " 1. Hence the following TABLE. 12 fourths make 1"'. 12 seconds make 1'. 12 thirds " 1". 12 inches or primes " Ift. ADDITION AND SUBTRACTION OF DUODECIMALS. ART. 272, Duodecimals are added and subtracted in the same manner as compound numbers. QUESTIONS. -A-rt. 271. What are duodecimals? In what ratio do duodecimals decrease from left to right? How are the different denominatic-ns distinguished from each other? - Art. 272. How are duodecimals added a-pd subtracted?

Page 267 SECT. XXXVI.] DUODECIMALS. 267 EXAMPLES FOR PRACTICE. 1. Add together 12ft. 6' 9", 14ft. 7' 8", 165ft. 11' 10". Ans. 193ft. 2' 3". 2. Add together 182ft. 11' 2" 4"', 127ft. 7' 8" 11"', 291ft. 5' 11" 10"'. Ans. 602ft. 0' 11" 1"'. 3. From 204ft. 7' 9" take 114ft. 10' 6T'. Ans. 89ft. 9' 3". 4. From 397ft. 9' 6" 11"' 7"" take 201ft. 11' 7" 8"' 10"". Ans. 195ft. 9' 11" 2"' 9"". MULTIPLICATION AND DIVISION OF DUODECIMALS. ART. 273. To find the denomination of the product of any two numbers in duodecimals, when multiplied together. Ex. 1. What is the product of 9ft. multiplied by 3ft.? Ans. 27ft. OPERATION. 9ft. X 3ft. - 27ft. 2. What is the product of 7ft. multiplied by 6'? Ans. 3ft. 6'. OPERATION. 6' - 6 of a foot; then 7ft. X -6 ft.-= 4 -42'; 42' * 12 -- 3ft. 6'. 3. What is the product of 5' multiplied by 4'? Ans. 1' 8". OPERATION. 5' -5, and 4' = 42; then -T X 4 — 2 4 20"; 20" + 12 --' 8". 4. What is the product of 9' multiplied by 11"'? Ans. 8"' 3"". OPERATION. 9' — ~,D and 11"' = T+1; then 92 X T+ h= Y - 199 99""; 99"": — 12 -- 8"' 3"". It will be observed in the examples above, that feet multiplied by feet produce feet; feet multiplied by primes produce primes; primes multiplied by primes produce seconds, &c.; and that the several products are of the same denomination as denoted by the sum of the indices of the numbers multiplied together. Hence, When two numbers are multiplied together, the sum of their indices annexed to their product denotes its denomination. ART. 274. To multiply duodecimals together. Ex. 1. Multiply 8ft. 6in. by 3ft. 7in. Ans. 30ft. 5' 6". QUEsTION. - Art. 273. Ilow is the deaomination of the product denoted when duodecimals are multiplied together?

Page 268 268 DUODEOIMALS. [SECT. XXXVI OPERATION. WAVe first multiply each of the terms. in the 8ft. 6' multiplicand by the 7' in the multiplier, thus, 7 3ftG 7' into 6' =42" - 3' and 6". Placing the 6" under its multiplier, we add the 3' to the product ot 4ft. 1 1' 6" 7' into 8ft. = 59'= 4ft. and 11', which we write 2 5ft. 6' down. We then multiply by the 3ft., thus: 3fa into 6'= 18' = Ift. and 6'. We write the 6 3 Oft. 5' 6" under its multiplier, and add the Ift. to the product of the 3ft. into 8ft., making 25ft., which we write down. The two products being added together, we obtain 30ft. 5' 6" for the answer. RULE. - Write the multiplier under the multiplicand, so that the same denominations shall stand in the same column. Beginning at the right hand, multiply each term in the multiplicana by each term of the multiplier, and write the first term of each partial product directly under its multiplier, observing to carry a unit for every twelve from each lower denomination to the next higher. The sum of the several partial products will be the product required. EXAMPLES FOR PRACTICE. 2. Multiply 8ft. 3in. by 7ft. 9in. Ans. 63ft. 11' 3". 3. Multiply 12ft. 9' by 9ft. 11'. Ans. 126ft. 5' 3". 4. My garden is 18 rods long and 10 rods wide; a ditch is dug round it 2 feet wide and 3 feet deep; but the ditch not being of a sufficient breadth and depth, I have caused it to be dug 1 foot deeper, and, outside, 1 ft. 6 in. wider. How many solid feet will it be necessary to remove? Ans. 7540. 5. I have a room 12 feet long, 11 feet wide, and 71 feet high. In it are two doors, 6 feet 6 inches high, and 30 inches wide, and the mop-boards are 8 inches high. There are 3 windows, 3 feet 6 inches wide, and 5 feet 6 inches high; how many square yards of paper will it require to cover the walls? Ans. 25- 2 9 square yards. ART. 275. To divide one duodecimal by another. Ex. 1. A certain aisle contains 68ft. 10' 8" of floor. The width of the floor being 2ft. 8', what is its length? Ans. 25ft. 10'. OPERATION. We first divide the 68ft. by 2ft. 8' ) 6 8ft. 1 0' 8" ( 2 5ft. 1 0' the divisor, and obtain 25ft. 6 6ft. 8' for the quotient. We multiply the entire divisor by the 2ft. 2' 8" 25ft., and subtract the prod-, 2ft. 2' 8" uct, 66ft. 8', from the corresponding portion of the QUESTION. - Art. 274. What is the rule for the multiplication of duodeci mals?

Page 269 SECT. XXXVII.] INVOLUTIOIT 269 dividend, and obtain 2ft. 2', to which remainder we bring down the 8", and dividing, we obtain 10' for the quotient. Multiplying the entire divisor by the 10', we obtain 2ft. 2' 8", which subtract, as before, leaves no remainder. Therefore, 25ft. 10' is the length of the aisle. RULE. - Find how many times the highest term of the dividend will contain the divisor. By this quotient multiply the entire divisor, and subtract the product from the corresponding terms of the dividend. To the remainder annex the next denomination of the dividend, and divide as before, and so continue till the division is complete. EXAIPLES FOR PRACTICE. 2. What must be the length of a board, that is 1ft.'9in. wide, to contain 22ft. 2in.? Ans. 12ft. 8in 3. I have engaged E. Holmes to cut me a quantity of wood, It is to be cut 4ft. 6in. in length, and to be " corded " in a range 256 ft. long. Required the height of the range to contain 75 cords. Ans. 8ft. 4in. ~ XXXVII. INVOLUTION. ART. 276. INVOLUTION is the method of finding any power of a given quantity. A power is a quantity produced by taking any given number, a certain number of times, as a factor. The factor, thus taken, is called the root of the power. The number denoting the power is called the index or exponent of the power, and is a small figure placed at the right of the root. Thus, the second power of 6 is written 62; the third power of 4 is written 43, and the fourth power of 2 is written (2)4. ART. 277. To raise a number to any required power. 3 3, the first power of 3, is written 31 or 3. 3 X 3 - 9, the second power of 3, is written 32. 3 X 3 X 3 - 27, the third power of 8, i 3'. 3 X 3 X 3 X 3 81, the fourth power of 3," " 34. 3 X 3 X 3 X 3 X 3 243, the fifth power of 3, It " 35 QUESTIONS. - Art. 275. What is the rule? - Art. 276. What is Involution? WVhat is a power? What is the number called that denotes the power? Where is it placed? - Art. 277. To what is the index in each power equal? 23:

Page 270 270 INVOLUTION. [SECT. XXXVIL By examining the several powers of 3 in the examples given, we see that the index of each power is equal to the number of times 3 is used as a factor in the multiplications producing the power, and that the number of times the number is multiplied into itself is one less than the power denoted by the index. Hence the RULE. - Multiply the given number continually by itself, till the number of multiplications is one less than the index of the power to be found, and the last product will be the power required. NOTE.- 1. A fraction may be raised to any power by this rule, by multiplying its terms continually together. Thus, the second power of 2 iS 25X = -5 5 ~-x Q —IT. NoTE.- -. A mixed number may be either reduced to an improper fraction, or the fractional part reduced to a decimal, and then raised to the required power. EXAMPLES FOR PRACTICE. 1. What is the 2d power of 6? Ans. 36. 2. What is the 3d power of 5? Ans. 125. 3. What is the 6th power of 4? Ans. 4096. 4. What is the 4th power of ~? Ans. 2T. 5. What is the 5th power of 32? Ans. 662 4. 6. What is the 3d power of.25? Ans..015625. 7. What is the 1st power of 17? Ans. 17. ART. 278. To raise a number to any required power with out producing all the intermediate powers. Ex. 1. What is the 8th power of 4? Ans. 65536. OPERATION. 1 2 3 5 + 3 + 2 8. 4, 16, 64; 64X64X16=65536. W7Ve raise the 4 to the 2d and to the 3d power, and write above each power its exponent. We then add the exponent 3 to itself, and, increasing the sum by the exponent 2, obtain 8, a number equal to the power required. We next multiply 64, the power belonging to the exponent 3, into itself, and this product by 16, the power belonging to the exponent 2, and obtain 65536 for the 8th power. Hence the following RULE. - Raise the given number to any convenient number of powers, and write above each of the respective powers its exponent. QUESTIONS. -What is the rule for raising a number to any required power? How may a vulgar fraction be raised to a required power? How a mixed number? —Art. 278. What are the numbers placed over the several powers of 4 called, and what do they denote?

Page 271 SECT. XXXVIII.] EVOLUTION. 271 Then add together such exponents as will make a number equal to the required power, repeating any one when it is more convenient, and the product of the powers belonging to these exponents will be the required answer. EXAMPLES FOR PRACTICE. 2. What is the 7th power of 5? Ans. 78125. 3. What is the 9th power of 6? Ans. 10077696. 4. What is the 12th power of 7? Ans. 13841287201. 5. What is the 8th power of 8? Ans. 16777216. 6. What is the 20th power of 4? Ans. 1099511627776. 7. What is the 30th power of 3? Ans. 205891132094649. ~ XXXVIII. EVOLUTION. ART. 279. EVOLUTION is the method of finding the root of a given power or number, and is therefore the reverse of Involution. A root of any power is a number which, being multiplied into itself a certain number of times, produces the given power. Thus 4 is the second or square root of 16, because 4 X 4 16; and 3 is the third or cube root of 27, because 3 X 3 X 3 =27. The root takes the name of the power of which it is the root. Thus, if the number is a second power, the root is called the second or square root; if it is a third power, the root is called the third or cube root; and if it is a fourth power, its root is called the fourth or biquadrate root. Those roots which can be exactly found are called rational roots; those which cannot be exactly found, but approximate towards the true root, are called surd roots. Numbers that have exact roots are called perfect powers, and all other numbers are called imperfect powers. QUESTIONS. -What is the rule for involving a number without producing all the intermediate powers? - Art. 279. What is Evolution? What is a root? From what does the root take its name? What are rational roots?. What surd roots?

Page 272 272 EVOLUTION. [SECT. XXXVIII. Roots are denoted by writing the character \/, called the radical sign, before the power, with the index of the root over it, or by a fractional index or exponent. The third or cube root of 27 is expressed thus, ~,/27, or 274; and the second or square root of 25 is expressed thus, /25, or 25~. NOTE. - The index 2 over A/ is usually omitted when the square root is required. Thus, %/64 denotes the square root of 64. EXTRACTION OF THE SQUARE ROOT. ART. 280. The Square Root is the root of any second power, and is so called because the square or second power of any number represents the contents of a square surface, of which the root is the length of one side. ART. 281, To extract the square root of any number is to find a number which, being multiplied by itself, will produce the given number. The following numbers in the upper line represent roots, and those in the lower line their second powers, or squares. Roots, 1 2 3 4 5 6 7 8 9 10 Squares, 1 4 9 16 25 36 49 64'81 100 It will be observed that the second power or square of each of the numbers above contains twice as many figures as the root, or twice 1s many wanting one. Hence, To ascertain the number of figures in the square root of any given number, it must be divided into periods, beginning at the right, each of which, excepting the last, must always contain two fiyures; and the number of periods will denote the number of figures of which the root uill consist. Ex. 1. I wish to arrange 625 tiles, each of which is 1 foot square, into a square pavement; what will be the length of one of the sides? Ans. 25 feet. OPERATION. It is evident, if we extract the square 6 2 5 ( 2 5, Ans. root of 625, we shall obtain one side of 4 the pavement, in feet. (Art. 280.) Beginning at the right hand, we divide 4 5 ) 2 2 5 the number into periods, by placing a point 2 2 5 over the right-hand figure of each period; and then find the greatest square nurlmber in QUESTIONS. -I-Iow are roots denoted? What is said of the index 2?Art. 280. What is meant by the square root, and why is it so called? - Art. 281. What is meant by extracting the square root? IIow do you ascertain the number of figures in the square root of any number?

Page 273 SECT. XXXVIII.] EXTRACTION OF THE SQUARE ROOT. 273 the left-hand period, 6 (hundreds) to be 4 (hundreds), and that its root is 2, which we write in the quotient. As this 2 is in the place of tens, its value is 20, and represents the side of a square, the area or superficial contents of which are 400 square feet, as seen in Fig. 1. We now subtract 400 feet from 625 feet, Fig. 1. and have 225 feet remaining, which must be added on two sides of Fig. 1, in order 20 feet. that it may remain a square. We therefore double the root 2 (tens) or 20, one side of the square, to obtain the length of the two a; D D sides to be enlarged, making 40 feet; and 2 0 so then inquire how many times 40, as a divi-20 sor, is contained in the dividend 225, and find it to be 5 times. This 5 we write in the 4 00 quotient or root, and also on the right of the divisor, and it represents the width of 20 feet. the additions to the square, as seen in Fig. 2. The width of the additions being multiplied by 40, the length of the two additions, makes 200 square feet, the contents of the two additions E and F, which are 100 feet for Fig. 2. each. The space G now remains to be filled, to complete the square, each side of 25 feet. which is 5 feet, or equal to the width of 20o 5 E and F. If, therefore, we square 5, we ET nu I G-5 have the contents of the last addition, G, tN equal to 25 square feet. It is on account D F )n of this last addition that the last figure a 2 0 2 0 of the root is placed in the divisor; for we 20 5 thus obtain 45 feet for the length of all -_00 1 the additions made, which, being multi400 0 1 00 plied by the width (5ft.), the last figure in the root, the product, 225 square feet, 25 feet. will be the contents of the three additions, E, F, and G, and equal to the feet remaining after we had found the first square. Hence, we obtain 25 feet for the length of one side of the pavement, since 25 X 25 = 625, the number of tiles to be arranged, and equal to the sum of the several parts of Fig. 2; thus, 400 + 100 -+- 100 + 25 _ 625. This illustration and explanation is founded upon the principle, QUESTIONS. - What is first done after dividing the number into periods? What part of Fig. 1 does this greatest square number represent? What place does the figure of the root occupy, and what part of the figure does it represent? Why do you double the root for a divisor? What part of Fig. 2 does the divisor represent? What part does the last figure of the root represent? Why do you multiply the divisor by the last figure of the root? What parts of the figure does the product represent? Why do you square the last figure of the root? What part of the figure does this square represent? What other way of finding the contents of the additions without multiplying the parts separately by the width?

