Chamberlain's Arithmetick being a plain and easie explanation of the most useful and necessary art of arithmetick in whole numbers and fractions, that the meanest capacity may obtain the knowledge thereof in a very short time : whereunto are added many rules and tables of interest, rebate, purchases, gaging of cask, and extraction of the square and cube roots / composed by Robert Chamberlain, accomptant and practitioner in the mathematicks.

Chamberlain's Arithmetick:

CHAP. I.

The Definition of Arithmetick.

ARithmetick is the Art of Numbers which teacheth to put the true value upon any figure or figures, and doth consist of these following Parts, which are as Branches.

• 1 Numeration.
• 2 Addition.
• 3 Subtraction.
• 4 Multiplication.
• 5 Division.

These are the chief general Parts of ••rithmetick, and Number is a company ••f Figures which are composed of many

Units, as two is composed of two Units, Three is composed of three Units, Four, of four Units, &c.

Therefore one unite is no number, but the beginning of number, for one doth neither multiply nor divide, for if you multiply or divide an unite by it self, it admitteth of neither increase nor decrease, but number admitteth of both, Increase by Addition or Multiplication, and Decrease by Subtraction or Division, as 6 is a num∣ber composed of six units, and admitteth of Increase first by Addition, as 1 unit and 6 make 7, secondly by Multiplication, as 3 times 6 make 18, and 6 admitteth like∣wise of Decrease, first by Subtraction, as, take 1 from 6 and there remains 5: secondly by Division, as how many 2 can I have in 6, the answer is 3 times.

Numeration.

NUmeration is that part of Arithme∣tick, by which is taught rightly and justly to value, declare and write any Number or Sum propounded. All Num∣bers and Sums are expressed by these fol∣lowing Characters or Figures, which are

in number ten, whose value if you con∣ceive each figure to be or stand by it self, you may take notice of by their several Sub∣scriptions.

• 1 one,
• 2 two,
• 3 three,
• 4 four,
• 5 five,
• 6 six,
• 7 seven,
• 8 eight,
• 9 nine,
• 0 cypher.

Nine of these figures, viz. 1, 2, 3, 4, 5, 6, 7, 8, 9, are significant figures, signify∣ing each figure so many Unites, but the Cypher is of no value nor signification, as as it standeth alone, but being joined with other figures it increaseth the value of those figures which stand to the left hand of him.

Now these nine significant figures 1, 2, 3, 4, 5, 6, 7, 8, 9, have a double Value, whereof the one is certain, according to its own signification as before. But the other value is uncertain, it being according to the place that the said Figure doth stand in and possess.

A Place is a seat or room that a figure stands in, and how many figures there are in any sum, so many places of value doth the said sum contain, and that is the first place of any Sum, which is next the right hand, every place being ten times the value of the next right hand place.

Numeration Table.

C. Millions	X. Millions	Millions	C. Thousands	X. Thousands	Thousands	Hundreds	Tens	Units	How each Num∣ber is to be read.
								1	One
							1	2	Twelve
						1	2	3	1 hundred twen∣ty three
					1	2	3	4	1 thousand 234
				1	2	3	4	5	12 thousand 345
			1	2	3	4	5	6	123 thous. 456
		1	2	3	4	5	6	7	1 Million 234567
	1	2	3	4	5	6	7	8	12 Mill. 345678
1	2	3	4	5	6	7	8	9	123 Mil. 456789

This Table needeth but little explana∣••ion, for thereby you may discern the first place next the right hand to be the place of Units; the second, the place of Tens; the ••hird, the place of Hundreds; the fourth, Thousands, &c. So admit the sum 6873 were given you to number, consisting of four places, because in it are contained four figures, which you must thus number, six thousand eight hundred seventy three. And in like manner for 784395, which you must thus number, seven hundred eighty four thousand three hundred ninety five.

CHAP. II.

Addition.

ADdition teacheth of several Sums to make one which shall contain as much as all those several Sums, as suppose, A oweth unto—

B	328	Pounds.
C	147
D	485
Total Sum	960

Set the several Sums as you see one under the other. Having drawn a Line, begin at the lower end of that row next your right hand; saying, 5 and 7 is 12, and 8 is 20, setting under that first row 0, and carry 2 for so many tens to the next row; saying, 2 as I carried and 8 is 10, and 4 is 14, and 2 is 16, setting down 6 under that row, and carry 1 for the 10 to the next row; saying, 1 as I carried and 4 makes 5, and 1 is 6 and 3 is 9, setting down 9 under that row; so you have under the Line these figures 960, which is the total Sum of those three. So that A oweth in all to B, C, and D, 960 Pounds.

When you have any Sum to set down which you are to add up, you must be sure to set down the several sums so orderly one under the other as the units may stand under themselves, tens under themselves, hundreds under themselves, and thousands under themselves, &c. According to these following Examples.

3842 l.	863219
428	76824
97	4386
9	327
	24
	6

Addition of Mony.

If you have any Sums which are com∣posed of several Denominations, as Pounds, Shillings, Pence and Farthings, or any other kind; you must first consider how many of the lesser make one of the next Denomina∣tion; as, in Sterling Mony

• 4 Farthings is 1 Penny
• 12 Pence is 1 Shilling
• 20 Shilling is 1 Pound Sterling.

And so carry the Pence from the Far∣things to the Pence, the Shillings from the Pence to the Shillings, and the Pounds from the Shillings to the Pounds; as you may plainly see in the working of this fol∣lowing Example.

	l.	s.	d.
Received of A	384	16	11	½
B	197	14	08	¾
C	79	12	03	¼
D	9	13	10	¾
	671	17	10	¼

Suppose I had received mony of four several persons, viz. of A 384 l. 16 s. 11 d. ½, of B 197 l. 14 s. 08 d. ¾, of C. 79 l. 12 s. 03 d. ¾, and of D I have re∣ceived 9 l. 13 s. 10 d. ¾; and I desire to know how much the four Sums make al∣together. Set down the said Sums in such order one under the other, as the farthings may stand under themselves, the pence under themselves, the shillings under them∣selves, and the pounds under themselves; and in such manner as Units, Tens and Hundreds may stand under one another, according as you see them here set down; then draw a line, and begin at the bottom of the least Denomination; saying, ¾ and ¼ is four farthings, and ¾ is seven farthings, and ½ is nine farthings, or two pence far∣thing, setting down ¼ under the farthings, and carrying 2 to the pence; saying, 2 as I carried and 10 is 12, and 3 is 15, and 8 is 23, and 11 is 34 pence, which is 2 s. 10 d. therefore set down 10 under the pence, and carry 2 to the shillings; saying, 2 as I carried and 3 makes 5, and 2 is 7, and 4 is 11, and 6 is 17; setting down 7 under the first row of shillings, and car∣carrying 1 to the next row of shillings;

saying, 1 as I carried and 1 is 2, and 1 is 3 and 1 is 4, and 1 makes 5; which being odd, I set 1 after the 7, but under the last row of shillings: now the half of 5 being 2, I carry 2 unto the pounds; saying, 2 as I carried and 9 is 11, and 9 is 20, and 7 is 27, and 4 is 31, set down 1 under that row and carry three to the next row; say∣ing, 3 as I carried and 7 is ••0, and 9 is 19, and 8 is 27: setting 7 under this row, and carry 2 to the next; saying, 2 as I carried and 1 is 3, and 3 is 6; setting down 6 under the last row, so I find I have received in all of A, B, C, and D 671 l. 17 s. 10 d. ¼.

	l.	s.	d.
	384	16	11	½
	197	14	08	¾
	79	12	03	¼
	9	13	10	¾
Total	671	17	10	¼
Proof	287	00	10	¾
671	17	10	¼

For the Proof of this Sum, whether it be cast up right or not, the common way is to draw a line under the first sum, and then sum up again, only leaving out the upper∣most row, which sum being added to the up∣permost row or line, then if the sum be the same as before, the work is right cast up. But it is as easie or better to cast up the whole sum again, and if you think by doing so, you may commit the same error again, or fall into the same mistake as before, you may cast it up the contrary way, viz. begin at top and go downwards, for the Accompt will amount to one and the same sum, if it be right done.

But Tradesmen and Merchants being most concerned herein, have another help, which is in writing out Bills by their Books, if they find the sums in both Accompts do not agree, they conclude there must be a mistake in one or the other, which they find out by casting the Accompts over again.

In casting up long Bills, where many shillings, are in the pence, it is a good plain way to make a prick or some small mark at every shilling, and so setting down the odd pen••e at last, counting over how many

pricks you have made, and carry so many Shillings to the place of Shillings.

But such as are ready and expert, count how many Pence are in the whole Column, then by getting without Book such a Table as here followeth, can readily tell how many Shillings and odd Pence are in any Number of Pence.

d.		s.	d.		s.	d.
12	are	1	20	are	1	08
24	2	30	2	06
36	3	40	3	04
48	4	50	4	02
60	5	60	5	00
72	6	70	5	10
84	7	80	6	08
96	8	90	7	06
108	9	100	8	04
120	10	110	9	02
132	11	120	10	00
144	12

Addition of Flemish Mony.

Flemish Mony is reckoned by Pounds, Shillings and Pence, as we do our Sterling. But frequently they do reckon by Stivers and Guilders, or Stivers and Ducatoons. 20 Stivers being 1 Guilder, and 63 Stivers is 1 Ducatoon.

Stiv.	Guild.	Stiv.	Ducat.	Stiv.
	323	16	321	23
	129	09	78	34
	14	18	69	57
	468	03	469	51

A Table for the Reducing Flemish Ducatoons into Guilders and Stivers.

〈◊〉〈◊〉	••uil.	Stiv.	Duc.	Guild.	Stiv.
1	3	3	100	315	00
2	6	6	200	630	00
3	9	9	300	945	00
4	12	12	400	1260	00
5	15	15	500	1575	00
6	18	18	600	1890	00
7	22	1	700	2205	00
8	25	4	800	2520	00
9	28	7	900	2835	00
10	31	10	1000	3150	00
20	63	00	2000	6300	00
30	94	10	3000	9450	00
40	126	00	4000	12600	00
50	157	10	5000	15750	00
60	189	00	6000	18900	00
70	220	10	7000	22050	00
80	252	00	8000	25200	00
90	283	10	9000	28350	00
100	315	00	10000	31500	00

Addition of French Mony.

4 Quarters make 1 Solx, 60 Solx make one French Crown.

Therefore for 4 Quarters carry 1 Solx to the Solxes, for 60 Solxes carry 1 French Crown. As you may plainly see in the following Example.

Cr.	solx.	qrs.
248	42	3
476	37	2
124	29	3
297	56	1
1147	46	1

Addition of Haverdupoize Weight.

TABLE.

• 16 Ounces is 1 Pound.
• 28 Pound is 1 Quarter.
• 4 Quarters is 1 Hundred weight.
• 20 Hundred weight is 1 Tun.

Instructions.

Therefore from your ounces for every 16 carry 1 pound to your pounds, from your pounds for every 28 carry 1 quarter to your quarters, and from your quarters for every 4 carry 1 hundred to your hun∣dreds, and from your hundreds for every 20 carry 1 Tun to your Tuns; as you per∣ceive by the following Example.

Example.

Tuns.	C.	qrs.	li.	oz.
27	18	2	24	12
15	16	3	26	14
9	12	1	18	09
7	08	3	12	07
60	16	3	26	10

Addition of Apothecaries Weights.

TABLE.

• 20 Grains is 1 Scruple ℈.
• 3 Scruples is 1 Drachm ʒ.
• 8 Drams is 1 Ounce ℥.
• 12 Ounces is 1 Pound lb.

Instructions.

From the Grains for every 20 you must carry 1 Scruple to the Scruples, from the Scruples for every 3 carry 1 Drachm to the Drachms, from the Drachms for every 8 carry 1 Ounce to the Ounces, and from the Ounces for every 16 carry 1 Pound to the Pounds; according as you may see done in the following Example.

Example.

lb	℥	ʒ	℈	gr.
4	12	6	2	16
3	15	5	1	10
1	10	3	1	15
10	07	0	0	01

Addition of Troy Weight.

TABLE.

• 〈◊〉〈◊〉 Grains is 1 Penny Weight.
• 〈◊〉〈◊〉 Penny Weight is 1 Ounce.
• 〈◊〉〈◊〉 Ounces is 1 Pound.

Instructions.

For 24 Grains carry 1 Penny Weight to the Penny Weights, for 20 Penny Weights carry 1 Ounce to the Ounces, for 12 Oun∣ces carry 1 Pound to the Pounds.

Example.

li.	oz.	p. weig.	gr.
19	08	16	21
14	09	17	17
9	11	18	22
44	06	13	12

Addition of Wine Measure.

TABLE.

• 2 Pints is 1 Quart.
• 2 Quarts is 1 Pottle.
• 2 Pottles is 1 Gallon.
• 63 Gallons is 1 Hogshead.
• 4 Hogsheads is 1 Tun.

Instructions.

From Pints for every 2 carry 1 Quart to the place of Quarts, from Quarts for every 2 carry 1 Pottle to the place of Pottles, from the place of Pottles for every 2 carry 1 Gallon to the place of Gallons, and from the place of Gallons for every 63 carry 1 Hogshead to the place of Hogsheads, and from the place of Hogsheads for every 4 carry 1 Tun to the place of Tuns; as by the following Example.

Example.

Tuns.	hogsh.	gal.	pottl.	qrts.	pints.
237	1	38	1	1	1
142	3	47	0	1	1
95	2	54	1	0	1
46	1	29	1	1	0
8	2	12	0	0	1
530	3	56	1	1	0

Addition of Beer Measure.

TABLE.

• 2 Pints is 1 Quart.
• 2 Quarts is 1 Pottle.
• 2 Pottles is 1 Gallon.
• 9 Gallons is 1 Firkin.
• 2 Firkins is 1 Kilderkin.
• 2 Kilderkins is 1 Barrel.

Example.

Bar.	kild.	firk.	gal.	pottl.	qrts.	pts.
54	1	1	7	1	0	1
32	1	0	6	0	1	0
25	1	1	4	1	1	1
12	1	1	3	0	1	0
9	0	1	5	1	0	1
135	1	1	0	1	0	1

Addition of dry Measure.

TABLE.

• 16 Pints is 1 Peck.
• 4 Pecks is 1 Bushel.
• 8 Bushels is 1 Quarter.
• 5 Quarters is 1 Wey.
• 2 Wey 1 Last.

Example.

Quar.	bush.	peck.	pin.
128	5	2	11
342	6	3	12
487	7	1	10
83	4	3	15
1043	1	0	00

Addition of long Measure.

TABLE.

• 3 Barly Corns make 1 Inch.
• 12 Inches 1 Foot.
• 3 Foot 1 Yard.
• 5 Yards ½ 1 Pole or Perch.
• 40 Perches 1 Furlong.
• 8 Furlongs 1 English Mile.

By the Table you may easily discern what to carry at in each Denomination, as by the following Example doth appear.

Example.

Mil.	Furl.	Poles.	Yards.	Feet.	Inch.	B. Cor.
42	5	32	4	2	11	2
36	3	28	5	2	09	1
18	7	16	0	1	07	2
98	0	38	0	1	04	2

Addition of Cloth Measure.

TABLE.

• 4 Nails is 1 Quarter.
• 4 Quarters is 1 Yard.
• 5 Quarters is 1 Ell English.
• 3 Quarters is 1 Ell Flemish.

1 Example.	2 Example.	3 Example.
Yard.	qrs.	nail.	Ell En.	qr.	nail.	Ell Fl.	qr.	na.
41	3	2	36	4	3	50	2	1
39	1	3	28	3	2	37	1	3
26	2	3	24	2	3	32	2	2
24	3	1	21	2	••	31	0	••
20	2	3	18	4	2	27	1	••
18	3	2	16	3	1	24	2	••
172	1	2	147	1	0	204	••	••

Addition of Time.

TABLE.

• 60 Minutes is 1 hour.
• 24 Hours is 1 Day.
• 365 Days is 1 Year.

Example.

Years.	days.	hours.	min.
29	268	21	45
27	172	18	28
21	163	16	24
18	157	12	22
9	126	08	16
106	59	05	15

I suppose I have explained this part of Arithmetick, that any capacity may easily understand how to make use of the said Rule, either in Addition of Mony, Weight, or any sort of Measure, wet or dry, &c.

CHAP. III.

Subtraction.

SƲbtraction teacheth how to take a less Number out of a greater, and likewise doth shew what remaineth of the greater Number when the lesser Number is taken out.

There be three Numbers in Subtraction to be observed. First, the Number from whence the Subtraction is to be made; se∣condly, the Number that is to be subtract∣ed; thirdly, the number that remaineth after the Subtraction is made. As if I would subtract 24 from 36, 36 is the Number out of which the Subtraction is made, 24 is the Number to be subtracted, and 12 is the Number that remaineth after you have ended your Subtraction.

When you are to subtract, you must first set down the greater Number, or Number from whence the Subtraction is to be made, and under that set down the lesser Number, or Number to be subtracted, in such man∣ner,

as one Denomination may stand under another, as Farthings under Farthings, Pence under Pence, Shillings under Shil∣lings, and Pounds under Pounds; Units under Units, Tens under Tens, and Hun∣dreds under Hundreds, according to their places, and then draw a right Line under those two Numbers.

Subtraction of one Denomination.

Having placed your Numbers one under the other as was directed, you must begin as▪ the right hand, and subtract the first Figure of the undermost Number from the first Figure of the uppermost Number over 〈◊〉〈◊〉 that which remaineth set underneath the Line, right under the Figure you have subtracted, then take the second Figure of your undermost Number, and take it from the second Figure of your uppermost Num∣ber, the third from the third, and so forth until you come to the end, putting always 〈…〉〈…〉 of every Figure under the ••ine in i••s due order and place.

Example.

	l.
Owing unto B C	2784
Paid unto B C	1552
Rest due	1232

According to the former Instructions, I do begin at the right hand; saying, 2 from 4 and there remains 2, which I set down under the Line, and right under the first Figure subtracted; then I go to the next Figure, saying, 5 from 8 and there remains 3, which I set under the 5, being the second Figure subtracted; then I go to the next Figure, and say, 5 from 7 and there doth remain 2, which 2 I set down under the third Figure subtracted; then I go to the next and say, 1 from 2 and there doth remain 1, which I set under the last Figure subtracted. So there stands under the Line this Number 1232 l. which doth remain due unto B C.

But sometimes it doth fall out to have 2 Figures of one Quantity and Likeness to be subtracted one from another. As, 5

and 5; then I say, 5 from 5 and there doth remain 0. And when you have a Cypher to be subtracted from any Figure. As 0 and 7; then you must say, 0 from 7 and there remains 7, and so for any Figure.

It will be requisite to give you one Ex∣ample more in sums of one Denomination, wherein the Figures of the lesser Number or Number to be subtracted, are some of them greater than the Figures of the greater Number, or Number from whence you must subtract; In such case you must bor∣row 10, and add to the uppermost Figure, and then subtract your Figure to be sub∣tracted from that Sum, and set down your remainer; always remembring when you borrow 10 to add 1 to the next Figure to be subtracted for paiment, as in this follow∣ing Example.

Example.

	l.
Due from A B	42317
Received	32140
Remains due	10177

Here is due from A B 42317 Pound, and there is received 32140 Pound, I desire to know what is due from the said A B. Having set the two Sums down according to the former Instructions; say∣ing, 0 from 7 and there remains 7, 4 from 1 I cannot, but 4 from 11 and there remains 7; 1 as I borrowed and 1 is 2, 2 from 3 and there remains 1; 2 from 2 and there remains 0, and 3 from 4 and there remains 1. So there doth remain due from A B 10177 Pound. And in like manner for any other Sum, wherein you have one or more Figures of the lower Number greater than the Figures under which they do stand to be subtracted.

Subtraction of Mony.

In Sums composed of several Denomi∣nations, you must begin to subtract at the least Denomination next your right hand, and subtract each Denomination from his same, and set down the remainer of each Denomination under his own proper De∣nomination; but if in the Denomination of Farthings, your lower Number be greater than your upper, you must borrow 1 Peny

which is 4 Farthings, to add to the upper∣most, and when you subtract the Pence, then add 1 Penny to the lower Number of Pence for paiment of the Penny or 4 Far∣things you borrowed. Likewise if in the Denomination of Pence the lower Number be greater than the upper, borrow 1 Shil∣ling or 12 Pence to add to the uppermost, remembring to add 1 Shilling to the lower Number of Shillings for paiment of 12 pence you borrowed. And in like manner, if the lower Number of Shillings exceed the upper, as you may easily discern by the following Example of Pounds, Shillings, Pence and Farthings.

Example.

	l.	s.	d.
Received	7836	14	7	¼
Paid	4524	17	9	¾
Remains	3311	16	9	½
Proof	7836	14	7	1/4

Here is received 7836 l. 14 s. 7 d. ¼, and out of it is paid 4524 l. 17 s. 9 d. ¾;

it is required what is left of the said Sum. I set the two Sums down as you see accord∣ing to the former directions, and begin at the least Denomination which is Farthings, saying, ¾ from ¼ I cannot, but 1 Penny as I borrowed which is 4 Farthings and 1 Farthing is 5 Farthings, 3 Farthings from 5 Farthings and there remains 2 Farthings, which is ½, which ½ I set under the Line, and right under the Denomination of Far∣things; then I go to the next Denomina∣tion, which is Pence, and because I bor∣rowed one Penny in the Farthings, I now pay it again, saying, 1 Penny as I borrowed and 9 Pence is 10 Pence, from 7 Pence I cannot, but 1 shilling which is 12 Pence, and 7 Pence is 19 Pence, and 10 Pence from 19 Pence and there remains 9 Pence, setting 9 under the Line and under the Denomination of Pence; then I proceed to the Shillings, and because I borrowed one Shilling in the Pence, I pay it again; say∣ing, 1 Shilling as I borrowed and 17 Shil∣lings is 18 Shillings, from 14 Shillings I cannot, but 1 Pound as I borrowed which is 20 Shillings and 14 Shillings is 34 Shil∣lings, 18 Shillings from 34 Shillings and there remains 16 Shillings, so I set down

16 under the Denomination of Shillings; then I go to the Pounds, and by reason I borrowed 1 Pound in the Shillings, I pay it, saying, 1 as I borrowed and 4 is 5, 5 from 6 and there remains 1; setting 1 under the first Figure of the Denomination of Pounds; so proceeding I say, 2 from 3 and there remains 1, 5 from 8 and there remains 3, and 4 from 7 and there doth remain 3, setting down each remainer un∣der his respective Figure as you see. So there doth remain 3311 l. 16 s. 9 d. ½. And in like manner for any other Sum of the same Denomination.

Subtraction of Haverdupoize Weight.

The several Denominations of Haverdu∣poize Weight, you may acquaint your self with by the Table of Haverdupoize in the second Chapter.

Suppose I have bought 45 C. 3 quarters, 16 l. of Sugar, and I have sold 28 C. 1 quarter, 24 l. of the same Sugar; it is required what is left of this parcel. I set down the Numbers according to the fore∣going Directions, finding therein three Denominations, viz. Hundreds, Quarters,

Pounds; therefore I begin saying, * 1.1 twenty four from sixteen I cannot, but one quarter as I borrowed which is 28 Pound and 16 is 44, 24 from 44 and there remains 20. Then one as I borrowed and 1 is 2, from 3 and there re∣mains 1. Then I come to the Hundreds; saying, 8 from 5 I cannot, but 8 from 15 and there doth remain 7, 1 as I borrowed and 2 makes 3, from 4 and there remains 1. So I find there doth remain unsold 17 C. 1 quarter, 20 l.

Sometimes it happeneth that there are many Sums or Numbers to be subtracted from one Sum or Number, as suppose I have bought a Butt of Currans weighing 18 C. 1 quarter, 21 l. and at several times I have sold several parcels; as, 4 C. 3 qu. 17 l. at one time; 3 C. 3 quarters, 20 l. at another; and 2 C. 1 quarter, 16 l. at an∣other time; 1 C. 2 quarters, 0 l. at another time; and 1 C 3 qu. 24 l. at another time; and 3 qu. 12 l. at another time. The question is what is remaining unsold in the Butt of Currans?

Example.

	C.	qu.	l.
Butt of Currans containeth	18	1	21
Sold out of it at several times	4	3	17
3	3	20
2	1	16
1	2	00
1	3	24
	3	12
	25	2	25
	2	2	24

In order to answer this question, I first set down the neat weight of the But of Currans, viz. 18 C. 1 qu. 21 l. drawing a Line underneath. I set down under that Line the several parcels sold out of this Butt of Currans, one under the other, and bring all these parcels under the Line into one Sum by Addition, as you see in the Example, which Sum amounts unto 15 C. 2 quarters, 25 l. which I subtract from the weight of the Butt of Currans first set

down by the foregoing Instructions of this Chapter for Subtraction of Averdupoize weight.

Subtraction of Troy Weight.

Example.

	l.	oz.	p. w••.	gr.
Bought	20	08	16	21
Sold	12	10	14	22
Remaineth	7	10	01	23

Here I begin to subtract; saying, 22 from 21 I cannot, but 1 as I borrowed from the Penny weights which is 24 Grains and 21 Grains is 45 Grains, and 22 from 45 and there remains 23 then 1 Penny weight as I borrowed and 14 Peny weights is 15, from 6 and there remains 1; then 10 from 8 I cannot, but 1 Pound as I bor∣rowed which is 12 ounces, and 8 is 20, and 10 from 20 and there remaineth 10; then 1 Pound as I borrowed and 2 is 3 Pounds, 3 from 0 I cannot, but borrow 1 and say, 3 from 10 and there remaineth 7; 1 as I borrowed and 1 is 2 from 2 and

there remains 0. So there remaineth in all 7 Pound, 10 Ounces, 1 Penny weight, 23 Grains.

Subtraction of Apothecaries Weights.

Example.

	lb.	℥.	ʒ.	℈.	gr.
Bought	12	9	5	1	12
Sold	7	6	7	2	15
Remains	5	2	5	1	17

Here in this Example you say, 15 from 12 I cannot, but 1 as I borrowed from the next Denomination being Scruples con∣taining 20 Grains, and 12 Grains is 32 Grains, then 15 from 32 and there remains 17; then proceed and say, one Scruple as I borrowed and 2 makes 3 from one Scruple I cannot, but 1 as I borrowed from the Drachms containing 3 Scruples and one is 4, three from 4 and there remains 1. Then I proceed to the next Denomination; say∣ing, 1 as I borrowed and 7 is 8 from 5 I cannot, but one as I borrowed from the Ounces containing 8 Drachms and 5 is 13,

8 from 13 and there remains 5. Then proceeding to the next Denomination; saying, 1 as I borrowed and 6 is 7 from 9 and there remains 2. Then proceeding to the last Denomination; I say, 7 from 12 and there remains 5. So there will remain 5 lb. 2 ℥. 5 ʒ. 1 ℈. 17 gr. which doth re∣main unsold of that parcel.

Subtraction of Wine Measure.

For the Contents of the several Denomi∣nations of Wine Measure, you may see by the Table in the second Chapter, and also the way of Subtraction in this Measure, by this Example.

Example.

	Tuns.	hogsh.	gal.	pottl.	qrt.	pt.
Bought	148	3	45	1	1	1
Sold	84	2	57	0	1	0
Remains	64	0	51	1	0	1

Subtraction of Beer Measure.

The Table of Beer Measure in the second Chapter will shew you the Content of the several Denominations, and the way of subtracting by the following Example.

Example.

	Bar.	kild.	firk.	gal.	pot.	qrt.	pt.
Bought	36	1	0	7	0	1	0
Sold	27	0	1	8	1	1	1
Remains	9	0	0	7	0	1	1

Subtraction of Dry Measure.

In the second Chapter by the Table of Dry Measure, you may find the Contents of all the Denominations. And by the fol∣lowing Example you may also see the manner of Working.

Example.

	Qu.	bush.	pec.	pin.
Bought	43	3	2	15
Sold	27	5	3	09
Remaineth	15	5	3	06

Subtraction of Long Measure.

The Table of Long Measure in the se∣cond Chapter, will inform you of the se∣veral Denominations and their Contents, and the way of subtracting by the following Example.

Example.

Engl.	Mile.	furl.	pole.	yard.	feet.	inch.	B. Corn.
	156	5	33	4	3	11	2
	98	7	17	5	2	07	1
	57	6	15	4½	1	04	1

Subtraction of Cloath Measure.

Here it will be convenient to let you know as by the said Table, in the second Chapter of Addition; you may discern, that in Cloth Measure there be three sorts of Measures. As, Broad Cloth is measured by Yards, Quarters and Nails; Linnen Cloth in England, by Ells, Quarters and Nails; and Linnen Cloth in Holland, by Flemish Ells, Quarters and Nails. In each of which it will be convenient to give an Example; and to note that whereas by the Table of Cloth measure in Addition you find 4 Quarters is 1 Yard, 5 Quarters 1 English Ell, and 3 Quarters 1 Flemish Ell, it is to be 5 such Quarters as one Yard doth contain, viz. 1 Quarter more makes an Ell English, and 3 Quarters of our English Yard makes 1 Ell Flemish.

	Example 1.	Example 2.
	Yard.	qrs.	na.	Ells.	qr.	n.
Bought in England	215	3	1	163	4	2
Sold	127	2	2	86	4	3
Remaineth	88	0	3	76	4	3

Example 3.

	Flem.	Ells.	qrs.	nails.
Bought	479	1	2
Sold	247	2	3
Remaineth	231	1	3

Subtraction of Time.

You will find by the Table of Time that 60 Minutes make an Hour, 24 Hours one Natural Day, 365 Days one Years; by the following Example of Subtraction you more plainly understand how to work all such like Examples.

Example.

Years.	days.	hours.	min.
348	340	16	45
179	285	12	25
169	55	04	20

Proof of Subtraction.

When your Subtraction is ended, if you desire to prove your work, whether you have done right or wrong; add the Re∣mainer to the Number that was to be sub∣tracted, and if the sum of those two Num∣bers be equal to the greater Number, or numbers from whence the Subtraction was to be made, then your Subtraction was true, otherwise false. Thus let us prove the se∣cond Example of this Chapter, where after Subtraction is ended, the Numbers stand as you see, the Remainder or difference being 3311 l. 16 s. 09 d. ½, now to prove the work I add this remainer 3311: 16: * 1.2 9: 2, to the num∣ber that was to be subtracted which is 4524: 17: 9: 3: and the Sum of these two Numbers is 7836 l. 14 s. 7 d. ½, being equal to the Number from whence the Subtraction was to be made. Therefore I do conclude the subtra∣ction is true.

CHAP. IV.

Multiplication.

MƲltiplication is a Rule which teach∣eth by two Numbers given to pro∣duce a third Number, which shall contain one of the given Numbers so often as there be Units in the other; and it serveth instead of many Additions, being a Compound of many equal numbers into one Sum, by multiplying one Sum into another. For it increaseth one Number so often as there are Units in the other.

2. Multiplication doth consist of three parts or numbers. That is to say, the Mul∣tiplicand, the Mul∣tiplicator, and the * 1.3 Product.

The Multiplicand is the Number to be multiplied. The Mul∣tiplicator is the Num∣ber given by which the Multiplicand is to

be multiplied. And the Product is the Num∣ber which cometh of the Multiplication; as you see in the Margent: 384 is the Multi∣plicand or Number to be multiplied; 23 is the Multiplicator or Number by which the Multiplicand is to be multiplied; and 8732 is the Product or Number that ariseth and cometh by the said Multiplication, contain∣ing the Multiplicand so often as there are Units in the Multiplier, that is 23 times. For the Product contains the Multiplicand as many times as there be Units in the Mul∣tiplicator.

It maketh no difference which of the two given Numbers you make the Multi∣plicand or the Multiplicator, for 8 multi∣plied by 6, is as much as 6 multiplied by 8; but commonly the greatest Number is made the Multiplicand, standing upper∣most; and the lesser Number the Multi∣plicator, standing under the Multiplicand.

The Learner must get the following Ta∣ble compleatly by heart, before he can pro∣ceed any further in Arithmetick, because there is no Rule in Arithmetick but hath a great dependance upon Multiplication and Division.

Multiplication Table.

1	2	3	4	5	6	7	8	9	10	11	12
2	4	6	8	10	12	14	16	18	20	22	24
3	6	9	12	15	18	21	24	27	30	33	36
4	8	12	16	20	24	28	32	36	40	44	48
5	10	15	20	25	30	35	40	45	50	55	60
6	12	18	24	30	36	42	48	54	60	66	72
7	14	21	28	35	42	49	56	63	70	77	84
8	16	24	32	40	48	56	64	72	80	88	96
9	18	27	36	45	54	63	72	81	90	99	108
10	20	30	40	50	60	70	80	90	100	110	120
11	22	33	44	55	66	77	88	99	110	121	132
12	24	36	48	60	72	84	96	108	120	132	144

The use of this Table is not difficult, in the top of the Table you have these Num∣bers, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; and likewise down the first Column to∣wards your left hand you have the same Figures, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; and in their respective Columns you have all those Numbers multiplied by one ano∣ther. So that if you would know the Pro∣duct of any two given Numbers from 1 to

12 multiplied together, look for either of the two given Numbers in the top or up∣permost Column, and for the other in the side Column, guiding your eye in the same until you come under the other Figure or Number in the uppermost Column, and in the common Angle or Meeting, you shall have the Product required. Example, I desire to know the Product of 9 multi∣plied by 7. In the uppermost Column I find 9, and in the side Column 7; and carrying my eye along the row or Column against 7, until I come under 9 in the uppermost Co∣lumn, I meet with 63, which is the Product of 9 and 7 multiplied together. And if you look for 7 in the top, and go down the same Column until you come against 9 in the side you will find 63 as before; so it mat∣ters not which you look for in the upper∣most Column, or which in the side Column. This Table being compleatly gotten with∣out Book, the Learner may proceed to the practice of Multiplication.

In Multiplication if the Multiplicand consist of many places, and the Multipli∣cator of one place, set down the Multi∣plicand, and under it in the place of Units set down the Multiplier, and draw a Line

underneath them, and multiply one into another, beginning at the place of Units, and proceeding towards the left hand, set∣ting down each particular Product as you go, and if any particular Product amount to ten or any Number of tens, set down a o, and for every ten carry so many Units to the next Product; and if any parti∣cular Product exceed ten or any Num∣ber of tens, set down the excess, and for every ten carry so many Units to the next Product; and when you come to the last Figure, set down the total Product, as by the following Example will appear; where it is required to multiply 7958 by 7. First, set down 7958 for the Multiplicand, and under it in the place of Units set down 7 for the Multiplicator, * 1.4 then drawing a Line I begin find∣ing by the former Multiplication Table, that 7 times 8 is 56, I set down 6 being the excess of 5 tens, and for those 5 tens I carry 5 to the next Product. Now proceeding towards the left hand to the next figure 5; saying, 7 times 5 is 35, and 5 as I carried is 40, set down 0, and for the 4 tens carry 4 to the next Product; saying, 7 times 9 is 63,

and 4 as I carried is 67, setting down 7, and carry 6 to the next Product for so many tens. Then I proceed; saying, 7 times 7 is 49, and 6 as I carried is 55; which being the last Product I set down the whole Product 55, having finished the said Multiplication and finding the Pro∣duct to be 55706 as was required.