Page 274 274 EXTRACTION OF THE SQUARE ROOT. [SECT. XXXVII1. That the square of the sum of two numbers is equal to the squares of the numbers, plus twice their product. Thus, 25 being equal to 20+ 5,its square is equal to the squares of 20 and of 5, plus twice the product of 20 and 5, or to 400 + 2 X 20 X 5 -- 25 625. RULE.- Separate the given number into periods of two figures each, by putting a point over the place of units, another over the place oj hundreds, and so on. Find the greatest square number in the left-hand period, writing the root of it at the right hand of the given number, after the manner of a quotient in division, for the first figure of the root. Subtract this square number from the first period, and to the remainder bring down. the next periodfor a dividend. Double the root already found for a divisor, and find how often the divisor is contained in the dividend, omitting the right-hand figure, ana annex the result to. the root for the second figure of it, and likewise to the divisor.* Multiply the divisor with the figure last annexed by the figure annexed to the root, and subtract the product from the dividend. To the remainder bring down the next period for a new dividend. Double the root already found for a new divisor, and continue th, operation as before, till all the periods have been brought down. NOTE 1. - It is evident, when the given number contains an odd number of figures, the left-hand period can contain but one figure. 2. If the dividend does not contain the divisor, a cipher must be placed in the root, and also at the right of the divisor; then, after bringing down the next period, this last divisor must be used as the divisor of the new dividend. 3. When there is a remainder after extracting the root of a number, periods of ciphers may be annexed, and the figures of the root thus obtained will be decimals. 4. If the given number is a decimal, or a whole number and a decimal, the root is extracted in the same manner as in whole numbers, except, in pointing off the decimals either alone or in connection with the whole number, we place a point over every second figure toward the right, from the separatrix, and fill the last period, if incomplete, with a cipher. 5. The square root of any number ending with 2, 3, 7, or 8, cannot be exactly found. EXAMPLES FOR PRACTICE. 2. What is the square root of 148996? * The figure of the root must generally be diminished by one or two units, on account of the deficiency in enlarging the square. QUESTIONS. -What is the rule for extracting the square root? What is to be done if the dividend does not contain the divisor? What must be done if there is a remainder after extracting the root? What do you do if the given number is a decimal? Of what numbers can the square root not be found?

Page 275 OIECT. XXXVIiI.) EXTRACTION O0 THE SQUARE ROOT. 275 OPERATION. 148996(386 9 6 8)5 89 544 766)4596 4596 S. What is the square root of 516961? Ans. 719. 4. What is the square root of 182329? Ans. 427. 5. What is the square root of 23804641? Ans. 4879. 6. What is the square root of 10673289? Ans. 3267. 7. WVhat is the square root of 20894041? Ans. 4571. 8. What is the square root of 42025? Ans. 205. 9. What is the square root of 1014049? Ans. 1007. 10. What is the square root of 538? Ans. 23.194+. 11. What is the square root of 71? Ans. 8.426+. 12. What is the square root of 7? Ans. 2.645+. 13. What is the square root of.1024? Ans..32. 14. What is the square root of.3364? Ans..58. 15. What is the square root of.895? Ans..946+. 16. What is the square root of.120409? Ans..347. 17. What is the square root of 61723020.96? Ans. 7856.4. 18. What is the square root of 9754.60423716? Ans. 98.7654. ART. 282, If it is required to extract the square root of a common fraction, or of a mixed number, the mixed number must be reduced to an improper fraction; and in both cases the fractions must be reduced to their lowest terms, and the root of the numerator and denominator extracted. NoTE. - When the exact root of the terms of a fraction cannot be found, it must be reduced to a decimal, and the root of the decimal extracted. EXAMPLES FOR PRACTICE. 1. What is the square root of — 9? Ans. 7I2. 2. What is the square root of 16o 6 Ans. a. 3. What is the square root of 752? Ans. Y-r. 4. What is the square root of 7 24 9? Ans. 6.1 5. What is the square root of 60I1? Ans. 7~. 6. What is the square root of 28-j? Ans. 53. QUESTION. — Art. 282. What do you do when it is required to extract the square root of a common fraction, or of a mixed number?

Page 276 276 APPLICATION O0' THE SQUARE ROOT. [SECT. XXXVIII 7. What is the square root of 471z? Ans. 6. 8. What is the square root of 42? Ans..858+. 9. What is the square root of 832? Ans. 9.14-. 10. What is the square root of 121 l? Ans. 11.042+. 11. What is the square root of 3394? Ans. $. 462' 12. What is the square root of 15579? Ans. W. APPLICATION OF THE SQUARE.ROOT. ART. 283. The square root may be applied to finding the dimensions and areas of squares, triangles, circles, and othei surfaces. 1. A general has an army of 226576 men; how many must he place rank and file to form them into a square? Ans. 476. 2. A gentleman purchased a lot of land in the form of a square, containing 640 acres; how many rods square is his lot? Ans. 320 rods. 3. I have three pieces of land; the first is 125 rods long, and 53 wide; the second is 621 rods long, and 34 wide; and the third contains 37 acres; what will be the length of the side of a square field whose area will be equal to the three pieces? Ans. 121.11+ rods. 4. W. Scott has 2 house-lots; the first is 242 feet square, and the second contains 9 times the area of the first; how many feet square is the second? Ans. 726 feet. 5. There are two pastures, one of which contains 124 acres, and the area of the other is to the former as 5 to 4; how many rods square is the latter? Ans. 157.48+ rods. 6. I wish to set out an orchard containing 216 fruit-trees, so that the length shall be to the breadth as 3 to 2, and the distance of the trees from each other 25 feet; how many trees will there be in a row each way, and how many square feet of ground will the orchard cover? - Ans. 18 in length; 12 in breadth; 116875sq. ft. ART. 284. A TRIANGLE is a figure having three sides and three angles. A right-angled triangle is a figure having three sides and three angles, one of which is a right angle. QUESTIONS. - Art. 283. To what may the square root be applied? - Art. 284. What is a triangle? What is a right-angled triangle? What is the longest side called? What the other two?

Page 277 "ECT. XXXVIII.] APPLICATION OF TIIE SQUARE ROOT. 277 The side A B is called the base of the tri- / ] angle A B C, the side B C the perpendicular, the side A C the hypothenuse, and the angle at B is a riyht angle. a A JB Base. ART. 285. In every right-angled triangle, the square of the hypothenuse is equal to the sum of the squares of the base and perpendicular, as shown by the following diagram. It will be seen, by examining this dia- C gram, that the large square, formed on the hypothenuse A C, contains the same number of small squares as the other two A counted together. Hence, the propriety A of the following rules. AnT. 286. To find the hypothenuse, the base and perpendicular being given. RULE. - Add the square of the base to the square of the perpendicular, and extract the square root of their sum. ART. 287. To find the perpendicular, the base and hypothenuse being given. RULE. - Subtract the square of the base from the square of the hypothenuse, and extract the square root of the remainder. ART. 288. To find the base, the hypothenuse and perpendicular being given. RULE. - Subtract- the square of the perpendicular from the square of the hypothenuse, and extract the square root of the remainder. EXAMPLES FOR PRACTICE. 1. What must be the length of a ladder to reach to the top of a house 40 feet in height, the bottom of the ladder being placed 9 feet from the sill? Ans. 41 feet. QUESTIONS. - Art. 285. ITow does the square of the hypothenuse compare with the base and perpendicular? How does this fact appear from Fig. 2?Art. 286. What is the rule for finding the hypothenuse? - Art. 287. What for finding the perpenik-lar? - Art. 288. What for finding the base? 24

Page 278 278 APPLICATION OF THE SQUARE ROOT. [SECT. XXXVIII. 2. Two vessels sail from the same port; one sails due north 360 miles, and the other due east 450 miles; what is their distance from each other? Ans. 576.'2-+ miles. 3. The hypothenuse of a certain right-angled triangle is 60 feet, and the perpendicular is 36 feet; what is the length of the base? Ans. 48 feet. 4. A line drawn from the top of the steeple of a certain meeting-house to a point at the distance of 50 feet on a level from the base of the steeple, is 120 feet in length; what is the height of the steeple? Ans. 109.08+ feet. 5. The height of a tree on an island in a certain river is 160 feet. The base of the tree is 100 feet on a horizontal line from the river, and is elevated 20 feet above its surface. A line extending from the top of the tree to the further shore of the river is 500 feet. Required the width of the river. Ans. 366.47+ feet. 6. On the edge of a perpendicular rock, whose base is 90 feet, on a level, from a certain road that is 110 feet wide, there is a tower 160 feet high; the length of a line extending from the top of the tower to a point on the opposite side of the road is 300 feet. rWhat is the elevation of the base of the tower above the road? Ans. 63.6+ feet. 7. John Snow's dwelling is 60 rods north of the meetinghouse, James Briggs's is 80 rods east of the meeting-house, Samuel Jenkins's is 70 rods south, and James Emerson's 90 rods west of the meeting-house; how far will Snow have to travel to visit his three neighbors, and then return home? Ans. 428.47+ rods. 8. A certain room is 24 feet long, 18 feet wide, and 12 feet high; required the distance from one of the lower corners to an opposite upper corner. Ans. 32.3+ feet. ART. 289. A CIRCLE is a plane figure bounded by a curved line, every part of which is equally distant from a point called the centre. The circumference or periphery of a circle is the line which bounds it. The diameter of a circle is a line drawn through A B the centre, and terminated by the circumference; as A B. QUESTIONS. - Art. 289. What is a circle? What is the circumference of a circle? What the diameter?

Page 279 SECT. XXXVIII.] APPLICATION OF THE SQUARE ROOT. 279 ART. 290. All circles are to each other as the squares of their diameters, semi-diameters, or circumferences. All similar triangles and other rectilineal figures are to each other as the squares of their homologous or corresponding sides. ART. 291. To find the side, diameter, or circumference, of any surface, which is similar to a given surface. RULE. - State the question as in Proportion, and square the given -sides, diameters, or circumferences, and the square root of the fourth term of the proportion will be the required answer. ART. 292. To find the area of any surface which is similar to a given surface. RULE. - State the question as in Proportion, and square the given sides, diameters, or circumferences, and the fourth term of the proportion is the required answer. EXAMPLES FOR PRACTICE. Ex. 1. I have a triangular piece of land containing 65 acres, one side of which is 100 rods in length; what is the length of the corresponding side of a similar triangle containing 32 acres? Ans. 70.71+ rods. OPERATION. 65:32.::1002: 5000; V 5000 = 70.71+ rods. 2. I have a board in the form of a triangle; the length of one of its sides is 16 feet. My neighbor wishes to purchase one half the board; at what distance from the smaller end must it be divided parallel to the base or larger end? Ans. 11.31+ feet. 3. There is a triangular piece of land, the length of one side of which is 11 rods; required the length of the corresponding side of a similar triangle containing three times as much. Ans. 19.05+ rods. 4. The diameter of a circle is 6 feet, and its area is 28.3 feet; what is the diameter of a circle whose area is 42.5 feet? Ans. 7.35+ feet. 5. If an anchor, which weighs 20001b., requires a cable 3 inches in diameter, what should be the diameter of the cable, when the anchor weighs 40001b.? Ans. 4.24+ inches. 6. A rope 4 inches in circumference will sustain a weight of 10001b.; what must be the circumference of a rope that will sustain 5000lb.? Ans. 8.94+ inches. 7. There is a triangle containing 72 square rods, and one of its QUEsrIoNs. - Art. 290. What proportion do circles have to each other? - Art. 291. What is the rule for finding the side, diameter, &c., of a surface similar to a given surface? - Art. 292. What is the rule for finding the area of a surface similar to a given surface?

Page 280 280 APPLICATION OF THE SQUARE ROOT. [SECT. XXXVII1 sides measures 12 rods; what is the area of a similar triangle whose corresponding side measures 8 rods? Ans. 32 rods. 8. A gentleman has a park, in the form of a right-anglea triangle, containing-950 square rods, the longest side or hypothenuse of which is 45 rods. He wishes to lay out another in the same form, with an hypothenuse ~ the length of the first; required the area. Ans. 105.55+ square fods. 9. If a cylinder 6 inches in diameter contain 1.178+ cubic feet, how many cubic feet will a cylinder of the same length contain that is 9 inches in diameter? Ans. 2.65+ feet. 10. If a pipe 2 inches in diameter will fill a cistern in 20{ minutes, how long would it take a-pipe that is 3 inches hi diameter? Ans. 9 minutes. 11. A tube ~ of an inch in diameter will empty a cistern in 50 minutes; required the time it will empty the cistern, when there is another pipe running into it ~-of an inch in diameter. Ans. 62-4 minutes. ART 293. To find the side of a square that can be inscribed in a circle of a given diameter. A square is said to be inscribed in a circle when each of its angles or corners touches the circumference. It may be conceived to be composed. of two right-angled triangles, the base and perpendicular of each being equal, and their hypothenuse the diameter of the circle, as seen in the diagram. Hence the RULE. - Extract the square root of half the square of the diameter, and it is the.side of the inscribed square. EXAMPLES FOR PRACTICE. 1. What is the length of one side of a square that can be inscribed in a circle, whose diameter is 12 feet? Ans. 8.48+ feet. 2. How large a square stick may be hewn from a round one, which is 30 inches in diameter? Ans. 21.2+ inches square. 3. A has a cylinder of lignum-vitse, 194 inches long and 1I inches in diameter; how large a square ruler may be made from it'? Ans. 1.06+ inches square. QUESTIONS. - Art. 293. When is a square said to be inscribed in a circle? Of what may the inscribed square be conceived to be composed? What part of the circle is the hypothenuse of the two triangles? What is the rule for finding the side of the inscribed square?