In Multiplication, if both the Multipli∣cand and the Multiplier do consist of divers places. Begin first with that Figure of your Multiplier as doth stand in the place of Units, multiplying it into all the Figures of the Multiplicand, placing the Product below the Line, according to the Instru∣ctions of the last Example. Then begin a new Product with the next Figure of the Multiplier, standing in the place of tens, multiplying it into all the places of the Multiplicand, and set the Product thereof under the first Product, but one Figure more towards the left hand. In like man∣ner multiply the third, fourth and fifth Figures, &c. until you have multiplied all the Figures of the Multiplier into all the Figures of the Multiplicand, remembring to set the place of Units of the second Product under the place of tens in the first

Product, and the place of units in the third Product, under the place of tens in the second Product, and the place of units of the fourth Product under the place of tens in the third Product; and in like manner for as many Products as there be Figures in the Multiplier, which will cause every Product to stand one Figure short of one another towards the right hand, according to the foregoing Instructions, the under∣most Product standing one Figure more towards the left hand than his undermost; as you see by the following Example.

It is required to multi∣ply 486739 by 4387. I set * 1.5 them down, the lesser un∣der the greater, according to the Instructions in the beginning of this Chapter, and drawing a Line under them, I begin with the first Figure of your Multi∣plier, standing in the place of Units which is 7; say∣ing, 7 times 9 is 63; I set down 3 and carry 6. Then 7 times 3 is 21, and 6 as I carried is 27; set down 7 and carry 2. Then 7 times 7 is 49, and 2

as I carried is 51, that is 1 and carry 5. Then 7 times 6 is 42, and 5 as I carried is 47; set down 7 and carry 4. Then 7 times 8 is 6, and 4 as I carried is 60; setting down 0 and carrying 6. Then 7 times 4 is 28, and 6 as I carried is 34; setting 4 and 3 after him as you see. So the first Product will be 3407173. Then I proceed to the next Figure, of the Multiplier which is 8, for a new Product. Saying, 8 times 9 is 72; now I must remem∣ber the former Instructions for placing the several Products, viz. to place the first Figure or place of Units of your second Product, under the second Figure or place of tens in the first Product. Therefore for 72 I set down 2 under the second Figure of the first Product being 7, and carry 7. Then 8 times 3 is 24, and 7 as I carried is 31; set down 1 and carry 3. Then 8 times 7 is 56, and 3 as I carried is 59; set down 9 and carry 5. Then 8 times 6 is 48, and 5 as I carried is 53, set down 3 and go 5. Then 8 times 8 is 64, and 5 as I carried is 69; set down 9 and carry 6. Then 8 times 4 is 32, and 6 as I carried is 38; I set down 8 and 3 behind being the last of that Product. Next I proceed to the third Figure of the Multi∣plier; saying, 3 times 9 is 27; setting

down 7 under the 1, which is the second Figure of the last Product, and I carry 2. then 3 times 3 is 9, and 2 as I carried is 11; set down 1 and carry 1. Then 3 times 7 is 21, and 1 as carried is 22; set down 2 and carry 2. Then 3 times 6 is 18, and 2 as I carried is 20; set down 0 and carry 2. Then 3 times 8 is 24, and 2 as I carried is 26, setdown 6 and carry 2. Then 3 times 4 is 12, and 2 as I carried is 14; I set down 14 being the last Figure multiplied in that Product. Then I next proceed to the fourth Figure of the Multiplier, which is 4. Saying, 4 times 9 is 36; setting down 6 under 1, according to the former Instru∣ctions, and I carry 3. Then 4 times 3 is 12, and 3 as I carried is 15; set down 5 and carry 1. Then 4 times 7 is 28, and 1 as I carried is 29: set down 9 and carry 2. Then 4 times 6 is 24, and 2 as I carried is 26: set down 6 and carry 2. Then 4 times 8 is 32, and 2 as I carried is 34, set down 4 and carry 3, then 4 times 4 is 16, and 3 as carried is 19: set down 19, being the last Figure of your Multiplicand.

You must observe the same order in pla∣cing the respective Products, and the Fi∣gures of each Product one under the ōther,

as you see in the Example, and draw a Line, and by the Instructions of the second Chapter, add those several Products up into one Sum under the Line, which will be 2135323993, the true Product required being equal to 4387 times 486739, or 486739 times 4387.

In the same manner are all Multiplica∣tions to be wrought, although there are more Figures either in the Multiplicand or Multiplier.

Another way of Multiplication.

The carrying of one Number is a little difficult to young Learners in this Art, or to such as do not frequently exercise them∣selves therein. Therefore for the help of such capacities, I shew you a plain and easie way for finding the several Products that are made by Multiplying the Multiplicand, by the several Figures of the Multiplier.

When you have two Numbers given to be multiplied one by another, multiply the greater Number by all the nine digits by continual Addition nine times over, setting the several Sums down Table-wise as you see in the following Example. The first Addition being as much as a Multiplication by 2, the second Addition being a Multi∣plication

by 3, the third Addition being a Multiplication by 4, the fourth a Multipli∣cation by 5, the fifth a Multiplication by 6, the sixth a Multiplication by 7, the seventh a Multiplication by 8, the eighth a Multi∣plication by 9. So have you the several Products of the Multiplicand, when it is multiplied by all the nine digits, cast up to your Hand by Addition, which you may choose out according to your Figures of the Multiplier, setting your several Pro∣ducts down one under another, according to the Instuctions of the last Example.

Example.

It is required to multiply 3968473 by 74698. First I set down the Multiplicand.

〈 math 〉〈 math 〉

The Multiplicand being doubled, which is multiplying by 2, the Product or Sum is 7936946, the Multiplicand added to that Sum, the Sum of them will be 11905419, which is the first Number multiplied by 3. The first Number added again giveth 15873892, being the Product of the first Number multiplied by 4. The next Addi∣tion giveth 19842365, or the first Number multiplied by 5. The next Addition giveth 23810838, or the Product of the first num∣ber multiplied by 6. The next Addition giveth 27779311, or the Product of the first Number multiplied by 7. The next Addition giveth 31747784, or the Pro∣duct of the first Number multiplied by 8. And the last Addition gives 35716257, or the Multiplication of the first Number by 9.

Here have you the first Number or Mul∣tiplicand multiplied by all the nine digits, which Products set one under the other as you see, your Multiplier being set under the Multiplicand, and a Line drawn, you take notice what figure stands in the first place of the Multiplier, which in this Ex∣ample is 8, and against 8 I find this Sum 31747784, which I set down under the

Line as you see. The next Figure of the Multiplier is 9, against which Figure stands this Sum 35716257, which I set under the other as you see. The next Figure of the Multiplier is 6, against it stands 23810838, which I set under the last Number as you see. The next Figure of the Multiplier is 4, against that stands 15873892, which must be set under the last Number as you see. And the last Figure of the Multiplier is 7, and against that stands 27779311, which must like wise be set under the last Number, observing the same order in set∣ting down the Sums as you see, according to the Instructions in the last Example but one of this Chapter, and drawing a Line, by adding up those Sums you will find the true product requir'd to be 296436996154 being equal to 74698 times 3968473, or unto 3968473 times 74698.

Abbreviations in Multiplication.

First, if you are to multiply any Number by 10, you need do no more but set a Cy∣pher 0 to the right hand of all the Figures. As, 5978 multiplied by 10, makes 59780. In the same manner if you were to multiply

any Number by 100, add the said Number two Cyphers thus 00, if by 1000 add three 000.

Secondly, if you are to multiply any Number by 20, 30, 40 or by 200, 300, 400, or any Number with more Cyphers at the end, set them as you see in the Example, as if you were * 1.6 only to multiply by Figures; and let the Cypher or Cyphers hang over, and add them to the Pro∣duct, as in this Example.

Thirdly, if you are to multiply any Number, that hath a Cypher or Cyphers at the end, by another Number which hath a Cypher or Cyphers in the end thereof, set them as you see in the Exam∣ple, in such manner as the first * 1.7 figure of the Multiplier may stand under the first Cypher of the Multiplicand, and cast up the Product as you see is here done.

Fourthly, if there be a Cypher in the middle of your Multiplier, you need do no more but put a Cypher under the se∣cond Figure of the foregoing Product, to fill up room, and set down the first▪

Figure of the next Product * 1.8 next to the Cypher in the same Line, as by the Example you may plainly perceive.

The Proof of Multiplication.

The surest way to prove Multiplication is by Division. But there is a way to prove Multiplication by the Cross, which is thus.

First, add up all the Fi∣gures of the Multiplicand, as * 1.9 in the Example of Multipli∣cation of four Figures.

Saying, 3 and 9 is 12, and 6 is 18, and 8 is 26, and 4 is 30, and 7 is 37, and 3 is 40. How many times 9 can I have in 40? 4 times 9 is 36, and there re∣mains 4; set down 4 by the Cross.

Secondly, do the same by the Multiplier. Saying, 7 and 4 is 11, and 6 is 17, and 9 is 26, and 8 is 34; 3 times 9 is 27 from 34 and there doth remain 7; which set down on the other side of the Cross.

Thirdly, multiply these two Numbers together. Saying, 4 times 7 is 28; 3 times 9 is 27 from 28 and there remains I; which I set over the Cross.

Fourthly, I do the like by the Product as I did by the Multiplicand and the Mul∣tiplier, and I find there will remain I; therefore I conclude the Multiplication is right.

CHAP. V.

Of Division.

DIvision is a Rule which teacheth to part any Number into any quantity of equal parts assigned, or to find how often any Number given doth contain a less Number. That is to say, by two Numbers given to find out a third Number, which third Number shall contain so many Units as the lesser of the two given Numbers is contained times in the greater.

Division hath four parts, viz. the Di∣vidend, the Divisor, the Quotient, the Remainer.

The Dividend is the Number given to be parted or divided.

The Divisor is the Number given, by which the Dividend is to be divided, and is always the lesser of the two given Num∣bers, or the Number that sheweth how many parts the Dividend is to be parted or divided into.

The Quotient is the Number that is produced of the parting or dividing the Dividend by the Divisor, and is the Num∣ber that tells you how many times the lesser Number is contained in the greater.

The Remainer is a Number always less than the Divisor.

Example.

There are two given Numbers 978462 for the Dividend, and 246 for the Divisor. Now if the first Figure of your Divisor towards your left hand be less than the first Figure of your Dividend, as in the follow∣ing Example it is set, the first Figure of the Divisor under the first Figure of your Dividend, as you see: but if it be greater, then set the first Figure of the Divisor under the second Figure of the Dividend, and the

rest in their order, keeping your Figures exactly one un∣der * 1.10 another.

Now beginning the said Example, I ask how many times 246 is contained in 978, but for the more easie working I need but say, how many times 2 is contained in 9, I find 4 times, therefore upon a piece of wast paper, prepared for that purpose, I multiply the Divisor by 4, the Product will be 984, which cannot be subtracted out of 978, therefore I find the Divisor will go but 3 times in 976; I set down 3 in the Quotient, making a separation as you see between the Dividend and the Quotient, then you begin to multiply your Divisor at the right hand end; saying, 3 times 6 is 18, cancel the 6. Then 8 from 8 and there remains 0, and 1 from 7, cancel the 7, and there remains 6. Then 3 times 4 is 12, cancel the 4, saying, 2 from 6 and there remains 4, and 1 from 9 and there remains 8, setting 8 over 9, and 4 over 6, cancelling the 9 and 6. Then 3 times 2 is 6, cancel the 2; and say, 6 from 8 and there remains 2, and the Division stands as you see.

Next to proceed for finding the next Figure of the Quotient, I remove my Di∣visor one Figure more towards the right hand, there standing over the Divisor these uncancelled Figures 2403. I ask how many times 2 the first Figure of the Divisor in 24 I find 9 times by multiplying the Divisor by on wast paper: there∣fore I set down 9 in the * 1.11 quotient after the 3. Then begin to multiply the Di∣visor; saying, 9 times 6 is 54, 4 from 3 I cannot, but borrowing 1 and say∣ing, 4 from 13 and there remains 9, cancelling the 6 and 3 and set 9 over it, and 1 as I borrowed and 5 is 6, from 0 I cannot, but borrowing 1, saying 6 from 10 and there remains 4, and 1 as I borrowed from 4 and there remains 3. Then 9 times 4 is 36, 6 from 4 I cannot, but 6 from 14 and there remains 8, cancel 4 at bottom and 4 above, setting 8 the remainer over head, than 1 as I borrowed and 3 is 4, from 3 I cannot, but 4 from 13 and there remains 9, and 1 as I borrowed from 2 and there remains 1. Then 9 times 2 is 18,

cancel the 2, and say, 8 from 9 there re∣mains 1, and 1 from 1 and there remains 0. So there is two Figures of the Quotient finished.

I remove the Divisor one Figure more towards the right hand, as by the Example doth appear, then over the first Figure of the Divisor towards the left hand which is 2 * 1.12 there stands 18. So I de∣mand how many times 2 in 18? answer is 8: so for experience I mul∣tiply the Divisor 246 by 8, the Product will be 1968 which doth ex∣ceed 1894 the Number standing over it: there∣fore I do conclude the Divisor must go but 7 times; I set down 7 for the third Figure of the Quotient, and proceed, saying, 7 times 6 is 42, 2 from 4 and there remains 2, and 4 from 9 and there remains 5, cancelling the 6, 4 and 9, setting the remainer over head. Then 7 times 4 is 28, cancel the 4, and 8 from 5 I cannot, but 8 from 15 and there remains 7, 1 as I borrowed and 2 is 3 from 8 and

there remains 5, cancelling your Figures as you subtract from. Then 7 times 2 is 14, 4 from 5 and there remains 1, and 1 from 1 and there remains 0, there being these Figures uncancelled 1722.

Now I must remove the Divisor one Figure more towards the right hand, as by the Example, and over the first Figure of your Divisor towards the left hand there will stand 17, I demand how many times 2 can I have in 17, I find but 7 times, so I set 7 in the Quotient and proceed in Multiplying the Divi∣sor thereby. Saying, * 1.13 7 times 6 is 42, cancel the 6; then 2 from 2 and there remains 0, and cancel the 2; and 4 from 2 the next Fi∣gure I cannot, but 4 from 12 and there re∣mains 8, and 1 as I borrowed from 7 and there remains 6, can∣celling 2 and 7 and set the remains over head. Then 7 times 4 is 28, 8 from 8 and there remains 0, and 2 from 6 and there remains 4, cancelling 4 in the Divisor,

and 8 and 6 in the Dividend. Then last of all 7 times 2 is 14, 4 from 4 and there remains 0, and 1 from 1 and there remains 0. The Division is now compleated, and I find by the Quotient that the Divisor 246 is contained 3977 times in the Dividend 978342, and there doth remain nothing.

A more easie and plainer way of Division.

There is a far plainer and better way of Division, than this cancelling and scratching of Figures, which I much admire that the Schoolmasters in these times do not pra∣ctise, the other way being too hard and difficult for a Learner, for I have known many that have learned so far as Division with a great delight and chearfulness, but with the hardness and difficulty of Division have been quite disheartned from proceed∣ing any further. But this way that I intend to shew the Learner is so plain, easie and delightful that the meanest and ordinariest capacity may attain the knowledge thereof by the help of the former Rules.

Example.

It is required to divide 8748 by 36. First, set down the Divisor 36, and some little distance in the same Line set down the Dividend, making a separation between them, as you see in the following Example. So I demand how many times 36 I can have in * 1.14 87, I find 2 times, I set 2 in the Quotient, and by it I multiply the Di∣visor, saying, 2 times 6 is 12, setting 2 under the 7, which is the second Figure of Dividend, then proceed∣ing to the next Figure of the Divisor; say∣ing, 2 times 3 is 6, and 1 as I carried is 7, setting down the 7 under 8, so there is 72 the Product of the Divisor multiplied by 2, under which I draw a Line and subtract one from the other by the third Chapter, and there will remain 15, which place under the Line, at which time the Operation will stand as you see.

Next you must make a prick under the next Figure of the Dividend being 4, and set 4 before the remainer 15, so you shall

have 154 for a new Dividend, as you see. Now I demand how ma∣ny times 3 in 15, I find * 1.15 4 times; so I set 4 for the second Figure in the Quotient, and I multiply the Divisor thereby, and the Product 144 I set under the new Dividend 154, and drawing a line, by the third Chapter I subtract one from the other and there will remain 10.

Lastly, I set a prick under the next Figure of the Dividend, which is the last, being 8, and drawing it down I set it before the Remainer 10, by which I have 108 for a new Dividend. Then I ask how many times * 1.16 3 I can have in 10? I find 3 times, therefore I set down 3 in the Quotient, and mul∣tiplying the Divisor thereby, I set the Pro∣duct which is 108 un∣der the last Dividend, and subtracting one from another there doth remain 0, and

the whole work of the Division is finished.

But if it doth fall, out that the first Figure towards your left hand of your Dividend be less than the first Figure of the Divisor towards the left hand, or if all the Figures of the Divisor be they more or less, do amount to more than so many Figures of the Dividend, set a prick one Figure nearer the right hand, as if in such case the Divisor doth consist of three Figures, then you must begin under the fourth Figure of your Di∣vidend from your left hand, and seek how often the first Figure on the left side of your Divisor is contained in the two first Figures to the left side of the Dividend, and place the answer in the Quotient; by which Fi∣gure multiply the Divisor, and place the Product thereof under the Dividend in order, and deduct one from another and proceed, always remembring in all the varieties of Division, that if you find by Multiplication the Product to be at any time greater than the Dividend, you must take a Figure lesser for your Quotient num∣ber by a Unite, or one.

Example.

It is required to divide 3865498 by 487. I set the Divisor and Dividend ac∣cording to the former Instructions, as you see in the following Example. And because the 3 first figures of your Dividend * 1.17 is less than the Di∣visor, I set a prick under the fourth figure from the left hand being 5, then ask how many times 4 in 38, answer is 8 times, now for experi∣ence sake, upon wast paper I multiply the Divisor 487 by 8, and the Product I find will be 3896, which is greater than my Dividend 3865, therefore it must go but 7 times, so I set 7 in the Quotient, and multiplying the Divisor by 7, the Product 3409 I set under the Dividend 3865, and by subtracting I find there will remain 456.

Now according to the former Instru∣ctions, I set a prick under the next figure which is 4, drawing it down and set it be∣fore the Remainer; so have I 4564 for a new

Dividend. Then do I ask how many times 4 will go in 45, answer is 9 times, and finding by Multiplication it will bear so much, I set down 9 in the Quotient, * 1.18 and multiply the Divisor thereby, the Product there∣of being 4383 I subtract from the Dividend, and there doth remain 181, the work standing as you may see.

I again proceed to the next figure of the Dividend, which is 9, and likewise accord∣ing to Rule I set a Prick under him, and drawing him down I set him before the last Remainer 181, so I have 1819 * 1.19 for a new Divi∣dend. Then I do ask how ma∣ny times 4 in 18? answer is 3 times, for the Product arising by 4, will be greater than the

Dividend 1819, therefore I set down 3 in the Quotient, and Multiplying the Divisor thereby, I set its Product 1461 under the last Dividend, and by subtracting I find there doth remain 358, as you see in the Example.

Lastly, I set a prick under the next and last figure of the Dividend which is 8, drawing it down I set the said figure before the last Remainer 358, which gives me 3588 for a new Dividend. I now de∣mand how ma∣ny times 4 the * 1.20 first Figure of the Divisor I can have in 35, by examinati∣on I find it will go but 7 times, therefore I set down 7 in the Quotient, and multiplying the Divisor by it, the Product of that Multipli∣cation I set under the last new Dividend 3588, and subtracting one from another I

find there will remain 179. Now the Divi∣sion is finished, and I find that in 3865498 there is contained 7937 times the Divisor 487, besides 179 which doth remain.

Here you must note that many times it happens that the Dividend cannot be ex∣actly divided by the Divisor; that is to say, after you have taken the Divisor as many times as you can out of the Dividend, yet there will remain some odd Number over and above.

But that Number which doth remain of your Division must always be less than the Divisor, or else your Division must in ne∣cessity be false; therefore you must endea∣vour to rectifie your work, and find out your errour, which in this plain and easie way of Division you may quickly find out, in your examination having but two things to have regard to, that is your Multiplica∣tion and Subtraction, both which in this way of Division lie so plain before your eye that they may be examined with ease and facility.

Another way of Division.

This way that I do intend now to ex∣plain doth not much differ from the latter way: but here you have all the Products of every Multiplication ready cast up to your hand by multiplying the Divisor by all the nine digits, as by the following Ex∣ample doth more plainly appear.

Example.

Let it be required to divide 27232072 by 3698. Multiply your Divisor which is 3698, by all the nine digits, which you may do by Addi∣tion, * 1.21 as you see in the Table, where against each respective digit you have the Product of the Divisor multiplied by that digit, as 2 times 3698 the Divisor is 7396, 3 times the said Divisor is 11094, 4 times the said Divisor is 14792, 5 times is 18490, &c.

Therefore by the former Rules having orderly placed your Divisor and Dividend

as you see, you do find the four * 1.22 first figures of your Dividend towards your left hand are less than the Divisor, and because the Di∣visor containeth four figures, you therefore set a prick under the figure from your left hand, which is 2, and demand how many times the Divisor 3698 will go in 27232, therefore I look over the Table prepared for that purpose, for the nearest Number to it that is less, which I find is 25886, which is the Product of the Divisor multi∣plied by 7, so I set 7 in the Quotient, and 25886 for a Product underneath 27232, and have drawn a Line, by Subtraction I find there doth remain 1346, but less than the Divisor 3698.

Next I proceed to the next figure of the Dividend, which is 0, setting a prick under it, and by drawing it down I place it be∣fore the Remainer 1346, and I shall have 13460 for a new Dividend, and I ask how many times the Divisor 3698, will go in 13460, I look in the Table of Pro∣ducts for the nearest less Number which

is 11094, be∣ing the Product * 1.23 of the Divisor multiplied by 3 wherefore I set 3 in the Quo∣tient, and the said Product 11094 under the new Dividend 13460, and by subtract∣ing one from another, I find there will remain 2366, but yet less than the Divisor 3698.

Next I proceed to the next Figure of the Dividend which is 7, * 1.24 and under it I set a prick, and by draw∣ing it down, I set it before the last Re∣mainer 2366, and there will be this num∣ber 23667 for a new Dividend, then to know how many times the Divisor 3698 is contained in that

Dividend 23667, I look in the former Table of Products, and there I do find 22188 being the nearest less Number, and the Product of the Multiplication of the Divisor 3698 by 6: so I set 6 in the Quo∣tient, and the said Product 22188 I set under the last Dividend 23667, and after Subtraction is made, there will remain 1479, but still you observe the Remainer is less than the Divisor 3698, and by that observation you may soon discern an error when it is committed.

Lastly, proceeding, I set a prick under the next and last Figure of the Divivend, and by drawing it down, set 2 before the last Re∣mainer * 1.25 1479, and you will have thenum∣ber 14792 for a new Divi∣dend, asking how many times the Di∣visor 3698, will go in the Dividend 14792 in the

Table of Products, and the nearest less num∣ber to it is 14792, which is the Product of the Divisor multiplied by 4, therefore I set down 4 in the Quotient, and the said Product 14792, under the last Dividend; and by Subtraction I find o remains, and now having finished the Division, I do hereby find that the Dividend 27232072 contains the Divisor 3698 just 7364 times, and nothing over nor under.

〈 math 〉〈 math 〉

If you divide 8743168495 by 67832, the Quotient will be 128894, and there doth remain 30687, as by the foregoing Example doth appear.

By this way of multiplying your Di∣visor by the nine digit Numbers, you may, First observe to be very useful, not only in this way of dividing downward, but in all other ways of Division, whereby you may certainly know by ocular inspection how many times your Divisor is contained in your Dividend, without running the danger of mistake, by reason whereof you may be constrained to begin your work again

Secondly, if you place the Numbers found in this manner as I have shewed you, di∣rectly under the Dividends, setting Figure under figure according to the several places, your work will be much more easie, for you then have nothing to do but to subtract the two Numbers one from another, without charging your memory.

Thirdly, this way you shall have a more plain and easie Proof of the truth of your work, than in any other way, for if you add together the several Sums to be sub∣tracted, and remain as they stand in the

order, which in the last Division you will find to be 30687. If your work be right the Sum will exactly agree with the first Sum of your Dividend, as by the Proof of the last Division here following doth appear.

〈 math 〉〈 math 〉

This being a far more easie and plain Proof of the work than by Multiplication, only you must observe this method of setting the Sums down, as you see here is done.

But because Multiplication is the com∣mon received way to prove Division, I will shew the Reader the way thereof.

When you are to prove any Division by Multiplication, you must set the Divisor under the Quotient, and multiply one by another according to the fourth Chapter,

and to the Product of that Multiplication add the Remainer if there be any, and the Sum thereof will be neither more nor less than the Number given to be divided. As for Example.

Let it be required to prove the last Divi∣sion but one of four Figures, whose Divisor is 3698, Dividend 27232072, and Quotient is 7364: there∣fore * 1.26 I set the Divisor 3698 under the Quotient 7364, and by the fourth Chapter I do multiply one by another, and being no Remainer of this Division the Product of the Multiplication is equal to the Dividend of the Division 27232072.

There be some will pretend to prove Division by casting out of nines, thus.

First, cast the nines out of the Divisor, as in the last Example of Division is 6, 7, 8, 3, and 2 which make 26, from which take the nines and there doth remain 8, which I set on one * 1.27 side of the Cross. Then I cast the nines out of the Quotient and there will remain 5, which set on the

other side of the Cross, and multiply the two Numbers on each side of the Cross together, and there remains of the Product 4, after the nines are taken out, which 4 I keep in mind, then join the Figures re∣maining of the Division together, viz. 3, 0, 6, 8 and 7, which make 24, and 4 kept in mind makes 28, nines from thence and there remains 1, which I set on the top of the Cross.

Lastly, I cast all the nines out of the Dividend, and there will remain 1 answer∣ing to 1 at the head of the Cross, and to be set at bottom, and when the two Figures at top and bottom of the Cross do agree, then the Division is well done, or else not.

Abbreviations in Division.

When a Number is given to be divided by any Number that hath one or more Cyphers on the right hand, for so many Cyphers cut as many Figures from the right hand of your Dividend, and also cut off those Cyphers of your Divisor, and let the remaining Numbers of the Dividend be divided by the remaining Numbers of the Divisor, and if any thing remain of your

Division, set the Figures cut off from your Dividend to the right hand of them, and this new Number is the true Remainer of the Division.

Example.

Let it be required to divide 86492 by 700, there being two Cyphers to the right hand of the Divisor, I cut off two Figures from the right hand of the Di∣vidend, * 1.28 viz. 9, 2, and likewise the two Cyphers from the Divi∣sor, and divide the remaining figures of the Di∣vidend 864 by the remaining Figures of the Divisor 7, and there doth remain of that Division 3, I therefore set 92 the Figures cut off from the Dividend to the right hand of 3, and there will be 392 for the just and exact Remainer, so that the Dividend 86492 doth contain the Divisor 700, 123 times, and 392 parts of 700 over and above, as by the Example doth appear.

And hence it doth by consequence fol∣low, that if you have any Number to divide by 10, 100, 1000, &c. you must cut off as many Figures from the right hand of the Dividend, as there are Cyphers in the Divisor, and those Figures remaining to the 〈◊〉〈◊〉 hand shall be the Quotient, and the Figures cut off shall be the Remainer, as thus; if 3784|2 be to be divided by 10, cut off the last Figure to the right hand, as you see 3784|2 the work being done, wherein 3784 is the Quotient, and 2 is the Remainer. Likewise if the same Number were to be divided by 100, then cut off two Figures from the right hand thus, 378|42, then 378 is the Quotient, and 42 is the Remainer. And if 1 were to divide the same Number by 1000, then cut off three Figures thus, 37|842, and 37 shall be the Quotient, and 842 the Remainer.

Having now explained the five Species of Arithmetick, viz. Numeration, Addition, Subtraction, Multiplication and Division; being the Foundation of all the following Rules, there being no Rule in Arithmetick but what doth depend upon them. I would in∣treat the Learner to be well acquainted with these Grounds before he proceed further.

CHAP. VI.

Of Reduction.

REduction is a Rule that teacheth to bring Numbers of several Denomi∣nations into one particular Denomination; whether the Denominations do signifie Mony English, Flemish or French, also Weights, Measures, &c.

Reduction is performed two ways, either by Multiplication or Division.

Reduction by Multiplication is a reducing of great Denominations into less, as Pounds into Shillings, Shillings into Pence, and Pence into Farthings. Likewise Hundreds into Quarters, Quarters into Pounds, and Pounds into Ounces, &c. which is done by Multiplication.

Reduction by Division is a reducing of lesser Denominations into greater, as Far∣things into Pence, Pence into Shillings, and Shillings into Pounds. Likewise, Oun∣ces into Pounds, Pounds into Quarters, and Quarters into Hundreds, &c. which is

done by dividing that Denomination by as much as make one of a greater, as in redu∣cing Farthings into Pence, I divide the Number propounded by 4, because 4 Far∣things make 1 Penny.

Reduction of Mony.

In this Reduction of Mony as well as in other Reductions, you must consider whe∣ther the Sum propounded be to be reduced into a greater or lesser Denomination, and next consider how many of the one Deno∣mination do make the other: by which Considerations you will know when to multiply and when to divide, and what to multiply or divide by. As for Example.

Quest. 1. In 374 Pounds it is required to know how many Shillings. Here I do consider that Pounds is the greater Deno∣mination, and Shillings is the lesser, and that in one Pound are contained 20 Shil∣lings, and that the Number of Shillings in 374 Pounds will be 20 times as much as 374, therefore I multiply 374 by 20, and the Product is 7480, and so many Shillings are contained in 374 Pound.

Quest. 2. In 526: 18, how many Shil∣lings and Pence?

In this Example you have two Denomi∣nations propounded to find out the Deno∣mination of Pence, here in multiplying your Pounds by 20, you must take in the 18 Shillings, setting 8 down first, because 0 cannot multiply, there∣fore multiply by the next * 1.29 Figure 2; saying, 2 times 6 is 12, and 1 is 13 set down 3 and carry 1, 2 times 2 is 4 and 1 is 5, set down 5, 2 times 5 is 10; setting 10 down. So the Product of this multiplicand and the odd Shillings make 10538 Shillings; now am I to reduce the Shillings into Pence, and know∣ing that 12 Pence make one Shilling, I therefore multiply the Number of Shillings by 12, and the Product of that Multipli∣cation 126456 will be so many Pence, and so many Pence doth 526 Pound 18 Shil∣lings contain.

In the work of Reduction many times you need not run through every particular

Denomination, but come to the thing re∣quired sooner. As for Example.

Quest. 3. In 398 l. 15 s. how many Farthings? First, multi∣ply 398 by 20, and ta∣king * 1.30 in the odd 15 Shil∣lings, the Product of Shillings will be 7975, then knowing that in one Shilling are 48 farthings, I do not reduce these Shillings into Pence as in the last Example, by mul∣tiplying them by 48. I do reduce them into Far∣things at once, and the Product of that Multiplication is 382800, and so many Farthings is there in 398 l. 15 s. as was required.

Rule 4. But if it be required to reduce Numbers of a lower Denomination into a greater, in such case you must divide the lower Denomination propounded, by so much as makes of that lower Denomina∣tion one Unite or Integer of the greater Denomination. As for Example.

In 1360 Shillings how many Pounds? and knowing that 20 Shillings of the lower

Denomination is contained in a Unite or Integer of the greater * 1.31 denomination of Pounds, I there∣fore divide the lower Denomina∣tion 1360 Shillings by 20, by cutting off the last Figure, and taking the half of the re∣maining figures, which half 68 is the Quo∣tient of that Division by 20, and the Num∣ber of Pounds contained in 1360 Shillings, as was required.

Rule 5. But if the question so happen in Reduction that there may be a denomination or denominations between the Denomina∣tion given, and the Denomination requi∣red, then you may reduce your Denomi∣nation given by Division into the next su∣perior or greater Denomination, and when it is so reduced, reduce it again into the next above that, and so until you have brought it into the Denomination required. As for Example.

Rule 6. Let it be demanded in 142080 farthings to know how many Pounds. First, I divide the given Number Farthings by 4, to bring them into Pence, because I Penny doth contain 4 Farthings, and the Quotient thereof will be 35520 Pence; then I di∣vide 35520 Pence by 12, because 12 Pence

make 1 Shilling, and the Quotient of that Division will be 2960 Shillings, and then I divide 2960 by 20, and the Quotient giveth 148 Pounds, which are equal in value to 142080 Farthings. According to the Example.

〈 math 〉〈 math 〉

Rule 7. When the Number given to be reduced doth consist of divers Denomina∣tions, as Pounds, Shillings, Pence and Farthings; Hundreds, Quarters, Pounds and Ounces, &c. Then you must reduce the highest or greater Denomination into

the next inferior or lower, taking in the odd Numbers standing in that Denomina∣tion which your greater Denomination is reduced to, then reduce that Sum into the next inferior Denomination, adding there∣to the odd Numbers standing in that De∣nomination, and so doing until you have brought the Number given into the Deno∣mination required. As for Example.

〈 math 〉〈 math 〉

Let it be required to Reduce 137: 16: 7: ½, into Farthings. First, I bring 137 Pounds into Shillings by multiplying by 20, and taking in the 16 Shillings, the

Product will be 2756 Shillings. Then I multiply 2756 Shillings by 12 to bring the Shillings into Pence, taking in the odd 7 Pence, and the Product will be 33079 Pence. Lastly, I multiply 33079 by 4, to bring Pence into Farthings, taking in 2 for the odd half Penny, and the Product is 132318 Farthings, being equal in value with 137: 16: 7: ½, or the Number of Farthings contained in the whole Sum, as was required.

Rule 8. If in Reduction any do remain of a Division, the Remainer is of the same Denomination as the Dividend is of.

There is a shorter way for reducing Pounds into Pence, and Pence into Pounds. As for Example.

Rule 9. Let it be required to know how many Pence is contained in 347 Pounds. Now because 20 Shillings is 1 Pound, and 12 Pence is 1 Shil∣ling, * 1.32 therefore 12 times 20 which is 240, is the Number of Pence contained in 20 Shillings or 1 Pound, and if you multiply your Pounds 347 by 240, the Product thereof 83280 are the Num∣ber of Pence contained in the

Sum of 347 Pounds, as was required.

Rule 10. It is required to know how many Groats are comprehended or con∣tained in 263 Pounds, the way of reducing Pounds into Groats is not difficult, for multiply the Pounds by 20 to bring them into Shillings, and because 1 Shilling doth contain 3 Groats, multiply those Shillings by 3, and that Product shall be the Num∣ber of Groats in so many Pounds.

But there is a quicker and shorter way than this, which is thus. Because 20 Shillings is 1 Pound, and 3 * 1.33 Groats is 1 Shilling, therefore 3 times 20 which is 60 is the Num∣ber of Groats in 20 or 1 Pound, and if you multiply the Number of Pounds 263 by 60, the Product will be 15870 the Number of Groats in 263 Pounds which is a shorter way than to reduce Pounds into Shillings, and Shillings into Pence by two Multiplications.

Rule 11. And if it were required to reduce Pence into Pounds, or Groats into Pounds. Divide by 240 for reducing Pence into Pounds, and by 60 for the reducing Groats into Pounds, and the Quotient shall be the Number of Pounds required.

〈 math 〉〈 math 〉

Rule 12. Let it be required to reduce 83280 Pence, and 15780 Groats into Pounds, for experience sake, being the Products of the two last Examples. Divide 83280 Pence into Pounds by 240, the Number of Pence in 20 Shillings or one Pound, and the Quotient will be 347, the Number of Pounds as before, also divide 15780 Groats by 60, the Number of Groats contained in 20 Shillings or one Pound, and the Quotient thereof will be 263 Pounds as before, by which you dis∣cern the infallibility of both Rules, for dividing or multiplying by 240, to reduce Pence into Pounds, and Pounds into Pence,

and by 60 for reducing Groats into Pounds and Pounds into Groats.