Page 281 .SECr. XXXVIII.] EXTRACTION OF THE CUBE ROOT. 281 EXTRACTION OF THE CLUBE ROOT. ART. 294. The CUBE ROOT is the root of any third power. It is so called, because the cube or third power of any number represents the contents of a cubic body, of which the cube root is one of its sides. ART. 295, To extract the cube root is to find a number which, being multiplied into its square, will produce the given number. The following numbers in the upper line represent roots, and those in the lower line their third powers, or cubes. Roots, 1 2 3 4 5 6 7 8 9 10 Cubes, I 8 27 64 125 216 343 512 729 1000 It will be observed that the cube or third power of each of the numbers above contains three times as many figures as the root, or three times as many wanting one, or two at most. Hence, to determine the number of figures in the cube root of a given number, Divide it into periods, beginning at the right, each of which, excepting the last, must always contain three Jigures; and the number of periods will denote the number of figures which the root will contain. Ex. 1. I have 17576 cubical blocks of marble, which measure one foot on each side; what will be the length of one of the sides of a cubical pile, which may be formed of them? Ans. 26 feet. OPERATION. 1 7 5 7 6 (2 6, Root. It is evident that the number of )8 blocks or feet on a 22 X 3 00= 12 00) 9 5 7 6, Ist dividend. side will be equal to the cube root of 7 2 0 0, 1st addition. 17576. (Art. 294.) 62 X 2 X 3 0 21 6 0, 2d "addition. Beginning at the 63 21 6, 3d addition. right hand, we divide the number 9 5 7 6, Subtrahend. into periods, by placing a point over the right-hand figure of each period. We then find the greatest cube number in the left-hand period, 17 (thousands), to be 8 (thousands), QUESTIONS. - Art. 294. What is the cube root, and why so called? - Art. 295. What is meant by extracting the cube root.? How many more figures in the cube of any number than in the root? How do you ascertain the number of figures in the cube root of any number? What is found by extracting the cube root of the number in the example? What is first done after separating the number into periods? 24*

Page 282 282 EXTRACTION OF TIHE CUBE ROOT. [SECT. XXXVIII and its root 2, which we place in the quotient or root. As 2 is in the place of tens, because there is to be another figure in the root, its value is 20, and it represents the side of a cube (Fig. 1) the contents of which are 8000 cubic feet; thus, 20 X 20 X 20= 8000. We now subtract the cube of 2 (tens) = 8 (thousands) from the first period, 17 Fi (thousands), and have 9 (thousand) feet 20 remaining, which,-being increased by the next period, makes 9576 cubic feet. This must. be added to three sides of the cube, 20 Fig. 1, in order that it may remain a cube. To do this, we must find the superficial 20 I i11!iiMii:,'lfqI contents of the three sides of the cube, to which the additions are to be made. Now, Im i iKiiii since one side is 2 (tens) or 20 feet square,'20 its superficial. contents will be 20 X 20 400 square feet, and this multiplied by 3 will be the superficial contents of three sides; thus, 20 X 20 X 3 - 1200, or, which is the same thing, we multiply the square of the quotient figure, or root, by 300; thus, 22 X 300 = 1200 square feet. Mtliking this number a divisor, we divide the dividend 9576 by it, and obtain 6, which we place in _j:b>. 2. the root. This 6 represents the thickness of each of the three additions to be: made to the cube, and their superficial contents being multiplied by it, we have cube, we find that there are three other deficiencies, n n, o o, and r r, the length of which is equal to one side of the adclitions, 2 (tens), or 20 feet; and their breadth and thickness, 6 feet, equal to the thickness of the additions. Therefore,-to find thie solid contents of the additions, necessary to supply these deficiencies, we multiply the product of their length, breadth, and thickness, by the number of additions; thus, 6 X 6 X 20 X 3 =2160, or, which is the same thing, we multiply the square of the last quotient figure by the former figure of the root, and that product by 30; thus, 62 X 2 QUESTIONS. - What is done with this greatest cube number, and what part of Fig. 1 does it represent? What is done with the root? What is its value, and what part of the figure does it represent? How are the cubical contents of the figure found? What constitutes the remainder after subtracting the cube number from the left-hand period? To how many sides of the cube must this remainder be added? How do you find the divisor? What parts of the figure does it represent? Hlow do you obtain the last figure of the root? What part of Fig. 2 does it represent? What parts of the figure does the product represent? What three other deficiencies in the figure?

Page 283 SECT. XXXVIII.] EXTRACTION OF THE CUBE ROOT. 283 X 30 -2160 cubic feet for the contents of the additions s s, u u, and v v, as seen in Fig. 3. These additions being made to the cube, Fig. 3. we still observe another deficiency of the 20 cubical space xx *x, the length, breadth, 6 and thickness of which are each equal 20 to the thickness of the other additions, 6 which is 6 feet. Therefore, we find the 6 [20dlil.l'!uilitli!' 2 contents of the addition necessary to l\ i supply this deficiency by multiplying its 20 length, breadth, and thickness together, or cubing the last figure of the root; thus, 6 X 6 X 6 = 216 cubic feet for the contents of the addition z z z, as seen in 20 6 Fi,. 4. The cube is now complete, and, if we add together the several additions that hgiy 4. have been made to it, thus, 7200 -+ 2160 + 216 = 956, we obtain the number 2 2 of cubic feet remaining after subtracting the first cube, which, being subtracted - from the dividend in the opieration, leaves no remainder. Hence, the cubical pile, I Ili formed is 26 feet on each side; since 26 26 I < 26 X 26 = 17576, the given number of blocks, and the sum of the several partsof Fig. 4. Thus, 8000 + 7200 - iii 2160 +- 216 = 17576. IIence the fol-'26 lowing RULE. - Separate the given number into as many periods as possible of three figures each, by placing a point over the unit fiyure, and every third figure beyond the place of units. Find the greatest cube in the left-hand period, and place its root on the right. Subtract the cube, thusfound, from this period, and to the remainde, bringL down the next periodfor a dividend. Multiply the square of the root alreadyfound by 300 for a divisor, by which divide the dividend, and place the quotient, usually diminished by one or two units, for the next figure of the root. Multiply the divisor by the last figure of the root, and write the product under the dividend; then multiply the square of the last figure of the root by its former figure or figures, and this product by 30, and p/lace the product under the last; under all set the cube of the last Afiure of the root, and call their sum the subtrahend. QUESTIONS. - How do you find their contents? What parts of Fig. 3 does the product represent? What other deficiency do you observe? To what are its length, breadth, and thickness, equal? How do you find its contents? What part of Fig. 4 does it represent? What is the rule for extracting the cube root?

Page 284 284 EXTRACTION OF THE CUBE ROOT. [SECT. XXXVIII. Subtract the subtrahend from the dividend, and to the remainder b6ing down the next period for a new dividend, with which proceed as before; and so on, till the whole is completed. NOTE 1.- In separating the given number into periods, when the num her of the figures is not divisible exactly by 3, the left-hand period will contain less than 3 figures. NOTE 2. The observations made in Notes 2, 3, and 4, under square root, are equally applicable to the cube root, except in pointing off decimals each period must contain three figures, and two ciphers must be placed at the right of the divisor when it is not contained in the dividend. EXAMPLES FOR PRACTICE. 1. What is the cube root of 78402752? Ans. 428 OPERATION. 7 84 0 7 5 2 (4 2 8 Root 64 4 8 0 0 )1 4 4 02 = 1st dividend. 9600 480 8 1 0 0 8 8 -- 1st subtrahend. 5 2 9 2 0 0) 4 3 1 4 7 5 2 = 2d dividend. 4233600 80640 512 4 3 1 4 7 5 2 - 2d subtrahend. 2. What is the cube root of 74088? Ans. 42. 3. What is the cube root of 185193? Ans. 57. 4. What is the cube root of 80621568? Ans. 432. 5. What is the cube root of 176558481? Ans. 561. 6. What is the cube root of 257259456? Ans. 636. 7. What is the cube root of 1860867? Ans. 123. 8. What is the cube root of 1879080904? Ans. 1234. 9. What is the cube root of 41673648.563? Ans. 346.7. 10. What is the cube root of 483921.516051? Ans. 78.51. 11. What is the cube root of 8.144x65728? Ans. 2.012. 12. What is the cube root of.075686967? Ans..423. QUESTION. - How many ciphers must be placed at the right of the divisor when it is not contained in the dividend?

Page 285 SECT. XXXVIII,1 EXTRACTION OF THE CUBE ROOT. 285 ART. 296, When it is required to extract the cube root of a common fraction, or a mixed number, it is prepared in the same manner as directed in square root. (Art. 272.) EXAMPLES FOR PRACTICE. 1. What is the cube root of 81T5? Ans. 4.334+ 2. What is the cube root of,2f9? Ans. 712. 3. What is the cube root of 49 8? Ans. 32 4. What is the cube root of 166-? Ans. 51. 5. What is the cube root of 85.6%? Ans. 42. APPLICATION OF THE, CUBE ROOT. ART. 297. THE cube root may be applied in finding the dimensions and contents of cubes and other solids. i. A carpenter wishes to make a cubical cistern that shall contain 2744 cubic, feet of water; what must be the length of one of its sides? Ans. 14 feet. 2. A farmer has a cubical box that will hold 400 bushels of grain; what is the depth of the box? Ans. 7.92+ feet. 3. There is a cellar, the length of which is 18 feet, the width 15 feet, and the depth 10 feet; what would be the depth of another cellar of the same size, having the length, width, and depth equal? Ans. 13.92+ feet. ART. 298. A SPHERE is a solid bounded by one continued convex surface, every part of which is equally distant from a point within, called the centre. A The diameter of a sphere is a straight line passing through the centre, and terminated by the l surface; as A B. B ART. 299. A CONE is a solid having a circle for its base, and its top terminated in a point, called the vertex. QUESTIONS. - Art. 296. How is a common fraction or a mixed number prepared for extracting the square root? - Art. 297. To what may the cube root be applied? - Art. 298. What is a sphere? What is the diameter of a sphere? - Art. 299. What is a cone?

Page 286 286 APPLIOATION OF THE CUBE ROOT. [SECT. XXXVIII: The altitude of a cone is its perpendicular height, or a line drawn from the vertex perpendicular to the plane of the base; as B 0. ART. 300. Spheres are to each other as the cubes of their diameters, or of their circumferences. Similar cones are to each other as the cubes of their altitudes, or the diameters of their bases. All similar solids are to each other as the cubes of their homologous or corresponding sides, or of their diameters. ART. 301. To find the contents of any solid which is similar to a given solid. RULE. - State the question as in Proportion, and cube the given sides, diameters, altitudes, or circumferences, and the fourth term of the proportion is the required answer. ART. 302. To find the side, diameter, circumference, or altitude, of any solid, which is similar to a given solid. RULE. - State the question as in Proportion, and cube the given sides, diameters, circumferences, or altitudes, and the cube root of the fourth term of the proportion is the required answer. EXAMPLES FOR PRACTICE. 1. If a cone 2 feet in height contains 456 cubic feet, what are the contents of a similar cone, the altitude of which is 3 feet? Ans. 1539 cubic feet. OPERATION. 23:33::456:1539. 2. If a cubic piece of metal, the side of which is 2 feet, is worth $6.25, what is another cubical piece of the same kind worth, one side of which is 12 feet? Ans. $1350. 3. If a ball, 4 inches in diameter, weighs 501b., what is the weight of a ball 6 inches in diameter? Ans. 168.7+-lb. QUESTIONS. - What is the altitude of a cone? Art. 300. What proportion do spheres have to each other? What proportion do cones have to each other? What proportion do all similar solids have to each other? - Art. 301. What is the rule for finding the contents of a solid similar to a given solid? - Art. 302. What is the rule for finding the side, diameter, &co., of a solid similar to a given solid?

Page 287 SEcT. XXIX.] ARITHMETICAL PROGRESSION. 287 4. If a sugar loaf, which is 12 inches in height, weighs 161b., how many inches may be broken from the base, that the residue may weigh 81b.? Ans. 2.5+ in. 5. If an ox, that weighs 8001b., girts 6 feet, what is the weight of an ox that girts 7 feet? Ans. 1270.31b. 6. If a tree, that is one foot in diameter, make one cord, how many cords are there, in a similar tree, whose diameter is two feet? Ans. 8 cords. 7. If a bell, 30 inches high, weighs 1000lb., what is the weight of a bell 40 inches high? Ans. 2370.31b. 8. If an apple, 6 inches in circumference, weighs 16 ounces, what is the weight of an apple 12 inches in circumference? Ans. 128 ounces. 9. A and B own a stack of hay in a conical form. It is 15 feet high, and A owns 2 of the stack; it is required to know how many feet he must take from the top of it for his share. Ans. 13.1+ feet. 9 XXXIX. ARITHMETICAL PROGRESSION. ART. 303. WHEN a series of numbers increases or decreases by a constant difference, it is called Arithmetical Progression, or Progression by Difference. Thus, 2, 5, 8, 11, 14, 17, 20, 23, 26, 29. 29, 26, 23, 20, 17, 14, 11, 8, 5, 2. The first is called an ascending series or progression. The second is called a descending series or progression. The numbers which form the series are called the terms of the progression. The first and last terms are called the extremes, and the other terms the means. The constant difference is called the common difference of the progression. Any three of the five following things being given, the other two may be found: QUEsTroNs. -Art. 303. What is arithmetical progression? What is an ascending series? What a descending series? What are the terms of a progression? What the extremes? What the means?