Rule 13. There are other Denominations of English Mony, besides Pounds, Shillings, Pence and Farthings. As the following Table doth shew.

Denominations.	Value.
	s.	d.
A Mark	13	04
An Angel	10	00
A Noble	6	08
A Crown	5	00
A Thirteen Pence half Penny	1	01 ½
A Nine Pence	00	09
A Four Pence half Penny	00	04
Pieces of	00	03
00	02

Rule 14. Let it be required in 783 Nobles to know how many Marks, and Pounds at 20 s. By this Table you find the value of a Mark to be thirteen Shillings and four Pence, and the value of a Noble to be half a Mark, that is 6 s. 8 d. and likewise 6 s. 8 d. is one third of 20 Shillings, there∣fore 20 Shillings or 1 Pound doth contain 3 Marks, therefore to resolve the question,

double the Number of Nobles given 783, that is to say, multiply 783 by 2, and the Pro∣duct * 1.34 1566 will be the number of Marks con∣tained, and because 3 Marks, viz 3 times 6 s. 8 d. is 20 Shillings, or 1 Pound, I therefore divide the Number of Marks 1566 by 3, and in the Quotient I do find 522, which tells me that in 1566 Marks there is 522 Pounds, therefore in 783 Nobles there is 1566 Marks, and 522 Pounds: which is the answer to the question re∣quired.

Rule 15. In 768 Pounds, how many Marks and Nobles? this question is but the reverse of the last, for if you multiply your Pounds * 1.35 given by 3, the Product 1566 will be Marks, and the half of that Number will be 783, the Number of Nobles, as was required.

Therefore in 522 Pounds there is con∣tained 1566 Marks, and 783 Nobles, as was required.

Rule 16. In 538 Pounds it is required to know how many Angels and Crowns? By the foregoing Table you find the value of an Angel to be 10 Shillings, and of a Crown 5 Shillings; therefore because twice 10 Shillings is 20 Shillings, and twice 5 Shillings is 10 Shillings; therefore I multiply the * 1.36 Number of Pounds given 538 by 2, the Product thereof 1076 is so many Angels, and that Product 1076 multiplied again by 2, the Product thereof 2152 are the Number of Crowns. Therefore in 538l. there are con∣tained 1076 Angels, and 2152 Crowns, as was required.

Rule 17. In 2152 Crowns how many Angels and Pounds?

This is the reverse of the last Rule, and whereas * 1.37 there you multiplied, here you must divide by, which is taking the half of it, as

for experience sake, the last Example of the last Rule, the half of 2152 Crowns is 1076 Angels, and the half of 1076 Angels is 538 Pounds, as before 2152 Crowns make 1076 Angels and 538 Pounds, as was re∣quired.

Rule 18. In 294 Pounds it is required to know how many Nine Pences and Four Pence half Pennies.

〈 math 〉〈 math 〉

For the resolving this question you must first reduce your Pounds into Shillings, then your Shillings into Pence by the se∣venth Rule, and the Product of the which in the Example you will find to be 70560, you must divide by 9 to bring them into Nine Pences, and the Quotient will give you 7840 which are Nine Pences, and the double thereof 15680, is the Number of Four Pence half Pennies, as was re∣quired.

Rule 19. In 15680 Four Pence half Pennies how many Nine Pences, Shillings, and Pounds?

This is the Reverse of the last Rule, and where in that you did multiply here you must divide, where you divided here you must multiply; there I have chosen the same Example for your better experience, which is thus resolved.

Because two 4 d. ½ make 9 Pence there∣fore take the half of your given Number of Marks 15680, being 7840 Nine Pences. Then by multiplying 7840 by 9, I reduce them into Pence and the Product there∣of is 70560 Pence, by the Instructions of the sixth Rule, and you will find the Product of shillings to be 5880 Shillings,

and the Pounds to be 294. So I find that 15680 Marks is 7840 Nobles, or in value 5880 Shillings, and 294 Pounds, as was required.

〈 math 〉〈 math 〉

Rule 20. In 6748 Thirteen Pence half Pennies how many Pence, Shillings and Pounds.

You must first compute how many half Pence there are in 13 d. ½, and you will find 27; therefore multiply 6748 by 27, and the Product 182196 will be all half pences, and the half thereof 91098 is so

much, which divided by 12 the Quotient will give you 7591 Shillings and 6 pence remaining, 7591 Shillings divided by 20 gives 379 Pounds, and 11 Shillings re∣maining. So that I find 6748 thirteen pence half pennies are in value 379l. 11s. 6d. as was required.

〈 math 〉〈 math 〉

Rule 21. In 436 l. 12 s. 9 d. how many pieces of 4 d. ½ per piece.

〈 math 〉〈 math 〉

By the seventh Rule reduce the given Number 436: 12: 09 into pence, and you will find the Product of pence to be as by the Example 104797 pence, which because there is an odd half penny in 4 d. ½ you must double to bring into half pence, and the Product then will be 209586 half pence, now by reason that in 4 d. ½ there are 9 half pence, you must divide these Number of half pence 209586 by 9, as you see in the Example, and the Quotient will be 23287 pieces of 4 d. ½, and of the Di∣vision there doth remain 3 which doth sig∣nifie so many half pence over and above 23287 pieces of 4 d. ½.

Rule 22. In 372 pounds how many pieces of 3 d. and 2 d. of each equal, that is what Number of 3 d. pieces, and what Number of 2 d. pieces will make up 372 pound, and the Numbers of each equal.

The way to resolve this question, and all other questions of this nature, is to add the pieces together, (into which the given Number is to be brought,) and to reduce the given Number into the same Denomi∣nation with their Sum, and to divide the given Number so reduced by their Sum, and the Quotient will give you the exact

Number of each piece, the same method being used in the Example foregoing, by which we find that 17856 three pences, and 17856 two pences are equal in value to 372 pounds.

〈 math 〉〈 math 〉

Rule 23. In 3748 pieces of eight Dollers or Duccats at 4s. 7d. ½, how many pounds, shillings and pence?

〈 math 〉〈 math 〉

First bring one single Doller or 4s. 7d. ½, into the lowest Denomination named, which is half pence, and it doth contain 111 half pence, by which I multiply my given Number of Dollers 3748, and the Product thereof is 416028 half pence, and by taking the half I reduce it into Pence, which is 208014, which by the sixth Rule you may reduce into Shillings and Pounds. So I find 3748 Dollers, Duccats, or pieces of Eight at 4 s. 7 d. ½ are worth 866l. 14s. 06 d. English Mony, as was required.

Reduction of Averdupoize Weight.

Averdupoize Weight is used for all such Goods and Commodities that are to be weighed, except Gold, Silver and Bread, which are weighed by Troy Weight.

For the Contents of the several Deno∣minations of Averdupoize Weight, the Learner may find in the Averdupoize Table in the second Chapter.

Rule 24. In 246 C. 3 qu. 16 l. It is re∣quired to know how many Ounces?

In the Averdupoize Table in the second Chapter I find 4 Quarter is one Hundred, therefore having set down the given Num∣ber

as in the Margent, 246 C. 3 qu. 16 l. and * 1.38 multiply the hundreds by 4, taking in the Quarters, and the Pro∣duct 987 is so many Quarters. Likewise in the said Table I find that 28 Pound makes one Quarter, therefore I multiply that Pro∣duct 987 by 28, and the Product thereof must be 27652, not forgetting in the mul∣tiplication by 28, to take in the odd pounds I likewise find in the Table that 16 Ounces makes one Pound: therefore I multiply the last Product of Pounds 27652 by 16 to bring them into Ounces, and I find the Product to be 442432, where I do conclude that in 246: 3: 16 there is 442432 Ounces contained, as was required.

Rule 25. In 442432 Ounces, how many Hundreds, Quarters and Pounds?

〈 math 〉〈 math 〉

This is no more but the reverse of the last Rule, for where in that question you did multiply, here you must divide, and by the same Figures, as the foregoing Example doth illustrate, in this manner.

First, you see the given Number of Ounces divided by 16, the quotient gives

27652 for Pounds, next I divide 27652 Pounds by 28, the quotient gives me 987 quarters, there remaining of that Division 16 Pounds, next I divide 987 quarters by 4, and the quotient thereof is 246 hundred weight, and of that Division there remains 3 which are quarters, by which work I find that 442432 Ounces is as much as 27652 pound, 987 quarters, and 16 pound; or which is all one but more Arithmetically expressed, 246 C. 3 qu. 16 l. as was requi∣red, perfectly agreeing with the given number of the last Rule, by which you clearly understand the infallibility of the method or manner of working.

Rule 26. A person is to send to his Chapmen in the Country 93 parcels of Tobacco of 12 Pound in a parcel, as many parcels of 8 pounds in a parcel, as many of 6 pound parcels, and as many of 4 pound parcels. The question is how many hun∣dred weight, quarters and pounds must be weighed off of the said Tobacco to comply with all these parcels?

First set down the several parcels one under another, as 12 pound, 8 pound, 6 pound, and 4 pound, the Sum whereof is 30 pounds; now because there is to be

93 of each parcel, I therefore multiply 93 by their Sum which is 30, the Product thereof gives me 2790 for the whole quan∣tity of pounds. Now by the directions of the last Rule I divide that Number of pounds 2790 by 28, the quotient gives me 99 quarters, and 18 pounds remaining, next I divide these quarters 99 by 4, and the quotient gives me 24 hundred, and there remains 3 quarters. So by this work I find the whole quantity doth contain, as by the foregoing Example doth appear 24 C. 3 qu. 18 l. as was required.

〈 math 〉〈 math 〉

Rule 27. A Chapman buyeth 24: 3: 18 of Tobacco, and orders it to be made up into parcels of 12 Pound, of 8 pound,

of 6 pound, and of 4 pound, and there must not be more of one parcel than ano∣ther, but all equal. I desire to know how many must there be of each parcel.

〈 math 〉〈 math 〉

This is the reverse of the last, therefore I have taken the answer of that question, and make it here the question upon which I desire the answer, which answer will come out the same as the question of the

last, if the work be truly done, else not.

The way to resolve this question is thus: Reduce your given Number into pounds by the twenty fourth Rule, and the Product will be 2790 pounds. Next add all your parcels together, as 12 pounds, 8 pounds, 6 pounds, and 4 pounds, and they make 30 pounds: now divide your Number of pounds before found 2790 by 30, the Sum of all the parcels added together, and the quotient gives me 93, by which I find there must be 93 of each parcel, as was required, and agreeing with the question of the last Rule.

Reduction of Troy Weight.

Troy Weight differs much from Averdu∣poize Weight, as by the said Table in the second Chapter, where you will find but 12 ounces Troy Weight to be in a pound; therefore I would have the Learner well experienced in the Table of Troy Weight.

Rule 28. In 243 l. 8 oz. 16 p. weight, how many grains.

By the Table of Troy Weight in the second Chapter you will find 24 Grains makes 1 Penny weight, 20 Penny weights

is 1 Ounce, and 12 Ounces one Pound. * 1.39 Therefore to reduce the given Number 243 l. 8 oz. 16 p. weig. into grains I do thus. First, multiply the pounds 243 by 12, and take in the odd 8 ounces, and the Pro∣duct gives you 2924 for ounces. Then multiply the ounces 2924 by 20, and take in the odd 16 penny weights, and the Pro∣duct will give you 58496 for penny weights. Next mul∣tiply the penny weights 58496 by 24, the Product will give you 1403904 which are grains, by which I find that 243 l. 8 oz. 16 p. weig. do contain 1403904 grains, as was required.

Rule 29. In 1403904 grains how many pounds, ounces and penny weights?

〈 math 〉〈 math 〉

This is no more but the inverse of the last Example, and is resolved by dividing here by the same numbers that there you multiplied by, which is taken backwards by 24, 20, and by 12, and the answer will be 243 l. 8 oz. 16 p. weig. as was re∣quired.

Reduction of Apothecaries Weights.

It will be requisite for the Learner to make himself acquainted with the several Denominations, and the notes or marks thereof, used by Apothecaries, Chirurgions and Doctors; which by the said Table in the second Chapter are these following.

• 20 Grains is 1 Scruple ℈.
• 3 Scruples is 1 Drachm ʒ.
• 8 Drachms 1 Ounce ℥.
• 12 Ounces 1 Pound lb.

The Learner getting these notes without Book readily to know them will be capable of reading any Doctors or Apothecaries bill.

Rule 30. In 18 lb. 9 ℥. 2 ʒ. 1 ℈. it is required to know how many Grains?

Which to interpret is thus, read in 18 pound, 9 ounces, 2 drachms and 1 scruple how many grains? Therefore you must multiply 18 the Pounds by 12, and take in the 9 ounces, the Product will give 225 for ounces. Then multiply those ounces 225 by 8 the Product will give 1802 for

Drachms, then multiply these Drachms by 3, and the Product gives 5407 for Scru∣ples. Next multiply those Scruples by 20, the Product gives you 108140 for Grains, as was required.

〈 math 〉〈 math 〉

Reduction of Wine Measure.

Rule 31. In 36 Tun of Wine how many Gallons?

Multiply by 4 and the Product is 144, then multiply 144 by 63, and the Product gives you 9072 Gallons, being so many contained in 36 Tun of Wine, as was re∣quired.

〈 math 〉〈 math 〉

But there is a shorter way to reduce Tuns into Gallons, which * 1.40 is this. There is in a Tun 4 times 63 Gallons, viz. 252 Gallons in a Tun, therefore multiply 36 by 252, or 252 by 36, the Product will be 9072 Gallons as before.

Rule 32. In 68 Tun how many Runlets? at 14 Gallons per Runlet.

Reduce your Tuns into Gallons * 1.41 bymultiplying the Tuns by 252, or 252 by the number of Tuns given 68, and the Product will give you how many Gallons are contained in 68 Tun, which if you divide by 14 the Number of Gallons in a Run∣let, the Quotient 1224 is the an∣swer, and so ma∣ny Runlets of 14 Gallons will there be in 68 Tun, as was required.

Reduction of Beer Measure.

Rule 32. In 54 Barrels of Beer how many Gallons and Quarts?

If you examine the Table of Beer Mea∣sure in the second Chapter you will find a

Barrel of Beer to contain * 1.42 36 Gallons, and one Gal∣lon 4 Quarts; therefore multiply your Barrels gi∣ven 54 by 36, the Product thereof gives you 1944 Gallons, which multipli∣ed by 4 gives you 7776 quarts, which are contain ed in 54 Barrels of Beer, as was required.

Rule 33. In 48 Kilderkins how many Gallons and Pints.

In the Table you will find that 9 Gallons is a Firkin, and 2 Firkins which is 18 Gallons is 1 * 1.43 Kilderkin; therefore mul∣tiply Kilderkins 48 by 18 and the Product will give you 864 Gallons. Now by the same Table I find 2 Pints is 1 Quart, 2 Quarts or 4 Pints 1 Pottle, and 2 Pottles or 8 Pints is 1 Gallon; therefore I multiply the Gallons 864 by 8 and there doth remain in the Product 6912 Pints, which are contained

in 48 Kilderkins of Beer, as was re∣quired.

Reduction of dry Measure.

Rule 34. In 72 Last of Corn how many Bushels and Pecks.

Because in the Table of dry Measure in the second Chapter you find 5 Quarters make a Wey, and 2 Wey * 1.44 1 Last; therefore I double the given number of Lasts 72, and make it 144, which I multiply by 5, and the Product is 720. There is a shorter way of redu∣cing Lasts into Weys thus: 5 quarter being 1 Wey, 2 Wey or 10 Quarters is 1 Last; so that multiplying 72 Last by 10, by adding thereto 0 makes 720 Quarters, as before. Now by the said Table I find that 8 Bushels is 1 Quar∣ter; I therefore multiply 720 Quarters by 8, and the Product gives me 5760 Bushels, and multiplying these 5760 Bushels by 4, the Product gives me 23040 Pecks, and so many Bushels and Pecks are contained in 72 Last of Corn, as was required.

Reduction of Long Measure.

Rule 35. I demand how many Yards, Feet, Inches and Barly Corns will reach from London to S. Davids, which is reck∣oned 202 Miles.

Here I multiply 202 Miles by 8 to bring them into Furlongs, and the product gives me 1616 Furlongs; then I multiply these Furlongs by 40 to bring them into poles, and they produce 64640 poles or perches. Then I multiply them poles by 11, and the product gives me 711040 half Yards.

The half whereof 355520 are Yards, which Yards I reduce into Feet by multi∣plying them by 3, and the product gives 1066560 Feet, then those Feet I reduce into Inches by multiplying them by 12, and the product gives 12798720 Inches. Lastly, I multiply those inches by 3, and the product thereof gives 38396160 Barly Corns.

〈 math 〉〈 math 〉

Rule 36. The greatest Circumference of the Globe of the Earth and Sea as all other greater Circles are, are divided into 360 Degrees, each Degree into 60 Minutes or Miles.

〈 math 〉〈 math 〉

Greatest Circumference in Inches	1368576000
Earths Diameter in Inches	435456000
Earths Semidiameter in Inches.	217728000

Reduction of Cloth Measure.

Rule 37. In 348 Yards how many quar∣ters and nalls?

〈 math 〉〈 math 〉

Rule 38. In 3462 Ells English, how many Ells Fle∣mish?

〈 math 〉〈 math 〉

Rule 39. In 5770 Ells Flemish, how many Ells En∣glish?

〈 math 〉〈 math 〉

This is the Reverse of the last Rule, but there is a shorter way than this to reduce Flemish Ells into English Ells, by multi∣plying the number of Flemish Ells given by 6, and from the Product cut off the last Figure towards the right hand, and the Remainer shall give you the number of English Ells.

Example.

Let it be required to reduce 678 Flemish into English, I multiply 678 by 6, and the Product is 4068, then I cut off the last Figure towards the * 1.45 right, which is 8, and there re∣mains 406, which is the true num∣ber of English Ells, and the half of 8 being 4, gives you four fifths of an English Ell, which is the same as eight tenths. Therefore 678 Ells Flemish containeth 406 Ells Eng∣lish, and ⅘ of an Ell, as was required.

CHAP. VII.

The Single Rule of Three Direct.

THe Rule of Three is sometimes called Golden Rule by reason of the great excellency of it, it being the chiefest Rule in Arithmetick, because the other Rules do make use of the Rule of Three.

This Rule doth teach to find out a fourth Proportional to three given numbers, in such manner that the fourth may be in proportion to the third given Number, as the second is to the first; for which reason it is also called the Rule of Pro∣portion.

Now Proportion is thus to be under∣stood, that is when the first given Number containeth or is contained by the second, as often as the third containeth or is con∣tained by the fourth, as you may see by these following Numbers.

As	is to	So is	to
3	6	4	8
2	8	3	12
4	20	7	35

These Numbers are all proportional, viz. as 3 is to 6, so is 4 to 8; Likewise as 2 is to 8, so is 3 to 12; and as 4 is to 20, so is 7 to 35: That is to say, as many times as 3 is contained in 6, so many times is 4 contained in 8: again, as often as 8 doth contain 2, so often doth 12 contain 3, &c.

The great difficulty is to know which of these three given Numbers are to be placed in the first place, which in the second, and which in the third. Now of these three given Numbers, there will be two of the said given Numbers of one Denomination, and the other is of the same Denomination as the fourth Number to be found will be of. Now when you have found out the two Numbers of one Denomination, put one in the first place, and the other in the third place, and in the second place set the Num∣ber which is the same Denomination with the fourth Number sought.

But perhaps it may be objected, that which of the two given Numbers of one

denomination must be put in the first place, and which in the third place. Now you must understand that upon one of these two Numbers there is a demand, and that Number upon which the demand is placed, must be always esteemed the third place; as in the following Example. If 3 Yards are worth 9 s. what shall 6 Yards be worth? here you see a demand made upon 6 Yards, for it is demanded what shall 6 Yards be worth? therefore 6 Yards must be in the third place, and 3 Yards in the first place, and then by consequence the number 9 must be in the second place, and they will stand in order thus.

Yards.	Shillings.	Yards.
3	9	6

Now you have prepared your Numbers as to order in standing, for the finding out the fourth Number, observe this Rule.

Rule. Multiply your second and third Number together, and divide by your first, the Quo∣tient is the fourth Proportional or Number sought.

〈 math 〉〈 math 〉

Having placed my given numbers accord∣ing to the former directions, as you see, I multiply the second number 9 by my third number 6, the Product will be 54 which by the fifth Chapter I do divide by the first number 3, and the Quotient giveth 18 for so many Shillings, because the second num∣ber was Shillings. So 18 Shillings is the price of 6 Yards, being the fourth Propor∣tional number, for the fourth number 18 doth contain the third number 6, as many times as the second number 9 doth contain the first number 3.

Therefore if 3 Yards cost 9 Shillings, 6 Yards will cost 18 Shillings.

Quest. 2. If 8 C. of Sugar cost 32 Pound, what will 27 C. of the same Sugar cost?

To observe the former Rules in placing the given numbers, I consider that the de∣mand is made upon 27 C. of Sugar, which must be the third number; next I consider 8C. to be of the same Denomination, there∣fore I put that in the first place, and by consequence I put 32 Pounds in the second place, being of the same Denomination as that number which is sought, as you see in the following Example.

Next I proceed and * 1.46 multiply according to the instructions before given, viz. 32 by 27, and the Product there∣of will be 864, which I likewise divide by the first number 8, and the Quotient will give 108 for the true value 27 of C. of Sugar, as was re∣quired, at the rate of 32 Pound for 8 C.

Now if any thing do remain after you have divided the Product of the Multipli∣cation of the second and third number by the first number, the remainder of your Division must be multiplied by the number of Units of the next lesser Denomination as are equal to a Unite or Integer of the second number in the question, the Product of this Multiplication divide by the first number in your question, and the quotient of this Di∣vision shall be the same denomination as the parts by which you last multiplied, and part of the fourth number sought, and if any thing remain of this Division, multiply such re∣mainer by the parts of the next inferior de∣nomination equal to a unite of the last quo∣tient, and divide the Product by the same Divisor, and this Quotient shall be of the same denomination as your last multiplier, and so proceed either till nothing remain of your division, or till you have reduced your remainer into your lowest denomination.

Quest. 3. If 17 C. of Tobacco cost 43 Pounds, what will 86 Pound cost?

By the following Example you will find according to the former Instructions the numbers, and also multiplied and divided, the Quotient of the Division gives 217 l.

because the second number is Pounds, but there remains 9, by which I understand the price or value of 86 C. of Tobacco to be something more than 217 l. and to know how much more I proceed according to the foregoing Instructions.

As by this Example, viz. I multiply the Remainer 9 by 20, because 20 Shillings is equal to an Integer of the second number, the Product of this Multiplication is 180, which I divide by the first number 17, and the Quotient is 10 Shilling, because the last Multiplier was of the Denomination of Shillings. But here yet doth remain 10, which I multiply by 12 Pence because the last quotient was Shillings, and the Pro∣duct of that Multiplication is 120, which I divide by 17 the first number, and the quotient is 7 Pence, because the last Multi∣plier was Pence. But here yet doth remain 1, which if I multiply by 4, cannot be di∣vided by 17, therefore I set it next after the Pence, and over the Divisor with a Line between thus, 1/17. By which I do find that 86 C. of Tobacco will cost 217 l. 10s. 7d. 1/17, if 17 C. cost 43 l. which Fraction is one seventeenth of a Penny, which is an exact answer to the question propounded.

〈 math 〉〈 math 〉

Another Example of this kind will not be amiss, wherein the second number will be of another kind, as Weight.

Quest. 4. If 32 Pound buy 53 Hundred Weight of Sugar, how much will 124 Pound buy.

As you may discern by the ensuing Ex∣ample. I have placed the numbers accord∣ing to the former Instructions, and likewise multiplied the second and third numbers together, and divided by the first number, of which Division there doth remain 12, and the Quotient is 205 C. weights, because the second number is of that denomination, viz. Hundred weights, now because there doth remain 12, I understand thereby I can buy more than 205 Hundred weight of Sugar for 124 Pound; which to find out I thus proceed according to the former In∣structions, viz. I multiply the remainer by 4, because the second number is Hundred weights, and 4 Quarters makes one Hun∣dred weight, and the Product is 48, which I divide by 32 the first number, and the Quotient will be 1 Quarter, because the Multiplier was so; but of this Division there doth remain 16. Next I multiply that remainer by 28 Pound, being the last

〈 math 〉〈 math 〉

Quotient was Quarters, and 28 Pounds make 1 Quarter, and the Product of this last Multiplication will be 448, which I divide by 32, and the Quotient giveth 14 Pounds, because the last Multiplier was Pounds, but of this last Division there doth remain nothing. Finding by the Operation that for 124 Pound I can buy 205 C. 1 qu. 14 l. if 53 Hundred weight cost 32 Pound, being the answer to the Question pro∣pounded.

It many times falleth out in questions of the Rule of Three, that notwithstanding the first and third numbers are of one Kind or Species, viz. Mony, Weight, Measure, &c. yet they may either or both consist of several denominations, in such case you are to reduce both numbers into the lowest deno∣mination, and likewise into one denomi∣nation; also if your second number be of several denominations, you must reduce it into the lowest denomination named, or lower if you please; which being done, then multiply the second and third number together, and divide by the first, according to the former directions, the answer to the question being of the same denomination as your second number is reduced unto.

Quest. 5. If 12 C. 3 qu. of Sugar cost 32l. 14 s. 6 d. what will 82 C. 1 qu. 22l. cost?

Having placed these several numbers in their several respective places, according to the former instructions, as you see in the following Example, 12 C. 3 qu. being in the first place, 32 l. 14 s. 6 d. in the second place, and 82 C. 1 qu. 22 l. in the third place.

Now I perceive my first and third num∣bers are of several denominations, my first number having two denominations, as Hundreds and Quarters, viz. 12 C. 3 qu. My third number hath three denomina∣tions, as Hundreds, Quarters and Pounds, viz. 82 C. 1 qu. 22 l. But yet the first and third numbers are of one kind, that is to say, their several denominations are of Weight. Now the first and third number must be both of them brought into one denomination, and that denomination must be the lowest named either in the first or third number, which here in this Example is Pounds weight. Now by the sixth Chapter for the Reduction of Averdupoize weight, you may reduce the first and third numbers into Pounds weight, the first number pro∣ducing 1428l. weight, and the third number produceth 9234 l. weight.

Next I consider the nature and quality of the second number which is Mony; but it likewise doth consist of several denomi∣nations, as Pounds, Shillings and Pence, viz. 32 l. 14 s. 6 d. which by the sixth Chapter of Reduction of Mony, must be reduced into Pence, being the lowest de∣nomination named, and it will produce 7854 Pence.

Having thus reduced the three given numbers, as you see in the following Ex∣ample, I proceed according to the former Instructions of this Chapter, multiplying the second number reduced 7854 by the third number reduced 9234, the Product of which Multiplication is 72523836, which I do divide by the first number re∣duced 1428 by the instructions of the fifth Chapter, and the Quotient maketh 50787, which are Pence, because the second num∣ber 7854 by Reduction was made Pence, which Quotient 50787 is the fourth Pro∣portional required, and the number of Pence answering to the Question, which by Reduction makes 211 l. 12 s. 3 d. And so much will 82 C. 1 qu 22 l. cost, if 12 C. 3 qu. cost 32 l. 14 s. 6 d. as was required.

〈 math 〉〈 math 〉

To prove the Rule of Three is to state the Question quite backward, by putting the fourth number found out, first, the third number put second, and the second number put third; so have you three new given numbers to find a fourth by the former In∣structions, which fourth number when it is found will be the same as the first number in the question first stated.

The Proof of the second Question of this whose fourth number so found is 108.

〈 math 〉〈 math 〉

Here you see the second and third num∣ber multiplied together, whose Product

864 I do divide by the first number 108, and the quotient 8 is the fourth number found out, which is the first number in the Example of the second question.

But there is a shorter way of proving any question in the Rule of Three thus.

Multiply the first number by the fourth number found; likewise multiply the se∣cond number by the third; if these two Products do agree, your Operation is without any errour, else not, as you may see by the Proof of the Example of the second question.

〈 math 〉〈 math 〉

By this you see that 8 the first number, and 108 the fourth number multiplied to∣gether make 864, and the second number 32 multiplied by the third number 27 pro∣duceth 864 as before.

The fifth Question proved both ways.

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

Here you multiply your first number 1428, and the fourth number found 50787 together, whose Product is 72523836; likewise multiply the second number 7854, and the third number 9234 together, and the Product is 72523836 as before.

CHAP. VIII.

The Single Rule of Three Backwards.

THIS Rule is so called, because it produ∣ceth and requireth a proportion quite backward, and contrary to the Rule of Three Direct, because in the Rule of Three Direct the greater the third number is, the greater will the fourth number be. But in this Rule the greater the third number is, the lesser will the fourth number be: You are to place your three given numbers in order according to the Directions of the last Rule. But in this Rule you must mul∣tiply the first and second numbers together, and the product of that Multiplication di∣vide by the third number, and the number that cometh in the Quotient shall be the fourth number required, or answer to the Question.

Quest. 1. If 12 Labourers will build a House in 28 days, how many days shall 16 men build such another in.

Here I set the three given numbers in or∣der,

as you see in the following Example, where you 〈 math 〉〈 math 〉 see 12 in the first place, 28 in the second place, 16 in the third place, and the first and third number both of one Denomina∣tion, and the fourth number required will be the same Denomination as the second number is of, according to the former In∣structions of the last Chapter.

These numbers being thus placed and disposed as you see, multiply the first num∣ber 12 by the second number 28, and the

Product thereof will be 336, which divide by 16 the third number, and the Quotient gives you 21 days for answer to the Questi∣on as was required: For if 12 men can do the work in 28 days, 16 men will do it in 21 days.

Quest. 2. If a Traveller performeth a Journey when the days are 12 hours long in 12 days, in how many days shall he per∣form the same Journey when the days are 15 hours long?

Here you may clearly perceive, the more hours are in a day, the fewer days will he be in performing his Journey, therefore set your numbers down as you see, 〈 math 〉〈 math 〉 Multiply the second number 12 days, and the first number 10 hours together, and the product 120 divide by 15 hours the third

number, and the Quotient will give you 8 days, in answer to the Quotient as was re∣quired.

Quest. 3. If I hang a room with 36 yards of stuff at 7 quarters wide, how many yards will hang the same Room at 4 quarters or yards wide.

〈 math 〉〈 math 〉

Place the numbers are you see, and mul∣tiply the first and second numbers together, the product thereof will be 252, which di∣vide by the third number 4, and the Quo∣tient will be 63 yards, and so many yards doth answer, or is equal to 36 yards of 7 quarters wide.

Quest. 4. If when the bushel of wheat is worth 3 s. a penny wheaten Loaf weigh∣eth 8 ounces: I demand, how many oun∣ces shall the same Loaf weigh, when the bushel is worth 5 shillings?

〈 math 〉〈 math 〉

Multiply the first and second numbers together, the product being 24 divide by 5 the third number, the Quotient will be 4 ounces ⅘ of an ounce.

I might set down many Examples more of this Rule, but I suppose these may be sufficient for instruction sake.

The Rule of three Backward may be proved after this manner, multiply the first and second numbers together, and likewise the third and fourth numbers together. If the products of these two several multipli∣cations do agree, the operation is right, else not.

〈 math 〉〈 math 〉

Here is the first Example of this Rule proved by multiplying the first and second numbers, and the third and fourth numbers together, the products of each multiplica∣tion agreeing in one sum 336.

For proving of such questions, wherein after you have divided something doth re∣main, such Remainder must in proving the said question, be added to the Product of the third and fourth numbers, for the Pro∣duct of the third and fourth numbers in such cases will fall out so much lesser than the Product of the first and second number, as there be Units remaining of the Division.

CHAP. IX.

The Double Rule of three Direct.

THE Rule of Three Double or Com∣posed is so called, because it is a Composing or Joyning of two Questions of operations of the common Rule of Three into one, which doth much abbre∣viate Work.

And as in the Rule of Three there is the Direct Rule and the Backward Rule, so here also the two Questions may require a Direct or Reverse Proportion.

The Double Rule of Three being either Direct or Reverse, which is Backward.

In the Double Rule of Three Direct there are 5 numbers given to find out a sixth Proportional, which sixth Propor∣tional may be found out by two single Questions of the Rule of Three Direct.

In these five given numbers of the Double Rule of Three Direct, are two Species or Kinds, first a supposition, secondly a Que∣stion or Demand.

The Supposition is contained in the three first of the five given numbers, the Demand in the two last.

The Question being stated, the five num∣bers given must be disposed in such order as the first and the fourth may be of one Denomination, the second and the fifth number of another Denomination, and the answer in the sixth will be the same with the third, and then the manner of operati∣on must be thus, multiply your first and se∣cond numbers, the Product shall be the Di∣visor, then multiply the other three num∣bers one by another, and the Product there∣of shall be the Dividend.

Quest. 1. If 100 l. in 12 months gain for Interest 7 l. what will 80 l. gain for 8 months?

In the following Example of this Que∣stion you see the three first numbers, viz. 100 l. 12 months, and 7 l. do contain a supposition, and the two last, viz. 80 l. 8 months do contain a Demand or Questi∣stion.

And these five numbers are so disposed and placed, that the first number 100 l. and the fourth number 80 l. are of one kind or Denomination, viz. Money, the

second number 12 months, and the fifth number 8 months, are both of another De∣nomination, viz. months, and the sixth number required will be the same kind or species as the third number, viz. mo∣ney.

Now according to the former Instructi∣ons of this Chapter, I multiply the first and second numbers together, and their Pro∣duct 1200 is the divisor, then I multiply the three other numbers 7, 80, and 8 one by another as you see, whose Product com∣eth to 4480 for the Dividend, which di∣vided by 1200 the Divisor, the Quo∣tient giveth 3 for so many pounds, because the third number was pounds, but of this Division there will remain 880, which I multiply by 20, and the Product giveth 7600 for a new Dividend, which divided by 1200, the Divisor, the Quotient gives 14 shillings, but there yet doth remain 800, which multipled by 12 gives 9600 in the Product for another new Dividend, which divided by the former Divisor 1200, the Quotient giveth 8 pence, as by the former Example, so the sixth number found is 3 l. 14 s. 8 d. being of the same Kind or Spe∣cies as the third number, viz. money,

〈 math 〉〈 math 〉

therefore by the foregoing Operation 80 Pound for 8 Months will gain 3 l. 14 s. 8 d. if 100 Pound for 12 Months gaineth 7 l. being the true answer to the question, as was required.

This question is done at two Operations by the Rule of Three Direct, but this is the quicker and shorter way by much.

First Operation. If 100 l. gain 7, what will 80 l? an∣swer 5: 12: 0.

Second Operation. If 12 Months give 5 l. 12 s. what will 8 Months? answer 3: 14: 8.

This Example will be sufficient for the explanation of this Rule, but to make it so plain, that the Learner may meet with nothing to obstruct his cheerful proceed∣ing in the Art of Arithmetick, I will exem∣plifie these two Operations, as followeth, according to the Rules and Instructions laid down in the seventh Chapter.

〈 math 〉〈 math 〉

Here you have the Work perspicuous to the eye, performed by two single questions in the Rule of Three Direct, exactly agree∣ing in the answer to the former way, which is a very great Argument to prove the in∣fallibility of the former, which is much the quicker and shorter way.

CHAP. X.

The Double Rule of Three Backward.

IN all questions of this Rule you have five numbers always given, in disposing of which numbers to their several proper places, observe the directions in the last Chapter, for this Rule is no other than the Reverse of the last Rule, when the num∣bers are placed in order, the Solution of any question at one Operation is thus.