Page 288 288 ARITHMETICAL PROGRESSION. [SECT. XXiX. 1st. The first term, or first extreme; 2d. The last term, or last extreme; 3d. The number of terms; 4th. The common difference; 5th. The sunm of the terms. ART. 304. To find the common difference, the firbt term, last term, and number of terms, being given. ILLUSTRATION. - In the following series, 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 2 and 29 are the extremes, 3 the common difference, 10 the number of terms, and the sum of the series 155. It is evident that the number of common differences in any series must be 1 less than the number of terms. Therefore, since the number of terms in this series is 10, the number of common differences will be 9, and their sum will be equal to the difference of the extremes; hence, if the difference of the extremes (29 - 2 - 27) be divided by the number of common differences, the quotient will be the common difference. Thus, 27 - 9 = 3, the common difference. Hence the following RULE. - Divide the difference of the extremes by the number of termless one, and the quotient is the common difference. EXAMPLES FOR PRACTICE. 1. The extremes of a series are 3 and 35, and the number of terms is 9; what is the common difference? Ans. 4. OPERATION. 35 3 = 4 common difference. 9-1 2. If the first term is 7, the last term 55, and the number of terms 17, required the common difference. Ans. 3. 3. If the first term is 4, the last term 14, and the number of terms 15, what is the common difference? Ans. -. 4. If a man travels 10 days, and the first day goes 9 milcs, QUESTIONS.- Art. 304. What is the common difference? What five things are named, any three of which being given the other two can be found. What is the rule for finding the common difference, the first term, last term, and number of terms, being given?

Page 289 SECT. XXXIX.] ARITHMETICAL PROGRESSION. 289 and the last 17 miles, and increases each day's travel by an equal difference, what is the daily increase? Ans. ~ miles. ART. 305. To find the sum of all the terms, the first term, last term, and number of terms, being given. ILLUSTRATION. - Let the two following series be arranged as follows: 2, 5, 8, 11, 14, 17, 20, - 77, sum of first series. 20, 17, 14, 11, 8, 5, 2, - 77, sum of inverted series. 22, 22, 22, 22, 22, 22, 22, =154, sum of both series. From the arrangement of the above series, we see that, by adding the two as they stand, we have the same number for the sum of the successive terms, and that the sum of both series is double the sum of either series. It is evident that, if 22 in the above series be multiplied by 7, the number of terms, the product will be the sum of bothl series; thus, 22 X 7 = 154; and, therefore, the sum of either series will be 154 + 2 - 77. But 22 is the sum of the extremes in each series; thus, 20 + 2 --- 22. Therefore, if the sum of the extremes be multiplied by the number of terms, the product will be double the sum of either series. Hence, RULE 1. - Multiply the sum of the extremes by the number of terms and half the product will be the sum of the series. Or, RULE 2.- Multiply the sum of the extremes by half the number oJ terms, and the product is the sum required. EXAMPLES FOR PRACTICE. 1. If the extremes of a series are 5 and 45, and the number of terms 9, what is the sum of the series? Ans. 225. OPERATION. ( 4 5 + 5 ) X 9 — 225, sum of the series. 2 2. John Oaks engaged to labor for me 12 months. For the first month I was to pay him $7, and for the last month $51. In each successive month he was to have an equal addition to his wages; what sum did he receive for his year's labor? Ans. $348. QUESTION. - Art. 305. What is the rule for finding the sum of all the terms, the first term, last term, and number of terms, being given?'25

Page 290 29.(0 ARITHMETIOAL PROG1RESSION. [SECT. XXXIX. 3. I have purchased from W. Hall's nursery 100 fruit-trees of various kinds, to be set around a circular lot of land, at the distance of one rod from each other. Having deposited them on one side of the lot, how far shall I have travelled when I have set out my last tree, provided I take only one tree at a time, and travel on the same line each way? Ans. 9801 rods. ART. 306. To find the number of termS, the extremes and common difference being given. ILLUSTRATION. - Let the extremes of a series be 2 and 29, and the common difference 3. The difference of the extremes will be 29 - 2 - 27. Now, it is evident that, if the difference of th9 extremes be divided by the common difference, the quotient will be the number of common differences; thus, 27. 3 ~ 9. It has been shown (Art. 303) that the number of terms is 1 more than the number of differences; therefore, 9 + 1 = 10 is the number of terms in this series. Hence the following RULE. - Divide the difference of the extremes by the common difference, and the quotient, increased by 1, will be the number of terms required. EXAMPLES FOR PRACTICE. 1. If the extremes of a series are 4 and 44, and the common difference 5, what is the number of terms? Ans. 9. OPERATION. 44-4 + 1 -=9, number of terms. 5 2. A man going a journey travelled the first day 8 miles, and the last day 47 miles, and each day increased his journey by 3 miles. How many days did he travel? Ans. 14 days. ART. 307. To find the sum of the series, the extremes and common difference being given. ILLUSTRATION. - Let the extremes be 2 and 29, and the common difference 3. The difference of the extremes will be 292 -27; and it has been shown (Art. 306) that if the difference of the extremes be divided by the common difference, the QUESTION.- Art. 306. What is the rule for finding the number of terms, the extremes and common difference being given?

Page 291 SECT. XXXIX.] ARITHMETICAL PROGRESSION. 291 quotient will be the number of terms less one. Therefore, the number of terms less one will be 27 3 -- 9, and the number of terms 9 + 1 = 10. It was also shown (Art. 305) that, if the number of terms be multiplied by the sum of the extremes, and the product divided by 2, the quotient will be the sum of the series. Hence the RULE. - Divide the difference of the extremes by the common difference, and to the quotient add 1; multiply this sum by the sum of the extremes, and half the product is the sum of the series. EXAMPLES FOR PRACTICE. 1. If the two extremes are 11 and 74, and the common difference 7, what is the sum of the series? Ans. 425. OPERATION. 74-11 (74+11) X 10 + I = 10; = 4 2 5, sum of series. 7 2 2. A pupil commenced Virgil by reading 12 lines the first day, 17 lines the second day, and thus increased every day by 5 lines, until he read 137 lines in a day. How many lines did he read in all? Ans. 1937 lines. ART. 308. To find the last term, the first term, the number of terms, and the common difference, being given. ILLUSTRATION. - Let the first term of a series be 2, the number of terms 10, and the common difference 3. It has been shown (Art. 304) that the number of common differences is always 1 less than the number of terms; and that the sum of the common differences is equal to the difference of the extremes; therefore, since the number of terms is 10, and the common difference 3, the difference of the extremes will be (10 - 1) X 3 =- 27; and this difference, added to the first term, must give the last term; thus, 2 + 27 = 2S. Hence the following RULE. - lultiply the number of terms less 1 by the common difference, and add this product to the first term for the last term. NOTE. - If the series is descending, the product must be subtracted from Phe first term. QUESTIONS. - Art. 307. What is the rule for finding the sum of the series, the extremes and common difference being given? -Art. 308. What is the rule for finding the last term, the first term, the number of terms, and common difference, being'given?

Page 292 292 ANNUITIES AT SIMPLE INTEREST. [SECT. XXXIX EXAMPLES FOR PRACTICE. 1. If the first term is 1, the number of terms 7, and the comnon difference 6, what is the last term? Ans. 37. OPERATION. 1 + (7- 1) X 6 - 3 7, last term. 2. If a man travel 7 miles the first day of his journey, and 9 miles the second, and shall each day travel 2 miles further than the preceding, how far will he travel the twelfth day? Ans. 29 miles. 3. If A set out from Portland for Boston, and travel 20~ miles the first day, and on each succeeding day 11 miles less than on the preceding, how far will he travel the tenth day? Ans. 6:~ miles. ANNUITIES AT SIMPLE INTEREST BY ARITHMETICAL PROGRESSION. ART. 309. AN ANNUITY is a sum of money to be paid annu ally, or at any other regular period, either for a limited time or forever. The present worth of an annuity is that sum which being put at interest will be sufficient to pay the annuity. The amount of an annuity is the interest of all the payments added to their sum. Annuities are said to be in arrears when they remain unpaid after they have become due. ART. 310, To find the amount of an annuity at simple interest. Ex. 1. A man purchased a farm for $2000, and agreed to pay for it in 5 years, paying $400 annually; but, finding himself unable to make the annual payments, he agreed to pay the whole amount at the end of the 5 years, with the simple interest, at 6 per cent., on each payment, from the time it became due till the time of settlement; what did the farm cost him? Ans. $2240. ILLUSTRATION. - It is evident the fifth payment will be $400, without interest; the fourth will be on interest 1 year, and will amount to $424; the third will be on interest 2 years, and will amount to $448; the second will be on interest 3 QUESTIONS. - Art. 309. What is an annuity? What is meant by the present worth of an annuity? By the amount? When are annuities said to be in arrears?

Page 293 SECT. XXXIX.] ANNUITIES AT SIMPLE INTEREST. 293 years, and will amount to $472; and thefirst will be on interest 4 years, and will amount to $496. Therefore, these several sums form an arithmetical series; thus, 400, 424, 448, 472, 496; of which the fifth payment, or the annuity, is the first term, the interest on the annuity for one year the com7nwnm difference, the time in years, the number of terms, and the amount of the annuity, the sum of the series. The sum of this series is found by Art. 305; thus,(400 + 496) X 5 = $2240. Hence the RULE. - First find the last term of the series (Art. 308), and then the sum of the series (Art. 305). NOTE. - If the payments are to be made semi-annually, quarterly, &c., these periods will be the number of terms, and the interest of the annuity for each period the common difference. EXAMPLES FOR PRACTICE. 2. What will an annuity of $250 amount to in 6 years, at 6 per cent. simple interest? Ans. $1725. 3. What will an annuity of $380 amount to in 10 years, at 5 per cent. simple interest? Ans. $4655. 4. An annuity of $825 was settled on a gentleman, January 1, 1840, to be paid annually. It was not paid until January 1, 1848; how much did he receive, allowing 6 per cent. simple interest? Ans. $7986. 5. A gentleman let a house for 3 years, at $200 a year, the rent to be paid semi-annually, at 8 per cent. per annum, simple interest. The rent, however, remained unpaid until the end of the three years; what did he then receive? Ans. $660. 6. A certain clergyman was to receive a salary of $700, to be paid annually; but, for certain reasons, which we fear were not very good, his parishioners neglected to pay him for 8 years; but he agreed to settle with them, and allow them $100 if they would pay him his just due with interest; required the sum received. Ans. $6676. 7. A certain gentleman in Boston has a very fine house, which he rents at $50 per month. Now, if his tenant shall omit payment until the end of the year, what sum should the owner receive, reckoning interest at 12 per cent.? Ans. $633. QUESTrONS. - Art. 310. What forms the first term of a progression in an annuity? What the common difference? What the number of terms? What the sum of the series? What is the rule for finding the amount of an annuity at simple interest? If the'payments are made semi-annually, quarterly, &c., what constitute the terms? What the common difference? 25*

Page 294 294 GEOMETRICAL PROGRESSION. [SECT. XL. ~ XL. GEOMETRICAL PROGRESSION. ART. 311. WHEN there are three or more numbers, and the same quotient is obtained by dividing the second by the first, the third by the second, and the fourth by the third, &c., these numbers are in Geometrical Progression, and may be called a Geometrical Series. Thus, 2, 4, 8, 16, 32, 64. 64, 32, 16, 8, 4, 2. The former is called an ascending series, and the latter a descending series. In the first series the quotient is 2, and is called the ratio; in the second, it is 2.. Hence, if the series is ascending, the quotient is more than unity; if it is descending, it is less than unity. The first and last terms of a series are called extremes, and the other terms means. Any three of the five following things being given, the other two may be found: 1st. The first term, or first extreme; 2d. The last term, or last extreme; 3d. The number of terms; 4th. The ratio; 5th. The sum of the terms, or series. ART. 312. One of the extremes, the ratio, and the number of terms. being given, to find the other extreme. ILLUSTRATION. - Let the first ternim be 2, the ratio 3, and the number of terms 7. It is evident that, if we multiply the first term by the ratio, the product will be the second termn in the series; and if we multiply the second term by the ratio, the product will be the third term; and, in this manner, we may carry the series to any desirable extent. By examining the following series, we find that 2 carried to the 7th term is 1458; thus, QUESTIONS. —Art. 311. When are numbers in geometrical progression? What is an ascending series? What a descending series? What is the ratio of a progression? Is the ratio greater or less than unity in an ascending series? In a descending series? What are the extremes of a series? What the means? What five things are mentioned,'any three of which being given, the other two may be found?

Page 295 ECT. XL.] GEOMETRICAL PROGRESSION. 293 1 2 3 4 5 6 7 2 6 18 54 162 486 1458 The factors of 1458 are 3, 3, 3, 3, 3, 3, and 2, the last of which is the first term of the series, and the others the ratio repeated a number of times one less than the number of terms. But multiplying these factors together is the same as raising the ratio to the sixth power, and then multiplying that power by the first term. Hence the following RULE. - Raise the ratio to a power whose index is equal to the number of terms less one; then multiply this power by the first term, and the product is the last term, or other extreme. NoTE. -This rule may be applied in computing compound interest, the principal being the first term, the amount of one dollar for one year the ratio, the time, in years, one less than the number of terms,.. and the amount the last term. EXAMPLES FOR PRACTICE. 1. The first term of a series is 1458, the number of terms 7, and tht vatio I; what is the last term? Ans. 2. OPERATION. Ratio (~)6 -=7 V; ~T X 14 5 8 = __ _ - 2, the last term. 2. If the first term of a series is 4, the ratio 5, and the numbers of terms 7, what is the last term? Ans. 62500. 3. If the first term of a series is 28672, the ratio i, and the number of terms 7, what is the last term? Ans. 7. 4. The first term of a series is 5, the ratio 4, and the number of terms is 8; required the last term. Ans. 81920. 5. If the first term of a series is 10, the ratio 20, and the number of terms 5, what is the last term? Ans. 1600000. 6. If the first term of a series is 30, the ratio 1.06, and the number of terms 6, what is the last term? Ans. 40.146767328. 7. What is the amount of $1728 for 5 years, at 6 per cent., compound interest? Ans. $2312.453798+. 8. What is the amount of $328.90 for 4 years, at 5 per cent., compound interest"? Ans. $399.78+. 9. A gentleman purchased a lot of land containing 15 acres, agreeing to pay for the whole what the last acre would come QUESTIONS. - Art. 312. What is the rule for finding the other extreme, one of the extremes, the ratio, and number of terms, being given? To what may this rule be applied?

Page 296 296 GEOMETRICAL PROGRESSION. [SECT. XL. to, reckoning 5 cents for the first acre, 15 cents for the second and so on, in a three-fold ratio. What did the lot cost him? Ans. $239148.45. ART. 313. To find the sum of all the terms, the first term the ratio, and the number of terms, being given. ILLUSTRATION. Let it be required to find the sum of the following series: 2, 6, 18, 54. If we multiply each term of this series by the ratio 3, the products will be 6, 18, 54, 162, forming a second series, whose sum is three times the sum of the first series; and the difference between these two series is twice the sum of the first series. Thus, 6, 18, 54, 162, the second series. 2, 6, 18, 54, the first series. 2, 0, 0, 0, 0, 162 - 2 - 160, difference of the two series Now, since this difference is twice the sum of the first series. one half this difference will be the sum of the first series; thus 160 + 2 - 80. It will be observed, by examining the operation above, that if we had simply multiplied 54, the last term of the first series, by the ratio 3, and subtracted 2, the first term, from it, we should have obtained 160; and this being divided by the ratio, 3 less 1, would have given 80, the same number as before, for the sum of the first series. Hence the RULE. 1. - Find the last term as in the preceding article, multiply it by the ratio, and from the product subtract the first term. Then divide this remainder by the ratio, less 1, and the quotient will be the sum of the series. Or, RULE. 2. - Raise the ratio to a power whose index is equal to the number of terms, from which subtract 1; divide the remainder by the ratio, less 1, and the quotient, multiplied by the given extreme, will be the sum of the series required. NoTE 1.- If the ratio is less than a unit, the product of the last term, multiplied by the ratio, must be subtracted from the first term; and, to obtain the divisor, the ratio must be subtracted from unity, or 1. QUESTIONS.- Art. 313. What is the rule for finding the sum of all the terms, the first term, ratio, and number of terms, being given? If the ratio is less than a unit, what must be done with the product of the last term multi. plied by the ratio? How is the divisor obtained when the ratio is less than 1?

Page 297 SEe. XL.] GEOMETRICAL PROGRESSION. 297 NoTE 2.- If the second rule is employed, when the ratio is less than 1, its power, denoted by the number of terms, must be subtracted from 1, and the remainder divided by the difference between 1 and the ratio. EXAMP'LES FOR PRACTICE. 1. If the first term of a series is 12, the ratio 3, and the number of terms 8, what is the sum of the series? Ans. 39360. OPERATION. Ratio 37 X 12 = 26244, the last term; 26244 X 3 = 78732; 78732 - 12 - 78720; 78720 -. (3 — 1) - 39360, the sum of the series. 2. The first term of a series is 5, the ratio', and the number of terms 6; required the sum of the series. Ans. 1316. OPERATION. Ra9tio (2)5 X 5 =1- 6, the last term; ~ X =2- 32S; 5. 2 4,2t-? (1 ~ 32 3325. 3325 (1 335 - 13 166 the sum of the series. 3. If the first term of a series is 8, the ratio 4, and the number of terms 7, required the sum of the series. Ans. 43688. 4. If the first term is 10, the ratio J, and the number of terms 5, what is the sum of the series? Ans. 30 T-68. 5. If the first term is 18, the ratio 1.06, and the number of terms 4, what is the sum of the series? Ans. 78.743+. 6. When the first term is $144, the ratio $1.05, and the number of terms 5, what is the sum of the series? Ans. $795.6909. 7. D. Baldwin agreed to labor for E. Thayer for 6 months. For the first month he was to receive $3, and each succeeding month's wages were to be increased by 2 of his wages for the month next preceding; required the sum he received for his 6 months' labor. Ans. $9177. 8. If the first term of a series is 2, the ratio 6, and the number of terms 4, what is the sum of the series? Ans. 518. 9. A lady, wishing to purchase 10 yards of silk for a new dress, thought $1.00 per yard too high a price; she, however, agreed to give 1 cent for the first yard, 4 for the second, 16 for the third, and so on, in a four-fold ratio; what was the cost of the dress? Ans. $3495.25. QUESTION.- When the second rule is employed, if the ratio is less than 1, what must be done?

Page 298 298 ANNUITIES AT COMPOUND INTEREST. [SECT. XL. ANNUITIES AT COMPOUND INTEREST BY GEOMETRICAL PROGRESSION ART. 314. WHVEN compound interest is reckoned on an annuity in arrears, the annuity is said to be at compound interest; and the amounts of the several payments form a geometrical series, of which the annuity is the first term, the amount of $1.00 for one year the ratio, the years the number of terms, and the amount of the annuity the sum of the series. Hence, ART. 315. To find the amount of an annuity at compound interest, we have the following RULE 1. - Find the sum of the series by either of the preceding rules. (Art. 313.) Or, RULE 2. - Multiply the amount of $1.00,for the given time, found in the table, by the annuity, and the product will be the required amount TABLE, Showing the amount of $1 annuity from 1 year to 40. Years. 5 per cent. 6 per cent. Years. 5 per cent. 6 per cent. 1 1.000000 1.000000 21 35.719252 39.992727 2 2.050000 2.060000 22 38.505214 43.392290 3 3.152500 3.183600 23 41.430475 46.995828 4 4.310125 4.374616 24 44.501999 50.815577 5 5.525631 5.637093 25 47.727099 54.864512 6 6.801913 6.975319 26 51.113454 59.156383 7 8.142008 8.393838 27 54.669126 63.705766 8 9.549109 9.897468 28 58.402583 68.528112 9 11.026564 11.491316 29 62.322712 73.639798 10 12.577893 13.180795 30 66.4388-47 79.058186 11 14.206787 14.971643 31 70.760790 84.801677 12 15.917127 16.869941 32 75.298829 90.88!9778 13 17.712983 18.882138 33 80.063771 97.343165 14 19.598632 21.015066 34 85.066959 104.183755 15 21.578564 23.275970 35 90.220307 111.434780 16 23.657492 25.672528 36 95.836323 119.120867 17 25.840866 28.212880 37 101.628139 127.268119 18 28.132385 30.905653 38 107.709546 135.904206 19 30.539004 33.759992 39 114.095023 145.058458 20 33.065954 36.785591 40 120.799774 154.761966 QUESTIONS. - Art. 314. When is an annuity said to be at compound interest? What do the amounts of the several payments form? What is the first term of the series? What the ratio? What the number of terms? What the sum of the series? - Art. 315. What is the first rule for finding the amount of an annuity? What the second? What does the table show?

Page 299 SECT. XL.] ANNUITIES AT COMPOUND INTEREST. 299 EXAMPLES FOR PRACTICE. 1. What will an annuity of $378 amount to in 5 years, at 6 per cent. compound interest? Ans. $2130.821+. OPERATION BY RULE FIRST. 1.0 6 1 X 37 8= $ 2 1 30.82 1. 1.06 -6 1 OPERATION BY RULE SECOND. 5.637093 X 37 8 _$21 3 0.821+. 2. What will an annuity of $1728 amount to in 4 years, at 5 per cent. compound interest? Ans. $7447.896 -+. 3. What will an annuity of $87 amount to in 7 years, at 6 per cent. compound interest? Ans. $730.263 +. 4. What will an annuity of $500 amount to in 6 years, at 6 per cent. compound interest? Ans. $3487.659 -. 5. What will an annuity of $96 amount to in 10 years, at 6 per cent. compound interest? Ans. $1265.356 +-. 6. What will an annuity of $1000 amount to in 3 years, at 6 per cent. compound interest? Ans. $3183.60. 7. July 4, 1842, H. Piper deposited in an annuity office, for his daughter, the sum of $56, and continued his deposits each year, until July 4, 1848. Required the sum in the office July 4, 1848, allowing 6 per cent. compound interest. Ans. $470.054 +. 8. 0. Greenleaf has two sons, Samuel and William. On Samuel's birth-day, when he was 15 years old, he deposited for him, in an annuity office, which paid 5 per cent. compound interest, the sum of $25, and this he continued yearly, until he was 21 years of age. On William's birth-day, when he was 12 years old, he deposited for him, in an office which paid 6 per cent. compound interest, the sum of $20, and continued this until he was 21 years of age. Which will receive the larger sum, when 21 years of age? Ans. $60.065+ — William receives more than Samuel. 9. I gave my daughter Lydia $10 when she was 8 years old, and the same sum on her birth-day each year, until she was 21 years old. This sum was deposited in the savings bank, which pays 5 per cent. annually. Now, supposing each deposit to remain on interest until she is 21 years of age, required the amount in the bank. Ans. $195.986 —.

Page 300 800 ALLIGATION. [ SECT. XLL ~ XLI. ALLIGATION. ART. 316. ALLIGATION is a rule employed in the solution of questions relating to the compounding or mixing of several ingredients. The term signifies the act -of connecting or tying together. It is of two kinds: Alligation Medial and Alligation Alternate. ALLIGATION MEDIAL. ART. 317. Alligation Medial is the method of finding the mean price of a mixture composed of articles of different values, the quantity and price of each being given. ART. 318. To find the mean price of several articles or ingredients, at different prices, or of different qualities. RULE. - Find the value of each of the ingredients, and divide the amount of their values by the sum of the ingredients. The quotient will be the price of the mixture. EXAMPLES FOR PRACTICE. Ex. 1. A grocer mixed 201b. of tea worth $0.50 a pound, with 301b. worth $0.75 a pound, and 501b. worth $0.45 a pound; what is 1 pound of the mixture worth? Ans. $0.55. OPERATION. $0.5 0 X 2 0 = $1 0.0 0 $0.7 5 X 3 0 = $2 2.5 0 $0.45 X50 =$2 2.50 Sum of ingredients, 1 00 $5 5.0 0, value. Then, $5 5.0 0. 1 0 0 = $0.5 5 per pound. Proof, $0.5 5 X 2 0 lb. = $1 1.0 0 $0.55 X 301b.- $1 6.50 $55.00. $0.55 X 5 0 lb. = $2 7.5 0 2. I have four kinds of molasses, and a different quantity of each, as follows: 30 gal. at 20 cents, 40 gal. at 25 cents, 70 gal. at 30 cents, and 80 gal. at 40 cents; what is a gallon of the mixture worth? Ans. $0.31-4T. 3. A farmer mixed 4 bush. of oats at 40 cents, 8 bush. of QUESTIONS. - Art. 316. What is alligation? What two kinds are there? - Art. 317. What is alligation medial? —Art. 318. What is the rule for finding the mean price of several articles at different prices? How does it appear that this process will give the mean price of the mixture?

Page 301 SECT. XL1.] ALLIGATION ALTERNATE. 301 corn at 85 cents, 12 bush. rye at $1.00, and 10 bush. of wheat at $1.50 per bushel. What will one bushel of the mixture be worth? Ans. $1.04?-. ALLIGATION ALTERNATE. ART. 319. Alligation Alternate is the method of finding what quantity of ingredients or articles, whose prices or qualities are given, must be taken, to compose a mixture of any given price or quality. ART. 320, To find what quantity of each ingredient must be taken to form a mixture of a given price. Ex. 1. I wish to mix spice, at 20 cents, 23 cents, 26 cents, and 28 cents per pound, so that the mixture may be worth 25 cents per pound. How many pounds of each must I take? FIRST OPERATION. PROOF. - lib. at 20cts. gain 5cts. lib. at 20cts. = 20cts. Mean price 25cts.~ llb. at 23cts. gain 2ets. Ans lib. at 23cts. - 23cts. lib. at 26cts. loss let. llb. at 26cts. - 26cts. Lllb. at 28cts. loss 3cts. 21b. a 28cts. -6cts. 51bs. whole val. $1.25 lib. at 28cts. loss 3cts. $1.25 5 =-25cts. per lb. Compared with the mean or average price given, by taking llb. at 20 cents there is a gain of 5 cents, by taking llb. at 23 cents a gain of 2 cents, by taking llb. at 26 cents a loss of. 1 cent, and by taking l1b. at 28 cents a loss of 3 cents; making an excess of gain over loss of 3 cents. Now, it is evident that the mixture, to be of the average value named, should have the several items of gain and loss in the aggregate exactly offset one another. This balance we can effect, in the present case, either by taking 31b. more of the spice at 26 cents, or llb. more of spice at 28 cents. We take the llb. at 28 cents, and thus have a mixture of the required average value, by having taken, in all, llb. at 20 cents, llb. at 23 cents, llb. at 26 cents, and 21b. at 28 cents. We prove the correctness of the result by dividing the value of the whole mixture, or $1.25, by the number of pounds taken, or 5, which gives 25 cents, or the given mean price per pound. SECOND OPERATION. Having arranged in a column (20cts. 31b.' the prices of the ingredients, 23cts.] Ilb' with the given mean price on 25 - Act s.m Ans. the left, we connect together 26cts. 21b. the terms denoting the price 28cts.- 51b. of each ingredient, so that a price less than the given QUESTIONS. - Art. 319. What is alligation alternate? How do you connect the prices? Explain the first operation. low is it proved to be correct? 26

Page 302 302 ALLIGATION ALTERNATE. [SECT. XLI. mean is united with one that is greater. We then proceed to find what quantity of each of the two kinds, whose prices have been connected, can be taken, in making a mixture, so that what shall be gained on the one kind shall be balanced by the loss on. the other. By taking 11b. of spice at 20 cents, the gain will be 5 cents; and by taking llb. at 28 cents, the loss will be 3 cents. To equalize the gain and loss in this case, it is evident we should take as many more pounds of that at 28 cents as the loss on lib. of it is less than the gain on llb. of that at 20 cents; or,.in other words, the ingredients taken should be in the inverse ratio (Art. 236) of the difference between their respective prices and the given mean price. Therefore, we take 5lbs. at 28 cents, and 31bs. of that at 20 cents, and the loss, 3cts. X 5 = 15 cents, on the former, exactly offsets the gain, 5cts. X 3 = 15 cents, on the latter. We write the 31b. against its price, 20 cents; and the 51hb. against its price, 28 cents. In like manner we determine the quantity that may be taken of the other two ingredients, whose prices are connected, by finding the difference between each price and the mean price; and, as before, write the quantity taken against its price. We obtain, as a result, 31b. at 20 cents, l1b. at 23 cents, 21b. at 26 cents, 51b. at 28 cents; this, in the same manner as the other answer, may be proved to satisfy the conditions of the question, since examples of this kind admit of several answers. RnLE. - Write the prices of the ingredients in a column, with the mean price on the left, and connect the price of each ingredient which is less than the given mean price with one that is greater. Write the difference between the mean price and that of each of the ingredients opposite to the price with which it is connected; and the number set against each price is the quantity of the ingredient to be taken at that price. NOTE.- There will be as many different answers as there are different ways of connecting the prices, and by multiplying and dividing these answers they may be varied indefinitely. EXAMPLES FOR PRACTICE. 2. A farmer wishes to mix corn at 75 cents a bushel, with rye at 60 cents a bushel, and oats at 40 cents a bushel, and wheat at 95 cents a bushel; what quantity of each must he take to make a mixture worth 70 cents a bushel? FIRST OPERATION. SECOND OPERATION. THIRD OPERATION. Ans. Ans. r40-25 440- 5 (40=. 25~+ 5=30 601 5 60- 25 601.- 5 +25 30 75J 10 75{ 30 7 75j 1 0-f30 40 t 95'- 30 95- 10 95- 30+10 40 QUESTIONS. - What is the rule for alligation alternate? How can you obtain different answers? Are they all true?