Multiply your third and fourth numbers together, the Product thereof is yoar. Di∣visor, then multiply your first, second and fifth numbers together, and their Product is your Dividend, which divide by your Divisor, and the quotient is the sixth num∣ber

required, and answer to the question propounded.

In the disposing of the five given num∣bers you must always remember to make the third and fifth numbers of one and the same Kind or Species.

Quest. 1. If 100 Pound in 12 Months gain 7 l. how much Mony in 8 Months will gain 3: 14: 8.

In the following Example of this que∣stion you find the three first numbers, 100, 12 and 7 do contain a supposition of the question, viz. suppose 100 l. in 12 months to gain 7 l. and the other two numbers 8 Months, and 3: 14: 8 contain the demand of the question upon those suppositions, viz. how much Mony in 8 Months will gain 3: 14: 8? which is according to the Instructions of the last Chapter.

And according to the Instructions of this Chapter, I have made the third number 7 l. and the fifth number 3 l. 14 s. 8 d. both of one Species or Kind, that is to say, they both do signifie Mony gained, although one of them doth contain three denomina∣tions, and the other but one.

Here you must note that if any of the five numbers given are of different denomi∣nations, you must reduce all the numbers of the same Kind or Species into the lowest denomination named, as here in the fol∣lowing Example, the fifth number con∣tains three denominations, viz. Pounds, Shil ings, Pence, as 3: 14: 8, therefore I do not only reduce that fifth number into Pence, but the rest of the five numbers, that are of the Species of Mony, 100 l. and 7 l. and afterwards I proceed according to In∣struction, as by the following Example.

Example.

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

The numbers given being orderly dis∣posed as you see, the next thing I find the first number, the third number, and the fifth number to be one of one Species, viz. Mony, but of different denominations, so I reduce them all three into the lowest de∣mination named which is Pence, the first number being reduced into 24000 d. the third number into 1680 d. and the fifth number into 896 d. at which time the five numbers are 24000, 12, 1680, 8, 896. Then I proceed according to the foregoing Instructions of this Chapter, and multiply the third number 1680 by the fourth num∣ber 8, and the Product 13440 is the Divi∣sor. Next I multiply the fifth number 896 by the second number 12, and the Product thereof is 10752, which I multiply again by the first number 24000, and the Pro∣duct

thereof is 258048000 for the Divi∣dend, which if you divide by the Divisor 13440, the Quotient will give you 19200d. for the sixth Proportional or answer to the Question propounded, which by Reduction you will find to be 80 l. which doth agree with the Question and Example of the last Chapter; the Question and Example of this Chapter being the Reverse of that Question and Example, and do prove one another, as you may perceive by their Examples.

This Question may also be resolved at two Operations, the first by the Rule of Three Reverse, the other by the Rule of Three Direct.

If 12 Months require 100 l. what will 8 Months? answer 150 l.

If 7 l. require 150 l. Principle, what will 3: 14: 8 require Principle? Answer 80 l.

The first Operation by the Rule of 3 Reverse. 〈 math 〉〈 math 〉

The second Operation by the Rule of Three. 〈 math 〉〈 math 〉

Here you see the two Operations worked according to the Rule of Three Reverse in the first Operation, and according to the Rule of Three Direct in the second Opera∣tion, and they do produce the same an∣swer as before, viz. 80.

And in like manner must you work in all other questions of this nature, suppo∣sing this question and Example sufficient to explain the meaning of this Rule.

CHAP. IX.

Reduction of Fractions.

BEFORE we treat of Reduction of Fractions, it will be convenient to shew the Learner what a Fraction is.

A Fraction or Broken Number is a part or many parts of any whole number, and it is usually expressed by two numbers in smaller figures, one above the other, with a little Line drawn between them, thus ¾ the figure or figures, standing above the Line, are called the Numerator, the other

underneath the Line are called the Deno∣minator, as by Example, 3 quarters is a Fraction, and must be thus expressed, whereof 3 standing above the Line, is cal∣led the Numerator, and 4 which stands un∣der the Line is called the Denominator, it being always convenient that the Numera∣tor be always less than the Denominator, for if the Numerator be either greater or equal to the Denominator, it is then an im∣proper Fraction.

Reducing of Fractions are according to the several Heads following.

• 1. To Reduce a mixt number into an improper Fraction equal thereunto.
• 2. To Reduce a whole number into an improper Fraction.
• 3. To Reduce an improper Fraction into its equivalent whole (or mixt) num∣ber.
• 4. To Reduce any Fraction into his low∣est equivalent term, to the Fraction given.
• 5. To find out the true value or worth of a Fraction in the known parts of Coyn, Weight, Measure, &c.
• 6. To Reduce Compound Fractions to a simple one of the same value.
• ...

• 7. To Reduce divers Fractions of seve∣ral unequal Denominators, to Fractions of equal Denominators, and of the same value.
• 8. To Reduce a Fraction of one Deno∣mination into another Fraction that is of the same value.

1. To Reduce mixt Numbers to an impro∣per Fraction.

A mixt Number is composed of two Parts, or Numbers, whole and broken In∣tegers and Fractions, the whole Numbers being composed of Integere, the broken being a Fraction annexed to the whole Number, as in the mixt Mumber 48 2/86, 48 is the whole number, and 12/16 is the bro∣ken number, shewing that one Integer of 48 being divided into 16 parts, this 12/16 doth signifie that there doth belong to 48, 12 of the 16 parts, the Rule is thus:

Multiply the whole number, by the De∣nominator of that Fraction annexed to him, and to the Product thereof, add the Numerator of the same Fraction, then set their sum, over the Denominator with a Line between them. And this new Fra∣ction,

though improper, shall be equal to the mixt Number given.

Example.

Let the mixt Number before-mentioned 48, 1/16 be required to be Reduced into an improper Fraction equal thereunto, mul∣tiply the whols Number 48 by the Denominator * 1.47 16, and the Product is 768, to which add the Numerator 12, and they do make 780, which is so many sixteens, therefore I set down 780, and under∣neath 16, drawing a Line between them as you see which is an improper Fra∣ction, but containing 780 sixteens is equal to the mixt number 48 12/16 as was required.

2. To Reduce a whole Number to an im∣proper Fraction.

Multiply the whole Number given by the intended Denominator, and place the

Product of the Multiplication over the De∣nominator, and draw a Line between.

Example.

Let it be required to Reduce the whole Number 24 into a Fraction whose Deno∣minator is 8, multiply the whole Number 24 by 8, and * 1.48 the Product is 192, which I set over the Denominator 8, and draw a Line between, so 192 is the Numerator, and 8 the Denominator, making 192/8 equal to 24 the whole Number.

3. To Reduce an improper Fraction into a whole or mixt Number equivalent there∣unto.

Divide the Numerator by the Denomi∣nator, and the Quotient will give the whole Number equal thereunto, that is equal to the improper Fraction given; and if any thing remain make it the Numerator, and make the Divisor Denominator, as you may see in the following Example.

Example.

Let it be required to Reduce the impro∣per Fraction 357/12 into a mixt Number equi∣valent thereunto, I divide the Numerator 357 by the * 1.49 Denominator 12, and the Quotient is 29, which is the whole Number, and there doth remain 9, which I make a Numera∣tor, setting 12 the Divisor underneath for a Denomi∣nator, so you have the mixt Number 29 9/12 equal to the improper Fraction 357/12 as was desired.

4. To Reduce a Fraction into its lowest terms equal thereunto.

If the Numerator and Denominator of the Fraction propounded be even numbers. Take the half of either as often as you can, and if either Numerator or Denominator be odd numbers, or falleth out to be so in the halfing, then divide them by any number that will divide both Numerator and De∣nominator

without any thing remaining in either, and when you have thus brought both Numerator and Denominator as low as you can, then this new Fraction shall be equal to the given Fraction.

Example.

Let it be required to Reduce the given Fraction 192/67 into the lowest terms. First I take the half of the Numerator 192,

192	96	48	24	12	6	2
672	336	168	84	42	21	7

which is 96. Likewise half the Denomi∣nator which is 336, so it is brought to 96/336, then I take half the Numerator 96, which is 48, and half the Denominator 336 which is 168, and they two numbers make the Fraction 48/18, and by the halfing 48/168 I bring it to 24/84 the half of which is 12/42, and the half of that 6/21, and because here the Denominator is an odd number 21, which will not admit of halfing, I try to divide them by 3, 4, 5, 6, &c. finding that 3 doth divide both the Numerator 6 and the De∣nominator

21 without any remains, bring∣ing the Fraction to 2/7 which is the lowest it can be brought, whereby I conclude 2/7 to be as much or equivalent with 192/672.

Another way of Abbreviation.

There is a more excellent and quicker way to Abbreviate or Reduce a Fraction into its lowest term, which is thus: First, divide the Denominator by his Numerator, and if any thing remain let your Divisor be divided by that number, and if any thing do still remain of this Division, di∣vide your last Divisor by this last Remain∣der, and so do until o remain, and the Divisor of that last Division shall be the number which shall Abbreviate your Fra∣ction by dividing both your Numerator and Denominator, and reduce them at once into the lowest terms.

Example.

It is required to Abbreviate 1152/2592 into its lowest terms, by a certain number which shall divide both Numerator and Denomi∣nator, without any Remainder, the Quoti∣ent

of which Divisions shall give the Nu∣merator and Denominator equivalent in the lowest terms.

Divide the Denominator 2592, by the Numerator 1152, and there will remain 288, therefore divide your Divisor 1152 by 288, and there will remain 0, the Di∣visor of this last Division was 288, which is the common number, by which if you divide the Numerator of your given Fra∣ction 1152, the Quotient gives 4 for a new Numerator, by which I likewise do divide the Denominator 2592, and the Quotient gives 9 for a new Denominator, and 0 remains of either of these Divisions, where∣by I find the new Fraction 4/9 is the same in signification as 1152/2592 being equivalent or equal thereunto in the lowest terms it can be reduced to, as was required.

Note.

If both Numerator and Denominator of any Fraction do end with a Cypher, or Cy∣phers, you may cast off as many Cyphers from one as from the other, and the re∣maining figure will give a Fraction of the same value as the given Fraction before was

of, Example 200/500 by cutting off the Cyphers will be reduced unto ⅖ which signifieth as much in value as the given Fraction 200/500.

V. To find the value of any Fraction in the known parts of Coyn, Weight, Mea∣sure, &c.

Multiply the Numerator by the number of Units in the next inferior Denominati∣on equal to an Unit of the same Denomina∣tion, as the Fraction given divide the Pro∣duct of that Multiplication by the Deno∣minator, and the Quotient giveth the value in such parts as you did multiply by; but if any number doth remain, multiply such remainder by the number of Units in the next inferior or lesser Denomination, and divide by your last Divisor, or Denomi∣nator of your given Fraction; do thus un∣til you have 0 remaining, or until you have brought it as low as you can, and the seve∣ral Quotients give the true value of the Fra∣ction in several Denominations, and if a∣ny thing remain of the last Division, make it a Numerator over the same Denominator in your Fraction given.

Example.

It is required to know or find out the true value of 283/487 of a Pound Sterling, the Fraction being given in a pound, the next inferior Denomination is shillings wherein 20 Units make an integer in pounds, there∣fore I multiply the Numerator of the given Fraction 283 by 20, the Product is 5660, which divide by 487 the Denominator, and the Quotient gives me 11 shilling, but there doth remain of this Division 303, which I multiply by 12 the number of Units in the next inferior Denomination, and the Product is 3636, which I likewise divide by 487, and the Quotient gives me 7 pence, and there doth remain 227, which I mul∣tiply by 4 the number of Units in the next inferior Denomination, and the Product 1008 I divide by 487 as before, and the Quotient gives me 2 farthings, and there still remains 34, which I put as a Nume∣rator over the old Denominator 487, so I find the just value of 283/487 of a pound ster∣ling is 11 s. 7 d. 2 f. 34/487 as you may see by the succeeding work.

〈 math 〉〈 math 〉

VI. To Reduce a Compound Fraction into a simple one of the same value.

A SimpleFraction consists but of one Nu∣merator and one Denominator; but a Compound Fraction doth consist of many Numerators, and as many Denominators ⅓ of ⅝ of 7/16 of a pound Sterling is a Com∣pound Fraction, because there are more Denominations than one represented in the Fraction.

When a Compound Fraction is given to be converted into a single Fraction, multi∣tiply all the Numerators together, and the Product thereof make a new Numerator, then Multiply all the Denominators toge∣ther, and the Product thereof make a new Denominator, which new Numerator and Denominator shall be a Single or Simple Fraction and equal to the Compound Fra∣ction given.

Example.

It is required to Reduce 9/16 of 3/9 of ⅝ of ¾ into a single or simple Fraction. Mul∣tiply the several Numerators, 9, 3, 5, 3,

one into an∣other as you * 1.50 see, the Pro∣duct there∣of will be 405, which make your Numerator, then multi∣ply all the Denomina∣tors in the samemanner and their Product will be 4608 which make your Denominator. So have you the single Fra∣ction 405/4608 equal to 9/16 of 3/9 of ⅝ of ¾ as was required, whereby you may know the va∣lue of a Compound Fraction: First by re∣ducing the Compound Fraction into a sim∣ple one as you are here taught, then by the Instructions of the fifth Rule of this Chap∣ter you may find its true value.

VII. To Reduce Fractions of unequal De∣nominators, into Fractions of the same va∣lue having equal Denominators.

Multiply all the Denominators into one Sum, and the Product shall be the common Denominator, then multiply each Nume∣rator into all the Denominators but his own, and the last Product put for a Nume∣rator over the Denominator before found. And this new Fraction is equal to the Fra∣ction whose Numerator you multiplied in∣to the Denominators; do so by the rest, and you have your desire fulfilled.

Example.

Reduce 243/578 and 7/16 into a common De∣nomination; do thus,

Multiply the Denominators 5, 7, 8 and 16 into one Sum, which will be 4480 for a common Denominator. Then multiply the Numerator 2 into all the Denominators but his own, viz. into 7, 8, and 16, and the Product thereof 1792 will be a Nume∣rator to the Denominator before found 4480, which new Fraction 1792/4480 is equal

to ⅖ the first given Fraction, next pro∣ceed for a new Numerator to a second Fraction by taking the next Numerator, which is 4, and multiply him by all the Denominators, except his, that is by 5, 8 and 16, and the Product is 2560, which is a new Numerator unto the Denominator before found 4480, and the new Fraction 2560/4480 is equal to the second given Fracti∣on, viz. 4/7. Likewise multiply the Numera∣tor 3 by the Denominators 5, 7, and 16, the Product gives 1680/4480 equal to the third gi∣ven Fraction which is 9/8, then multiply the last Numerator 7, by 5, 7, and 8, and the Product gives 1960/4480 equal to 7/16. So instead of 243/578 and 7/16, I have found 1792/4480 2560/4480 1680/4480 and 1960/4480 which have all equal Deno∣minators, and each Fraction equal to his respective given Fraction as was requi∣red.

VIII. To Reduce Fractions of one Denomi∣nation into another.

Reduction of Fractions of one Denomi∣nation to another, is two-fold, viz. Ascen∣ding or Descending. Ascending when a Fraction is to be reduced from a lesser to a

greater Denomination, as Fractions of pence into Fractions of pounds; Descen∣ding when a Fraction of a greater Deno∣mination is brought into a lesser, as Fra∣ctions of pounds into Fractions of pence.

A Fraction being given to be reduced from a lower Denomination to a greater Denomination, make of it a compound Fraction, by adjoining the parts of the se∣veral Denominations, between that and the Denomination you would reduce it to for Denominators, and a Unit for a Nume∣rator, then by the 6th hereof reduce your Compound Fraction to a Simple one.

Example.

It is required to know what part of a pound 7/9 of a farthing is?

To reduce this, you must make it a Com∣pound Fraction, by considering that 1 far∣thing is ¼ of a penny, a penny is the 1/12 of a shilling, and a shilling the 1/20 of a pound, therefore 7/9 of a farthing is 7/9 of ¼ of 1/12 of 1/20, which by the 6th hereof, having already made a Compound Fraction of it, you may bring it to 7/8640 of a pound as was re∣quired.

To reduce Fractions of a greater Deno∣mination into a lesser; you must multiply the given Numerator by the parts contain∣ed in the several denominations between the denomination your Fraction is of, and the denomination you would reduce it to, then place the Product over your given De∣nominator.

Example.

I would reduce ⅝ of a pound to the Fra∣ction of a farthing.

Multiply the Numerator 5 by 20, by 12 and 4 into one Sum, the Product is 4800, which put over the Denominator 8, and it makes 4800/8 of a farthing equal to ⅝ of a pound as was required.

And if the Fraction propounded to be reduced be in weight, you must then have respects to the parts contained in the seve∣ral denominations of weight.

CHAP. XII.

Addition of Fractions.

ADdition of Fractions is the putting or joyning together of several broken numbers, thereby to make one principal Fraction of them all, and in this and the next, observe that all Fractions proper or improper, must be of one denomination, or by the former Rules must be reduced thereunto, and when you have so reduced them, add all the Numerators together, and subtract their total sum for a new Nu∣merator over the common Denominator.

Example.

It is required to know the sum of 17/36 19/36 12/36 18/36, here you have one common Denominator, which is 36, and the sum of Numerators, 17, 19, 12, 18, is 66, which put as a Numerator over the Deno∣minator thus 66/3, and by the former Rules of the last Chapter you may reduce them to the mixt number 1: 30/36.

But if the Fractions that are required to be added together be of unequal and diffe∣rent Denominators, bring them a com∣mon Denominator by the 7th Rule of the last Chapter, and having so reduced them follow the foregoing instructions.

CHAP. XIII.

Subtraction of Fractions.

THE instruction given in Addition of Fractions are likewise here to be fol∣lowed in Subtraction of Fractions, in re∣ducing the Fractions given to one Deno∣mination, and before any Subtraction is to be made the Fractions must be reduced to a common Denominator, then deduct one Numerator from the other, and set the remainder over the common Denominator, which Fraction is the difference between the given Fractions.