Page 303 SECT. XLI.] ALLIGATION ALTERNATE. 30: 3. I have 4 kinds of salt,worth 25, 30, 40, and 50 cents per bushel; how much of each kind must be taken, that a mixture might be sold at 42 cents per bushel? Ans. 8 bushels at 25, 30, and 40 cents, and 31 bushels at 50 cents. ART. 321. When the quantity of one ingredient is given to find the quantity of each of the others. Ex. 1. How much sugar, that is worth 6, 10, and 13 cents a pound, must be mixed with 201b. worth 15 cents a pound, so that the mixture will be worth 11 cents a pound? OPERATION. f 6 2 44 10 2 A Then, 5: 1::20: 4 11 13_ H 5: 2 20: 8 Ans. 15 —-Ja 5: 4 20 16) By the conditions of the question we are to take 20tb. at 15 cents a pound; but by the operation we find the difference at 15 cents a pound to be only 51b., which is but 4 of the given quantity. Therefore, if we increase the 51b. to 20, the other differences must be increased in the same proportion. Hence the propriety of the following RULE. - Find the difference between each price and the mean price; then say, As the difference of that ingredient whose quantity is given is to each of the differences separately, so is the quantity yiven to the several quantities required. EXAMPLES FOR PRACTICE. 2. A farmer has oats at 50 cents per bushel, peas at 60 cents, and beans at $1.50. These he wishes to mix with 30 bushels of corn at $1.70 per bushel, that he may sell the whole at $1.25 per bushel; how much of each kind must he take? Ans. 18 bushels of oats, 10 bushels of peas, and 26 bushels of beans. 3. A merchant has two kinds of sugar, one of which cost him 10 cents per lb., and the other 12 cents per lb.; he has also 001b. of an excellent quality, which. cost him 15 cents per lb. Now, as he ought to make 25 per cent. on his cost, how much of each quantity must be taken that he may sell the mixture at 14 cents per lb.? Ans. 38341b. at 10 cents, and l00b. at 12 cents. QUESTION.- Art. 321. What is the rule for finding the quantity of each of the other ingredients when one is given?

Page 304 804 ALLIGATION ALTERNATE. [SECT. XLI. ART. 322. When the sum of the ingredients and their mean price are given, to find what quantity of each must be taken. Ex. 1. I have teas at 25 cents, 35 cents, 50 cents, and 70 cents a pouiir, with which I wish to make a mixture of 1801b. that will be worth 45 cents a pound. How much of each kind must I take? OPERATION. ( 25 25 Then, 60: 25: 180: 75 35- 5 60: 5::180:15 50J 10 60: 10: 180: 30 70- 20 60:20::180:60 Sum of differences, 60 Proof, 180 By the conditions of the question, the weight of the mixture is 1801b., but by the operation we find the sum of the differences to be only 601b., which is but i of the quantity required. Therefore, if we increase 60b. to 180, each of the differences must be increased in the same proportion, in order to make a mixture of 1801b., the quantity required. IHence the RULE. - Find the differences as before; then say, As the sum of the difjerences is to each of the differences separately, so is the given quantity to the required quantity of each ingredient. EXAMPLES FOR PRACTICE. 2. John Smith's " great box " will hold 100 bushels. He has wheat worth $2.50 per bushel, and rye worth $2.00 per bushel. How much chaff, of no value, must he mix with the wheat and rye, that, if he fill the box, a bushel of the mixture may be sold at $1.80? Ans. 40bu. each of wheat and rye, and 20 bushels of chaff. 3. I have two kinds of molasses, which cost me 20 and 30 cents per gallon; I wish to fill a hogshead, that will hold 80 gallons, with- these two kinds. How much of each kind must be taken, that I may sell a gallon of the mixture at 25 cents per gallon, and make 10 per cent. on my purchase? Ans. 58-5L of 20 cents, and 21-9T of 30 cents. 4. I have sugars at 10 cents and 15 cents per pound. How much of each must be taken, that a mixture containing 60 pounds shall be worth $7.20? Ans. 36 pounds at 10 cents, and 24 pounds at 15 cents. QUESTION. - Art. 322. How do you find what quantity of each ingredient must be taken when the sum and mean price are given?

Page 305 SECT. XII.] PERMUTATIO0N. 305 O XLII. PERMUTATION. ART. 323. PERMUTATION is the process of finding the different orders in which may be arranged a given number of things. ART. 324. To find the number of different arrangements that can be made of any given number of things. Ex. 1. How many different numbers may be formed from the figures of the following number, 432, making use of three figures in each number? Ans. 6. FIRST OPERATION. In the 1st operation, 432, 423, 342, 324, 24 3, 234. we have made all the different arrangements SECOND OPERATION. that can be made of the 1 X 2 X 3 = 6. given figures, and find the number to be 6. In the second operation, the same result is obtained by simply multiplying together the first three of the digits, a number equal to the number of figures to be arranged. Hence the following RULE. - Multiply together all the terms of the natural series of numbers,from 1 up to the given,number, and the last product will be the answer required. EXAMIPJLES FOR PRACTICE. 2. My family consists of nine persons, and each person has his particular seat around my table. Now, if their situations were to be changed once each day, for how many days could they be seated in a different position? Ans. 362880 days, or 994 years 70 days. 3. On a certain shelf in my library there are 12 books. If a person should remove them without noticing their order, what would be the probability of his replacing them in the same position they were at first? Ans. 1 to 479001600. 4. How many words can be made from the letters in the word "Embargo," provided that any arrangement of them may be used, and that all the letters shall be taken each time? Ans. 5040 words. QUESTIONS. - Art. 323. What is permutation? - Art. 324. What is the rule for finding the number of arrangements that can be made of any given number of things? 26*

Page 306 306 MENSURATION OF SURFACES. [SECT. XLIII. ~ XLIII. MENSURATION OF SURFACES. ART. 325. A SURFACE is a magnitude, which has length and breadth without thickness. The surface or superficial contents of a figure are called its area. ART. 326. An ANGLE is the inclination or opening of two lines, which meet in a point. A right angle is an angle formed by one line A falling perpendicularly on another, and it contains 90 degrees; as A B C. i - An acute angle is an angle less than a right angle, or less than 90 degrees; as E B C. B An obtuse angle is an angle greater than a F right angle, or more than 90 degrees; as F B C. B THE TRIANGLE. ART. 327. A TRIANGLE is a figure having three sides and three angles. It receives the particular names of an equilateral triangle, isosceles triangle, and scalene triangle. It is also called a right-angled triangle when it has one right angle; an acute-angled triangle, when it has all its angles acute; and an obtuse-angled triangle, when it has one obtuse angle. The base of a triangle, or other plane figure, is the lowest side, or that which is parallel to the horizon; as C D. The altitude of a triangle is a line drawn from one of its angles perpendicular to its opposite side or base; as A B. A An equilateral triangle is a figure which has its three sides equal. / B QUESTIONS. - Art. 325. What is a surface? What are the superficial contents of a figure called? - Art. 326. What is an angle? What is a right angle?'An acute angle? An obtuse angle? - Art. 327. What is a triangle? What particular names does it receive? When is it called a right-angled triangle? When an acute-angled triangle? When an obtuse-angled triangle? What is the base of a triangle? What the altitude? What is an equilateral triangle?

Page 307 bECT. XLIII.] MENSURATION OF SURFACES. 307 An isosceles triangle is a figure which has two of its A sides equal. A scalene trzangle is a figure which has its threee sides unequal. A right-angled triangle is a figure having three sides and three angles, one of which is a right angle. ART. 328. To find the area of a triangle. RULE 1.- Multiply the base by half the altitude, and the product will be the area. Or, RULE 2. -Add the three sides together, take half that sum, andfrom this subtract each side separately; then multiply the half of the sum and these remainders together, and the square root of this product will he the area. 1. What are the contents of a triangle, whose base is 24 feet,.nd whose perpendicular height is 18 feet? Ans. 216 feet. 2. What are the contents of a triangular piece of land, whose sides are 50 rods, 60 rods, and 70 rods? Ans. 1469.69+ rods. THE QUADRILATERAL. ART. 329. A QUADRILATERAL is a figure having four sides, and consequently four angles. It comprehends the rectangle, square, rhombus, rhomboid, trapezium, and trapezoid. ART. 330, A PARALLELOGRAM is any quadrilateral whose opposite sides are parallel. It takes the particular names of rectangle, square, rhombus, and rhomboid. The altitude of a parallelogram is a perpendicular line drawn between its opposite sides; as C D in the rhomboid. QUESTIONS. - What is an isosceles triangle? A scalene triangle? A rightangled triangle? - Art. 328. What is the first rule for finding the area of a triangle? What the second?- Art. 329. What is a quadrilateral? What figures does it, comprehend? —Art. 330. What is a parallelogram? What particular names does it take? What is the altitude of a parallelogram?

Page 308 308 MENSURATION OF SURFACES. [SECT. XLlli A rectangle is a right-angled parallelogram, whose opposite sides are equal. A square is a parallelogram, having four [ eq:id sides and four right angles. A rhomboid is an oblique-angled parallelogram, whose opposite sides are equal. A rhombus is an oblique-angled parallelogram, /7 having all its sides equal. ART. 331. To find the area of a parallelogram. RULE. - Multiply the base by the altitude, and the product will be the area. 1. What are the contents of a board 25 feet long and 3 feet wide? Ans. 75 square feet. 2. What is the difference between the contents of two floors; one is 37 feet long and 27 feet wide, and the other is 40 feet long and 20 feet wide? Ans. 199 square feet. 3. The base of a rhombus is 15 feet, and its perpendicular height is 12 feet; what are its contents? Ans. 180 square feet. ART. 332, A TRAPEZOID is a quadrilateral, which has only one pair of its opposite sides parallel. ART. 333. To find the area of a trapezoid. RULE. -Multiply half of the sum of the parallel sides by the altitude, and the product is the area. 1. What is the area of a trapezoid, the longer parallel side QUESTIONS. - What is a rectangle? A square? A rhomboid? A rhombus? - Art. 331. What is the rule for finding the area of a parallelogram? -Art. 332. What is a trapezoid?-Art. 333. What is the rule for finding the area of a trapezoid?

Page 309 bECT. XLIIL] MENSURATION OF SURFACES. 309 being 482 feet, the shorter 324 feet, and the altitude 216 feet? Ans. 87048 square feet. 2. ]What is the area of a plank, whose length is 22 feet, the width of the wider end being 28 inches, and of the narrower 2(, inches? Ans. 44 square feet. ART. 331, A TRAPEZIUM is a quadrilateral, E which has neither two of its opposite sides parallel. A diagonal is a line joining any two opposite r angles of a quadrilateral; as E F. ART. 335. To find the area of a trapezium. RULE. - Divide the trapezium into two triangles by a diagonal, and hen find the areas of these triangles; their sum will be the area of the trapezium. 1. What is the area of a trapezium, whose diagonal is 65 feet, and the length of the perpendiculars let fall upon it are 14 and 18 feet? Ans. 1040 square feet. 2. What is the area of a trapezium, whose diagonal is 125 rods, and the length of the perpendiculars let fall upon it are 70 and 85 rods? Ans. 9687.5 square rods. THE POLYGON. ART. 336. A POLYGON is any rectilineal figure having more than four sides and four angles. It takes the particular names of pentagon, which is a polygon of five sides; hexagon, one of six sides; heptagon, one of seven sides; octagon, one of eight sides; nonagon, one of nine sides; decagon, one of ten sides; undecagon, one of eleven sides; and dodecayon, one of twelve sides. ART. 337, A REGULAR POLYGON is a plane rectilineal figure, which has all its sides and all its angles equal. The perimeter of a polygon is the sum of all its sides. ART. 338, To find the area of a regular polygon. QUESTIONS. — Art. 334. What is a trapezium? What is a diagonal? - Art. 335. What is the rule for finding the area of a trapezium?- Art. 336. What is a polygon? What particular names does it take?- Art. 337. What is a regular polygon?

Page 310 310 MENSURATION OF SURFACES. [SECT. XLIIL RULE. - Multiply the perimeter by half the perpendicular let fait from the centre on one of its sides, and the product will be the area. 1. What is the area of a regular pentagon, whose sides are each 35 feet, and the perpendicular 24.08 feet? Ans. 2107 square feet. 2. What is- the area of a regular hexagon, whose sides are each 20 feet, and the perpendicular 17.32 feet? Ans. 1039.20 square feet. -THE CIRCLE. ART. 339. A CIRCLE is a plane figure bounded by a curved line, every part of which is equally distant from a point, called its centre. G H The circumference or periphery of a circle is the line which bounds it. The diameter of a circle is a line drawn through the centre, and terminated by the circumference; as G H. ART. 340. To find the circumference of a circle, the diameter being given. RULE. - fultiply the diameter by 3.141592, and the product is the circurrference. 1. What is the circumference of a circle, whose diameter is 50 feet? Ans. 157.0796-+- feet. 2. A gentleman has a circular garden whose diameter is 100 rods; what is the length of the fence necessary to enclose it? Ans. 314.15+ rods. ART. 341. To find the diameter of a circle, the circumference being given. RULE. - Multiply the circumference by.318309, and the product wilo be the diameter. 1. What is the diameter of a circle, whose circumference is 80 miles?. Ans. 25.46+ miles. 2. If the circumference of a wheel is 62.84 feet, what is the diameter? Ans. 20+ feet. QUESTIONS. - What is the perimeter of a polygon? - Art. 338. What is the rule for efnding the area of a regular polygon?-Art. 339. What is a circle? What is the circumference of a circle? The diameter of a circle:? - Art. 340. What is the rule for finding the circumference of a circle, the diameter being given?- Art. 341. What is the rule for finding the diameter of a circle, the circumference being given?

Page 311 SECT. XLIII.] MENSURATION OF SURFACES. 311 ART. 342. To find the area of a circle, the diameter, the circumference, or both, being given. RULE 1. - Multiply the square of the diameter by.785398, and the product is the area. Or, RULE 2.- Multiply the square of the circumference by.079577, and the product is the area. Or, RULE 3. - Multiply hayf the diameter by half the circumference, and the product is the area. 1. If the diameter of a circle be 200 feet, what is the area? Ans. 31415.92 square feet. 2. There is a certain farm, in the form of a circle, whose circumference is 400 rods; how many acres does it contain? Ans. 79A. 2R. 12+p. ART. 343, To find the side of a square equal in area to a given circle. The square in the figure is supposed to have the same area as the circle. RULE 1.- Multiply the diameter by.886227, and the product is the side of an equal square. Or, RULE 2.- Multiply the circumference by.282094, and the product is the side of an equal square. 1. We have a round field 40 rods in diameter; what is the side of a square field that will contain the same quantity? Ans. 35.44+ rods. 2. I have a circular field 100 rods in circumference; what must be the side of a square field that shall contain the same area? Ans. 28.2+ rods. ART. 344, To find the side of a square inscribed in a given circle.,A square is said to be inscribed in a circle when each of its angles touches the circumference or periphery of the circle. QUESTIONS. -Art. 342. What is the rule for finding the area of a circle, when the diameter is given? When the circumference is given? When the diameter and circumference are both given?- Art. 343. What is the first rule for finding the side of a square equal in area to a given circle? What the second? - Art. 344. When is a square said to be inscribed in a circle? What is the first rule for finding the side of a square inscribed in a circle? The second?

Page 312 312 MENSURATION OF SOLIDS. [SECT. XLIV RULE 1. - Multiply the diameter by.707106, and the product is the side of the square inscribed. Or, RULE 2. - Multiply the circumference by.225079, and the product is the side of the square inscribed. 1. What is the thickness of a square stick of timber that may be hewn from a log 30 inches in diameter? Ans. 21.21+ inches. 2. How large a square field may be inscribed in a circle whose circumference is 100 rods? Ans. 22.5+ rods square. THE ELLIPSE. ART. 345. An ELLIPSE is an oval figure having two diameters, or axes, the longer of which is called the transverse and the shorter the conjugate diameter. ART. 346. To find the area of an ellipse. RULE. - Multiply the two diameters together, and their product by.785398; the last product is the area. 1. What is the area of an ellipse whose transverse diameter is 14 inches, and its conjugate diameter 10 inches? Ans. 109.95+ square inches. 2. What is the area of an elliptical table, 8 feet long and 5 feet wide? Ans. 31 square feet, 59+ square inches. ~ XLIV. MENSURATION OF SOLIDS. ART. 317, A SOLID is a magnitude which has length, breadth. and thickness. Mensuration of solids includes two operations: first, to find their superficial contents, and, second, their solidities. THE PRISM. ART. 348. A PRISM is a solid whose ends are any plane figures which are equal and similar, and whose sides are parallelograms. It takes particular names, according to the figure QUESTIONS. -Art. 345. What is an ellipse? What is the longer diameter called? The shorter? - Art. 346. What is the rule for finding the area of an ellipse? - Art. 347. What is a solid? What two operations does mensuration of solids include? - Art. 248. What is a prism? What particular names does it take?

Page 313 SECT. XLIV.] MENSURATION OF SOLIDS. 313 of its base or ends, namely, triangular prism, square prism, pentagonal prism, &c. The base of a prism is either end; and of solids in general, the part upon which they are supposed to stand.. All prisms whose bases are parallelograms are comprehended under the general name parallelopipedons or parallelopipeds. A triangular prism is a solid whose base is a triangle. A square prism is a solid whose base is a square, and when all the sides are squares it is called a cube. A pentayonal prism is a solid whose base is a pentagon. ART. 349. To find the surface of a prism. RULE. - Multiply the perimeter of its base by its height, and to this product add the area of the two ends; the sum is the area of the prism. 1. What are the superficial contents of a triangular prism, the width of whose side is 3 feet, and its length 15 feet? Ans. 142.79+ square feet. 2. What is the surface of a square prism, whose side is 9 feet wide, and its length 25 feet? Ans. 1062 square feet. ART. 350. To find the solidity of a prism. RULE. - Multiply the area of the base by the height, and the product is the solidity. QUESTIONS. - What is the base of a prism and of solids in general? What is a parallelopiped or parallelopipedon? What is a triangular prism? A square prism? A pentagonal prism? - Art. 349. What is the rule for finding the surface of a prism?- Art. 350. What is the rule for finding the solidity of a prism? 27

Page 314 314 MENSURATION OF SOLIDS. [SECT. XLIV. 1. What are the contents of a triangular prism, whose length is 20 feet, and the three sides of its triangular end or base 5, 4 and 3 feet? Ans. 120 cubic feet. 2. How many cubic feet are there in a cube, whose sides are 8 feet? Ans. 512 cubic feet. 3. What is the number of cubic feet in a room 30 feet long, 20 feet wide, and 10 feet high? Ans. 6000 cubic feet. THE CYLINDER. ART. 351. A CYLINDER is a round solid, of uniform diameter, with circular ends. r The axis of a cylinder is a straight line drawn through- i it, from the centre of one end to the centre of the other. ART. 352. To find the surface of a cylinder. RULE. - Multiply the circumference of the base by the altitude, and to the product add the areas of the two ends; the sum will, be the whole surface. 1. WVhat is the surface of a cylinder, whose length is 4 feet, and the circumference 3 feet? Ans. 13.43+ square feet. 2. John Snow has a roller 12 feet long and 2 feet in diameter; what is its convex surface? Ans. 75.39+ square feet. ART. 353. To find the solidity of a cylinder. RULE.- Multiply the area of the base by the altitude, and the prod. uct will be the solidity. 1. What is the solidity of a cylinder 8 feet in length and 2 feet in diameter? Ans. 25.13+ cubic feet. 2. What is the solidity of a cylinder, whose diameter is 5 feet, and its altitude 20 feet? Ans. 392.69+ cubic feet. THE PYRAMID AND CONE. ART. 354. A PYRAMID is a solid, standing on a triangular, square, or polygonal base, with its sides tapering uniformly to a point at the top, called the vertex. The slant height of a pyramid is a line drawn from;1il \ the vertex to the middle of one of the sides of the base. QUESTIONS. -Art. 351. What is a cylinder? What is the axis of a cylin der? -Art. 352. What is the rule for finding the surface of a cylinder?Art. 353. What is the rule for finding the solidity of a cylinder? -Art. 354. What is a pyramid? What is the slant height of a pyramid?

Page 315 SECT. XLIV.] MENSURATION OF SOLIDS. 315 ART. 355. A CONE is a solid, having a circle for its base, and tapering uniformly to a point, called the vertex. The altitude of a pyramid and of a cone is a line drawn from the vertex perpendicular to the plane of the base; as B O. The slant height of a cone is a line drawn from the vertex to the circumference of the base; as A C. ART. 356. To find the surface of a pyramid and of a colle. RULE. - Multiply the perimeter or the circumference of the base by half its slant height, and the product is the convex surf/ce. 1. How many yards of cloth, that is 27 inches wide, will it require to cover the sides of a pyramid whose slant height is 100 feet, and whose perimeter at the base is 54 feet? Ans. 400 yards. 2. Required the convex surface of a cone, whose slant height is 50 feet, and the circumference at its base 12 feet. Ans. 300 square feet. ART. 357. To find the solidity of a pyramid and of a cone. RULE. - Multiply the area of the base by one third of its altitude, and the product will be its solidity. 1. The largest of the Egyptian pyramids is square at its base, and measures 693 feet on a side. Its height is 500 feet. Now, supposing it to come to a point at its vertex, what are its solid contents, and how many miles in length of wall would it make, 4 feet in height and 2 feet thick? Ans. 80,041,500 cubic feet; 1894.9 miles in length. 2. What are the solid contents of a cone, whose height is 30 feet, and the diameter of its base 5 feet? Ans. 196.3+ feet. ART. 358. A FRUSTUM OF A PYRAMID is the part that remains after cutting off the top, by a plane parallel to the base. QUJESTTONS.- Art. 35.5. What is a cone? What is the altitude of a pyramid and of a cone? What is the slant height of a cone?-Art. 356. What is the rule for finding the surface of a pyramid and of a cone? —Art. 357. What is the rule for finding the solidity of a pyramid and of a cone? - Art. 358. What is the frustum of a pyramid?

Page 316 316 MENSURATION OF SOLIDS. [SECT. XLIV. ART. 359. A FRUSTUM OF A CONE is the part that remains after cutting off the top, by a plane parallel to the base. Qis-7 ART. 360. To find the surface of a frustum of a pyramid or of a cone. RULE. -Add the perimeters or the circumferences of the two ends together, and multiply this sum by half the slant height. Then add the areas of the two ends to this product, and their sum will be the surface. 1. There is a square pyramid, whose top is broken off 20 feet slant height from the base. The length of each side at the base is 8 feet, and at the top 4 feet; what is its whole surface? Ans. 560 square feet. 2. There is a frustum of a cone, whose slant height is 12 feet, the circumference of the base 18 feet, and that of the upper end 9 feet; what is its whole surface? Ans. 194.22+ square feet. ART. 361. To find the solidity of a frustum of a pyramid or of a cone. RULE. - Find the area of the two ends of the frustum; multiply these two areas together, and rvtract the square root of the product. To this root add the tuwo areas, and multiply their sum by one third of the altitude of the frustum; the product will be the solidity. 1. What is the solidity of the frustum of a square pyramid, whose height is 30 feet, and whose side at the bottom is 20 feet, and at the top 10 feet? Ans. 7000 cubic feet. 2. What are the contents of a stick of timber 20 feet long, and the diameter at the larger end 12 inches, and at the smaller end 6 inches? Ans. 9.162+ feet. THE SPHERE. ART. 362. A SPHERE is a solid, bounded by one continued convex surface, every part of which is equally distant from a point within, called the centre. The axis or diameter of a sphere is a line passing through the centre, and terminated by the surface. QUESTIONS. - Art. 359. What is the frustum of a cone? - Art 360. What is the rule for finding the surface of a frustum of a pyramid or of a cone?Art. 361. What is the rule for finding the solidity of a frustum of a pyramid or of a cone? - Art. 362. What is a sphere? What is the diameter or axis of a sphere?

Page 317 SECT. XLIV.] MENSURATION OF SOLIDS. 317 AR;T. 363. To find the surface of a sphere. RULE. - AlUtiply the diameter by the circumference, and the product will be the surface. 1. What is the convex surface of a globe, whose diameter is 20 inches? Ans. 1256.6+ square inches. 2. If the diameter of the earth is 8000 miles,'what is its convex surface? Ans. 201061888 square miles. ART. 361, To find the solidity of a sphere. RULE. - Multiply the cube of the diameter by.523598, and the product is the solidity. 1. What is the solidity of a sphere, whose diameter is 20 inches? Ans. 4188.7+ inches. 2. If the diameter of a globe or sphere is 5 feet, how many cubic feet does it contain? Ans. 65.44+ cubic feet. ART. 365, To find how large a cube may be cut from any given sphere, or be inscribed in it. RULE. - Square the diameter of the sphere, divide the product by 3, and extract the square root of the quotientfor the answer. 1. How large a cube may be inscribed in a sphere 10 inches in diameter? - Ans. 5.773+ inches. 2. What is the side of a cube that may be cut from a sphere 30 inches in diameter? Ans. 17.32+ feet. THE SPHEROID. ART. 366. A SPHEROID is a solid, generated by the revolution of an ellipse about one of its diammeters. If the ellipse revolves about its longer.or transverse diameter, the spheroid is prolate, or oblong; if about its shorter or conjugate diameter, the spheroid is oblate, or flattened. ART. 367, To find the solidity of a spheroid. RULE 1. - Multiply the square of the shorter axis by the longer axis, and this product by.523598, if the spheroid is prolate, and the product will be its solidity. QUESTIONS. - Art. 363. What is the rule for finding the surface of a sphere? - Art. 364. What is the rule for finding the solidity of a sphere?Art. 365. What is the rule for finding how large a cube can be cut from a given sphere? - Art. 366. What is a spheroid? What is a prolate spheroid? What an oblate spheroid? - Art. 367. What is the rule for finding the solidity ef a spheroid? 27*

Page 318 318 MENSURATION OF LUMBER, ETC. [SECT. XLV. RULE 2. —If it is oblate, multiply the square of the longer axis by the shorter axis, and this product by.523598; the last product will be the solidity. 1. What is the solidity of a prolate spheroid, whose transverse axis is 30 feet, and the conjugate axis 20 feet? Ans. 6283.17+ cubic feet. 2. What is the solidity of an oblate spheroid, whose axes are 30 and 10 feet? Ans. 4712.38+ cubic feet. ~ XLV. MENSURATION OF LUMBER AND TIMBER. ART. 368. ALL rectangular and square lumber and timber, as planks, joists, beams, &c., are usually surveyed by board measure, the board being considered to be 1 inch in thickness. Round timber is sometimes measured by the ton, and sometimes by board measure. ART. 369. To find the contents of a board. RULE. —Multiply the length of the board, taken in feet, by its breadth, taken in inches, and divide this product by 12; the quotient is the contents in square feet. 1. What are the contents of a board 18 inches wide and 16 feet long? Ans. 24 feet. 2. What are the contents of a board 24 feet long and 30 inches wide? Ans. 60 feet. ART. 370, To find the contents of joists, beams, &c. RULE. - Multiply the depth, taken in inches, by the thickness, and this product by the length, in feet; divide the last product by 12, and the quotient is the contents in feet. 1. What are the contents of a joist 4 inches wide, 3 inches thick, and 12 feet long? Ans. 12 feet. 2. What are the contents of a square stick of timber 25 feet long and 10 inches thick? Ans. 2081 feet. ART. 371. To find the contents of round timber. RULE. - Multiply the length of the stick, taken in feet, by the square of one fourth the girt, taken in inches; divide this product by 144, ana the quotient is the contents in cubic feet. QUESTIONS.- Art. 368. By what measure are planks, joists, &c., usually surveyed? What is the usual thickness of a board? How is round timber measured? - Art. 369. What is the rule for finding the contents of a board? - Art. 370. What is the rule for finding the contents of joists, &e.?

Page 319 SECT. XLVI.] MISCELLANEOUS QUESTIONS. 319 NoTE 1. — The girt is usually taken about one third the distance from the larger to the smaller end. NOrE 2. - A ton of timber, estimated by this method, contains 50l'2o0 cubic feet. 1. How many cubic feet of timber in a stick, whose length is 50 feet, and whose girt is 60 inches? Ans. 78- cubic feet. 2. What are the contents of a stick, whose length is 30 feet, and girt 30 inches? Ans. 11.7+ solid feet. ~ XLVI. MISCELLANEOUS QUESTIONS. 1. WHAT number is that, to which if ~ be added, the sum will be 72? Ans. 73. 2. What number is that, from whiclh if 3 be taken, the remainder will be 41? Ans. 7 3. 3. What number is that, to which if 3 be added, and the sum divided by 53, the quotient will be 5? Ans. 23$. 4. From 7T of a mile take; of a furlong. Ans. 4fur. 12rd. 8ft. 8in. 5. John Swift can travel 7 miles in ~ of an hour, but Thomas Slow can travel only 5 miles in ]TT of an hour. Both started from Danvers at the same time for Boston, the distance being 12 miles. Howmuch sooner will Swift arrive in Boston than Slow? Ans. 12A6 seconds. 6. If I of a ton cost $49, what will 1cwt. cost? Ans. $3.92. 7. How many bricks 8 inches long, 4 inches wide, and 2 inches thick, will it take to build a wall 40 feet long, 20 feet high, and 2 feet thick? Ans. 43200 bricks. 8. Itow many bricks will it take to build the walls of a house, which is 80 feet long, 40 feet wide, and 25 feet high, the wall to be 12 inches thick; the brick being of the same dimensions as in the last question? Ans. 159300 bricks. 9. How many tiles, 8 inches square, will-cover a floor 18 feet long, and 12 feet wide? Ans. 486 tiles. 10. If it cost $18.25 to carry llcwt. 3qr. 19lb. 46 miles, how much must be paid for carrying 83cwt. 2qr. lllb. 96 miles? Ans. $266.702#oa-o. "4 57T7 11. A merchant sold a piece of cloth for $24, and thereby lost 25 per cent.; what would he have gained had he sold it for $34? Ans. 61 per cent.

Page 320 OV20 MISCELLANEOUS QUESTIONS. [SECT. XLVI 12. Bought a hogshead of molasses, containing 120 gallons for $30; but 20 gallons having leaked out, fbr what must I sel' the remainder per gallon to gain $10? Ans. $(0.40. 13. Bought a quantity of' goods fbr $128.25, and having kept them on hand 6 nmonths, for what must I sell them to gain 6 per cent.? Ans. $140.02. 14. If a sportsman spends ~ of his time in smoking, v in "gunning," 2 hours per day in loafing, and 6 hours in eating, drinking, and sleeping, how much remains for useful purposes? Ans. 2 hours. 15. If a lady spend I of her time in sleep, I in making calls, at her toilet, 1 in reading novels, and 2 hours each day in receiving visits, how large a portion of her time will remain for improving her mind, and for domestic employments? Ans. 327. hours per day. 16. If 5] ells English cost $15.16, what will 713 yards cost? Ans. $155.39. 17. If a staff 4 feet long cast a shadow 5 feet, what is the height of a steeple whose shadow is 150 feet? Ans. 107+ feet. 18. Borrowed of James Day $150 for six months; afterwards I lent him $100; how long shall he keep it to compensate him for the sum he lent me? Ans. 9 months. 19. A certain town is taxed $6045.50; the valuation of the town is $293275.00; there are 150 polls in the town, which are taxed $1.20 each. What is the tax on a dollar, and what does A pay, who has 4 polls, and whose property is valued at $3675? Ans. $0.02. A's tax $78.30. 20. D. Sanborn's garden -is 23$ rods long, and 134 rods vide, and is surrounded by a good fence 7- feet high. Now, if he shall make a walk around his garden within the fence, 752 feet wide, how much will remain for cultivation? Ans. 1A. 3R. 7p. 85 lIft. 21. J. Ladd's garden is 100 feet long and 80 feet wide.; he wishes to enclose it with a ditch, to be dug outside, 4 feet wide; how deep must it be dug, that the soil taken from it may raise the surface one foot? Ans. 54 feet. 22. How many yards of paper, that is 30 inches wide, will it require to cover the walls of a room that is 15 feet long, 11' feet wide, and 7' feet high? Ans. 55' yards. 23. Charles Carleton has agreed to plaster the above room, walls and ceiling, at 10 cents per square yard; what will be his bill? Ans. $ 6.54-.

Page 321 tCT. XLVI.] MISCELLANEOUS QUESTIONS. 321 24. What is the interest of $17.86, from Feb. 9, 1850, to Oct. 29, 1852, at 7- per cent.? Ans. $3.52-+-. 25. Required the superficial surface of the largest cube that can be inscribed in a sphere 30 inches in diameter. Ans. 1800 inches. 26. What is due, on the following note, at compound interest, Oct. 29, 1862? $1000. Salem, V. H., Oct. 29, 1856. For value received, I promise to pay Luther Emerson, Jr., or order, on demand, one thousand dollars with interest. EMERSON LUTHER Attest, ADAMS AYER. On this note are the following endorsements: Jan. 1, 1857, was received $125.00, June 5, 1857, do. $316.00, Sept. 25, 1857, do. $417.00, April 1, 1858, do. $100.00, July 7, 1858, do. $ 50.00. Ans. $53.79. 27. How many cubic inches are contained in a cube that may be inscribed in a sphere 40 inches in diameter? Ans. 12316.8+ inches. 25. The dimensions of a bushel measure are 184 inches wide and 8 inches deep; what should be the dimensions of a similar measure that would contain 4 quarts? Ans. 9- inches wide, 4 inches deep. 29. A gentleman willed 4 of his estate to his wife, and 4 of the remainder to his oldest son, and 4 of the residue, which was $151.33~, to his oldest daughter; how much of his estate is left to be divided among his other heirs? Ans. $756.6623. 30. A man bequeathed X of his estate to his son, and - of the remainder to his daughter, and the residue to his wife; the difference between his son and daughter's portion was $100; what did he give his wife? Ans. $600.00. 31. Sold a lot of shingles for $50, and by so doing I gained 12 per cent.; what was their value? Ans. $44.44-. 32. If 1T of a yard cost $5.00, what quantity will $17.50 purchase? Ans. 2a yard. 33. John Savory and Thomas Hardy traded in company; Savory put in for capital $1000; they gained $128.00; HIardy received for his share of the gains $70; what was his capital? Ans. $1206.89g.

Page 322 322 MISCELLANEOUS QUESTIONS. [SEcT. XLVI. 34. E. Fuller lent a certain sum of money to C. Lamson, and at the end of 3 years, 7 months, and 20 days, he received inter. est and principal $1000; what was -the sum lent? Ans. $820.7925i. 35. Lent $88 for 18 months, and received for interest and principal $97.57; what was the per cent.? Anls. 7, per cent. 36. When 3 of a gallon cost $87, what cost 7- gallons? Ans. $1051.25. 37. When $71 are paid for 183 yards of broadcloth, what cost 5 yards? Ans. $19.?264,6 38. How many yards of cloth, at $4.00 per yard, must be given for 18 tons 17cwt. 3qr. of sugar, at $9.50 per cwt.? Ans. 897k- yards. 39. How much grain, at $1.25 per bushel, must be given for 98 bushels of salt, at $0.45 per bushel? Ans. 351-' bushels. 40. A person, being asked the time of day, replied that 1 of the time passed from noon was equal to -I- of the time to midnight. Required the time. Ans. 40 minutes past 4. 41. On a certain night, in the year 1852, rain fell to the depth of 3 inches in the town of Eaverhill; the town contains about 20,000 square acres. Required the number of hogsheads of water fallen, supposing each hogshead to contain 100 gallons, and each gallon 282 cubic inches. Ans. 13346042hhd. 55gal. 1qt. Opt. 2oggi. 42. If the sun pass over one degree in 4 minutes, and the longitude of Boston is 71.4' west, what will be the time at Boston, when it is 1lh. 16m. A. M. at London? Ans. 6h. 31m. 44sec. A. M. 43. When it is 2h. 36m. A. M. at the Cape of Good Hope, in longitude 180 24' east, what is the time at Cape Horn, in longitude 670 21' west? Ans. 8h. 53m. P. M. 44. Yesterday my longitude, at noon, was 160 18' west; today I perceive by my watch, which has kept correct time, that the sun is on the meridian at llh. 36m.; what is my longitude? Ans. 10~ 1' west. 45. Sound, uninterrupted, will pass 1142 feet in I second; how long will it be in passing from -Boston to London, the distance being shabout 3000 nlilcs? Ans. 3h. 51mr. 10 —sec. 46. The time which elapsed between seeing the flash of a gun and hearing its report was 10 seconds; what was the dis;tance? Ans. 2 miles 860 feet. 47. J. Pearson has tea, which he barters with M. Swift, at

Page 323 SECT. XLVI.] MISCELLANEOUS QUESTIONS. 323 10 cents per lb. more than it costs him, against sugar, which costs Swift 15 cents per pound, but which he puts at 20 cents per pound; what was the first cost of the tea? Ans. $0.30 per lb. 48. Q and Y barter; Q makes of 10 cents 121- cents; Y makes of 15 cents 19 cents; which makes the most per cent., and how much? Ans. Y makes 13 per cent. more than Q. 49. A certain individual was born in 1786, September 25, at 23 minutes past 3 o'clock, A. M.; how many minutes old will he be July 4, 1844, at 30 minutes past 5 o'clock, P. M., reckoning 365 days for a year, excepting leap years, which have 366 days each? Ans. 30,386,287 minutes. 50. The longitude of a certain star is 3s. 14~ 26' 14", and the longitude of the moon at the same time is 8s. 190 43' 28"; how far will the moon have to move in her orbit to be in conjunction with the star? Ans. 6s. 240 42' 46". 51. From a small field, containing 3A. 1R. 23p. 200ft., there were sold 1A. 2R. 37p. 30yd. 8ft.; what quantity remained? Ans. 1A. 2R. 25p. 21yd. 5ft. 36in. 52. What part of 4 of an acre is - of an acre? Ans. 2 53. A thief was brought before a certain judge, and it was proved that he had stolen property to the value of 1~. 19s 113d. He was sentenced either to one year's imprisonment in the county jail, or to pay 1C. 19s. 113d. for the value of every pound he had stolen; required the amount of the fine. Ans. 3~. 19s. 11d. 0 Mqr. 54. My chaise having been injured by a very bad boy, I am obliged to sell it for $68.75, which is 40 per cent. less than its original value; what was the cost? Ans. $114.583. 55. Charles Webster's horse is valued at $120, but he will not sell him for less than $134.40; what per cent. does he intend to make? Ans. 12 per cent. 56. Three merchants, L. Emerson, E. Bailey, and S. Curtiss, engage in a cotton speculation. Emerson advanced $3600, Bailey $4200, and Curtiss $2200. They invested their whole capital in cotton, for which they received $15000 in bills on a bank in New Orleans. These bills were sold ta a Boston broker at 15 per cent. below par; what is each man's net gain? Ans. Emerson $990.00, Bailey $1155.00, Curtiss $605.00. 57. Bought a box made of plank, 31 inches thick. Its length on the outside is 4ft. 9in., its breadth 3ft. 7in., and its height

Page 324 324 MISCELLANEOUS QUESTIONO. [SECT. XLVL 2ft. 11in. How many square feet did it require to make the box, and how many cubic feet will it hold? Ans. 70254 square feet, 29{ cubic feet. 58. How many bricks will it require to construct the walls of a house, 64 feet long, and 32 feet wide, and 28 feet high? The walls are to be Ift. 4in. thick, and there are also three doors 7ft. 4in. high, and 3ft. 8in. wide; also 14 windows 3 feet wide and 6 feet high, and 16 windows 2ft. 8in. wide and 5ft. 8in. high. Each brick is to be 8 inches long, and 4 inches wide, and 2 inches thick. Ans. 167,480 bricks. 59. John Brown gave to his three sons, Benjamin, Samuel, and William, $1000, to be divided in the proportion of ~, -4, and 1, respectively; but William, having received a fortune by his wife, resigns his share to his brothers. It is required to divide the whole sum between Benjamin and Samuel. Ans. Benjamin $571.429; Samuel $428.57+. 60. Peter Webster rented a house for l year to Thomas Bailey, for' $100; at the end of four months Bailey rented one half of the house to John Bricket, and at the end of eight months it was agreed by Bricket and Bailey to rent one third of the house to John Dana. What share of the rent must each pay? Ans. Bailey $61, Bricket $277 and Dana $111. 61. I have a plank 424 feet in length, 24 inches wide, and 3 inches thick; required the side of a cubical box that can be made from it. Ans. 48 inches. 62. D. Small purchased a horse for 10 per cent. less than his value, and sold him for 16 per cent. more than his value, by which he gained $21.84; what did he pay for the horse? 63. Minot Thayer sold broadcloth at $4.40 per yard, and by so doing he lost 12 per cent.; whereas, he ought to have gained 10 per cent.; for what should the cloth have been sold per yard? 64. A gentleman has five daughters, Emily, Jane, Betsey, Abigail, and Nancy, whose fortunes are as follows. The first two and the last two have $19,000; the first four, $19,200; the last four, $20,000; the first and the last three, $20,500; the first three and the last, $21,300. What was the fortune of each? Ans. Emily has $5,000; Jane, $4,500; Betsey, $6,000; Abigail, $3,700 r and Nancy, $5,800. 65. I have a fenced garden, 12 rods square. How many trees may be set on it, whose distance from each other shall be one -rod, and no tree to be within half a rod of the fence? Ans. 152 trees. THE END.