Example.

It is required to subtract 3/7 from 7/9 by the 7th of the 11th Chapter, you may re∣duce

the said given Fractions into a com∣mon Denominator, which will be 49/63 and 27/63, subtract the lesser Numerator 27 from the greater 49, and there will remain 22, which place over the Denominator 63 makes the Fraction 22/63 which is the diffe∣rence between the two given Fractions as was required.

Secondly, If you are to subtract a bro∣ken number from a whole number, you must borrow a Unit and convert it into a Fraction by making both Denominator and Numerator equal to the Denominator of the Fraction given, then take the lesser Nu∣merator out of the greater, and place the remainder as a Numerator over the given Denominator, and for one you borrowed of the whole number, pay a Unit by sub∣tracting a Unit from the whole number and the remainder set down before the Fraction found as before, and the mixt number is the remainer or difference be∣tween the two given Fractions, when they are taken one from the other.

Example.

It is required to subtract 7/16 from 12, here 7/16 is a fraction or broken number, and 12 is the whole number, from which whole number I take a Unit and make a Fraction of, so the number is 11 16/16, then I deduct the Fraction 7/16 from 11 16/16, taking the Numerator 7 out of the Numerator 16, and there remains 9, which I make a Numerator over the Denomina∣tor given 16, and set down 11 * 1.51 before the said Fraction as you see, So shall the difference or remainer between the two given Fractions be 11 9/16 as was required.

Thirdly, If it be required to subtract a Fraction from a mixt number, or one mixt number from another, reduce the Fractions into a common Denominator by the 7th of the 11th, and when you have so done, if the Numerator of the Fraction to be subtracted, be less than the Numerator of the Fraction from whence you subtract, then take one from the other, and the remainer is the Nu∣merator to that common Denominator, then subtract the less whole number from the greater, and that remainer with the Fracti∣on

annexed, is the difference between the two given numbers.

Example.

It is required to subtract the mixt num∣ber 7 ⅜ from the mixt number 9 5/9 by the 7th of the 11, the Fractions are reducted unto 27/72 and 40/72 therefore I subtract the Numerator of the number to be subtracted being lesser, viz. 27/72 from * 1.52 40/72 and the remainer is 13 which I set over the common Denominator 72 as a Numerator. Likewise I subtract the whole number 7 from 9, and there remains 2, by which I find that the mixt number 2 13/72 is the remainer or difference between the two given mixt numbers 9 5/9 and 7 ⅜ as was required.

But if the Fraction to be subtracted be greater than the Fraction from which you are to subtract, reduce the Fractions as you were taught by 7th of the 11th into a com∣mon Denominator, then subtract the grea∣test Numerator from the common Denomi∣nator, add the remainer to the lesser Nu∣merator, and their Sum make a new Nu∣merator to the common Denominator, and

carry one to the whole number to be sub∣tracted, and subtract it from the greater whole number, and the remainer with the Fraction annexed, being a mixt number, shall be the difference of the two mixt num∣bers given.

Example.

It is required to subtract 24 3/4 from 28 2/6. By the 7th of the 11th the Fraction 24 ¾ is reduced into 18/24, and the other Fraction 28 2/6 into 8/24, now because the numerator 18 cannot be subtracted from 8, therefore I subtract the Numerator 18 from 24 the common Denomina∣tor, and there remains 6, * 1.53 which I add to the Nu∣merator 8, and it makes 14, which makes a new Numerator to the com∣mon Denominator 24, then I carry 1 to the whole number 24, say∣ing 25 from 28, and there remains 3, so the mixt number 3 14/24 is the remainer and difference between the two Fractions given 28 2/6 and 24 ¾ as was required.

Fourthly, If you are to subtract Fracti∣on

of Fractions from Fraction of Fracti∣ons, that is one Compound Fraction from another, you must by the 6th of the 11. re∣duce the two Compound Fractions given into two Simple or Single Fractions, then by the 4th of the 11, you abbreviate those two Fractions, and by the first Example of this Chapter subrtact them.

Example.

Let it be required to subtract 2/7 of 3/16 of ⅝ from ⅜ of 5/7 of 8/12 by the 6th of the 11, you reduce 2/7 of 3/16 of ⅝ into the Simple Fraction 30/896, and ⅜ of 5/7 of 8/12 into 120/672, now these two single Fractions 30/896 and 120/672, which two Fractions by the 4th of the 11th you may abbreviate to 15/448 and 60/336, then by the 7th of the 11. reduce these two last Fractions into a common Denomi∣nator, and you will make the Fractions 26880/161528 5040/161528 subtract the lesser Nume∣rator 5040 out of the greater 26880, and there remains 21840, which put as a Numerator over the common Denomina∣tor, and you will have the Fraction 21840/161528 which is the difference between the two Compound Fractions 2/7 of 3/16 of ⅝ and ⅜ of 5/7 of 8/12 as was required.

CHAP. XIV.

Muliplication of Fractions.

IF the Fractions you are to multiply are Simple Fractions, multiply the Deno∣minators one by another, the Product is a new Denominator, also multiply the Numerators one by another, and their Pro∣duct is a new Numerator, and this new Fraction is the Product of the Multiplica∣tion required.

Example.

It is required to multiply 2/6 by ⅜.

The Product of the two Denominators 6 and 8, multiplied together is 48, and the Product of the two Numerators 2 and 3 is 6, therefore the new Fraction 6/48, which by abbreviation is ⅛, is the Product of the two Fractions 2/6 and ⅜ multiplied one by another as was required.

If the given Fractions to be multiplied be mixt numbers, by the first of the 11th,

you must first put either the whole number into his broken by reducing them into im∣proper Fractions, and then multiply Nu∣merator by Numerator, and Denominator by Denominator, their Products will give you a new Numerator and Denominator for a new Fraction, which new Fraction shall be the Product required, but this Fra∣ction will be an improper Fraction, there∣fore if you divide the Numerator by the Denominator, the Quotient gives the whole numbers, and the remainer is the Numerator, and the Divisor the Denomi∣nator of the proper Fraction, and this mixt number found shall be the true Product as before.

Example.

It is required to multiply 36 5/9 by 18 ⅖, by the 1st of the 11, the mixt number 36 5/9 is 329/9 and the mixt number 18 ⅖ is 92/5, now multiply the two Numerators 329 and 92 together, the Product is 30268, which is your new Numerator, and the Product of the two Denominators 9 and 5 is 45 the new Denominator, so have you the improper Fraction 30268/45 being the

Product required; but if you divide the Numerator 30268 by the Denominator 45, the Quotient gives the whole number 672, and the remainer 28 is Numerator to the Divisor 45, so the mixt number 672 28/45 is the Product of 36 5/9 multiplied by 18 ⅖ as was required.

If a Compound Fraction be given to be multiplied by a Simple or Single Fraction, reduce that Compound Fraction given into a Simple one by the 6. of the 11th, and pro∣ceed according to the instructions of the first Example, in multiplying one by another.

Example.

Let it be required to multiply 3/7 by ⅜ of 2/9 of 3/12, and this Compound Fraction ⅜ of 2/9 of 3/12 is by the 6th of the 11th redu∣ced to the Simple or Single Fraction 18/864, which by abbreviation is 1/48, which multi∣plied by 3/7, the Product is 3/326 as was requi∣red.

If the Fractions given to be multiplied are both Compound Fractions, reduce them into Simple ones both, by the 6th of the 11th, and proceed as before, and you have the Product for answer.

Example.

It is required to multiply 7/9 of ⅗ of ⅚ by 6/7 of ¾ of 2/8 by the 6th of the 11th, the Com∣pound Fraction 7/9 of ⅗ of ⅚ is reduced unto 105/270 and 6/7 of ¾ of ⅔ unto 36/84, multiply these two simple Fractions Numerator by Nume∣rator, and Denominator by Denominator, and you will have the new Fraction 3750/22680, which by abbreviation is 125/756, both which last Fractions are either of them equal to the Product of the two given Compound Fractions, multiplied one by another as was required.

If it be required to multiply a Fraction by a whole number, make that whole num∣ber a Numerator, and set 1 under him for a Denominator, so the Fraction will be im∣proper, but multiply their Numerators and Denominators together as before.

Example.

It is required to multiply 16 by ther e∣fore I make a Fraction of the whole num∣ber thus 16/1, which multiplied by 3/7, produ∣ceth 48/7, or 6 6/7 as was required.

CHAP. XV.

Division of Fractions.

IF the Dividend and Divisor the two Fra∣ctions given are both Simple or Single Fractions, set the Divisor next the left hand, and the Dividend even with it to∣wards the right hand, then multiply them cross-wise thus: Multiply the Numerator of the Dividend, by the Denominator of the Divisor, and the Product is the new Numerator, then multiply the Denomina∣tor of the Dividend by the Numerator of the Divisor, and the Product is a new De∣nominator, which new Fraction is the Quotient you do desire.

Example.

It is required to divide ⅘ by ¾, to effect it I proceed thus, ⅘ being the Dividend, and ¾ the Divisor, * 1.54 I place them as you see, the Divisor to the left hand, and the Dividend

to the right, so I multiply the Numerator of the Dividend 4 by the Denominator of the Divisor 4, the Product is 16, for a new Numerator, then I multiply the Denomi∣nator of the Dividend 5 by the Numera∣tor of the Divisor 3, and the Product 15 is the new Denominator, so I find 16/13 or 1 1/15 is the Quotient sought.

Secondly, if you divide a Simple Fra∣ction by a Compound, or a Compound by a Simple, or Compound by Compound, reduce such Compound Fractions to simple Fractions, then proceed according to the last Example.

Example.

It is required to divide 7/9 of 12/16 by ⅔ of ⅝, you must reduce 7/9 of 12/6, which makes the Simple or Single Fraction 84/144, and of ⅔ of ⅝ 10/24, so 84/144 is the Dividend, and 10/24 is the Divisor, which mul∣tiplied cross-wise, as * 1.55 in the last Example, doth produce the Fraction 4216/4410 for the Quotient, which in a proper Fraction is 1 576/1440.

If the Dividend or Divisor, or both be mixt numbers, reduce them to improper Fractions, and perform the Division as in the two last Examples.

Example.

It is required to divide 22 3/9 by 7 ⅔ the improper Fraction of 22 3/9 is 201/9 and of 7 ⅔ is 21/3, so the improper Fraction 23/3 is the Divi∣sor, * 1.56 and 201/9 is the Di∣vidend, which if you do multiply cross-wise, you will have the improper Fraction 603/207, now if you divide the Numerator of this Fraction 603 by the Denominator 207, the Quotient gives 2, and there will remain 189, which is the Numerator, and the Di∣sor 207 the Denominator, so the mixt num∣ber 2 189/207 is the Quotient required.

And in like manner must you work if you divide a Fraction by a whole number, or a whole number by a Fraciion, as you see in this last Example.

CHAP. XVI.

The Rule of Three Direct in Fractions.

IN the Direct Rule of Three in Fractions there are four things to be observed and practised.

1. That the Fractions of the first and third places be of one Denomination, as Mony, Weight, or Measure.

2. That if any of the given Fractions be Compound, they must be reduced to simple of the same value.

3. If any of the Fractions given should be mixt numbers, reduce it into an impro-Fraction.

4. If any of the three terms given should be a whole number, make it an im∣proper Fraction by putting a Unite for his Denominator.

The Solution of the Rule of Three Di∣rect in Fractions is performed the same way as in whole numbers, that is to say, mul∣tiply the second and third Fractions toge∣ther by the 14th Chapter, and the Product

thereof by the first Fraction, according to the fifteenth Chapter, and the Quotient is the fourth Fraction required.

Another Way.

Multiply the Numerator of the first Fra∣ction by the Denominators of the second and third Fractions, the Product thereof is a new Denominator; then multiply the Denominator of the first Fraction by the Numerators of the second and third Fra∣ctions, and the Product is the new Nu∣merator, and this Fraction is the fourth proportional Fraction required, and an∣swer to the Question, which if it should fall out to be an improper Fraction, must be reduced into a whole or mixt number by the third of the eleventh.

Quest. 1. If ⅜ of a Pound weight cost 5/12 l. what will 9/21 of a Pound weight cost?

〈 math 〉〈 math 〉

Having disposed of the given Fractions according to the four first Rules of this

Chapter and the seventh, I proceed to the Solution thus. Multiply the Numerator of the first Fraction which is 3 by 12, and 12 the Denominators of the second and third Fractions, the Product will be 432 for a new Denominator. Then multiply 8 the Denominator of the first Fraction by 5 and 9 the Numerators of the second and third Fractions; the Product thereof is 360, which is a new Numerator. So I have found the fourth proportional Fra∣ction 360/432 for answer to the Question pro∣pounded, and by abbreviation is found equal to ⅚ or 16 s. 8 d.

CHAP. XVII.

The Rule of Three Backward in Fractions.

FOr disposing of the given Fractions fol∣low the four Rules in the last Chapter, and the way of resolving is not different from the Instructions of the eighth Chapter, for it was there shewed how to find out a fourth proportional to three given numbers.

When you have a Question to be re∣solved by this Rule, then the third Term is the Divisor, then having exactly reduced the Terms given according to the Rules of the last Chapter, then multiply the Nume∣rator of the third Fraction by the Denomi∣nators of the second and first Fractions, and the Product gives a new Denominator, Then multiply the Denominator of the third Fraction by the Numerators of the second and first Fraction, and the Product gives a new Numerator, and this new Fra∣ction is the fourth proportional Fraction and answer to the Question propounded.

Quest. If ⅞ of a Yard of Cloth, that is 1 Yard ¾ will make a Garment, how much of another piece of Cloth that is 1 Yard ⅖ wide will make the same Garment.

〈 math 〉〈 math 〉

Here you see in this Question the second and third Fractions are mixt Fractions, there you see they are reduced to improper Fractions, so the three given Fractions are ⅞, 7/4 and 7/5. Now according to the former Instructions of this Chapter, I multiply the Numerator of the third Fraction being 3, by the Denominators of the second and first Fractions, which is 4 and 8, the Product thereof is 224, which is a new Denomi∣nator. Then I multiply the Deuomina∣tor of the third Fraction 5 by the Nume∣rators of the second and first Fractions, which is 7 and 7, the Product thereof will be 245, which is a new Numerator. So I have found out the improper Fraction 245/224, which reduced is 1 21/224 equal to 1 3/32, which is the answer to the propounded Question.

CHAP. XVIII.

The Rule of Practice.

THE Rule of Practice is so called be∣cause it is the Practice of the other Rules, and chiefly of the Rule of Three, in a far more compendious and shorter way, being by some called the brief Rules, be∣cause thereby many Questions may be ope∣rated with more expedition than by the Rule of Three, and because this Rule is so expeditious in finding a resolution to any Question propounded, it is therefore of infinite great Use amongst Merchants, Tradesmen and others.

It is very conve∣nient * 1.57 for the Lear∣ner to get perfectly by Heart the an∣nexed Table con∣taining the Pro∣duct of 12 multi∣plied by all the nine Digits, for the

speedy reducing Pence into Shillings, and Shillings into Pence; and for reducing Shillings into Pounds, I refer the ingenious Reader to the sixth Chapter, in Reduction of Mony.

	s.	f.
The even Parts of a Shilling.	6		is the	½
4		⅓
3		¼
2		⅙
1	½	⅛
1		1/12
	¾	1/16
	½	1/24
	¼	1/48

	s.	d.
The even Parts of a Pound.	10		is the	½
6	8	⅓
5	0	¼
4	0	⅙
3	4	⅛
2	6	⅛
2	0	1/10
1	8	1/12
1	0	1/20

I would likewise advise the Learner to get the two foregoing Tables by Heart, the first containing the even parts of a Shilling, the second the even parts of a Pound.

When a Question is propounded to be resolved, you must consider whether the price of the Integer be parts of a Shilling, or parts of a Pound.

And whether the said price be either the even part of a Shilling, or the even part of a Pound; if it be neither the even part of a Shilling, nor the even part of a Pound, you may divide the said price into so many parts that each part may be the even part either of a Shilling or Pound, as 5 d. ¾ may be divided into three even parts of a Shil∣ling, viz. 3 d. which is ¼ of a Shilling, 2 d. which is ⅙, and ¾ which is 1/16.

Proposition 1.

If the price of a Unite of the Sum pro∣pounded be a Farthing, I find by the Table a Farthing to be the 1/48 of a Shilling, but because the 48. part of any Sum is not so soon taken as the ½, the ⅓, the ¼, the ⅙, the ⅛, and the like, therefore because 8 Farthings make 2 d. I divide the given

number by 8, that is to say, I take the eighth part, and if any thing remain they are all Farthings. Then by the Table I find that 2 d. is ⅙ of a Shilling, therefore take the sixth part of them for Shillings, and what remains is so many two Pences; now these Shillings reduce into Pounds, by cutting off the first figure to the right hand, with a small stroak with your pen; then draw a line underneath, and take half of the re∣maining figures, which I set under the line for so many Pounds; but if the last figure you are to halve be an odd figure, take the half and add 10 to the figure cut off, for that figure and the 10 added together makes so many Shillings.

Example. Let it be required to know how much 9783 l. at a Farthing per Pound cometh to. First, as you see in the follow∣ing Example, take the eighth part of 9783, and it is 1222 two Pences, and 7 Farthings remaining, which is 1: ¾ I set a little distance to the right hand, next by the Table of the even parts of a Shilling, I find 2 d. to be the 1/0 part, therefore I draw a line under 1222, and take the ⅙ part thereof, which is 203 Shilling, and 4 remaining which is so many two pences; therefore I

set down 8 under the last Remainer, I draw a line under these Shillings 203, and reduce them into Pounds, according to the Instructions hereof, and of the sixth Cha∣pter of Reduction of Mony, by cutting off the first figure towards your right hand, which is 3, and drawing a line I take the half of the Remainer 20, which is 10 Pounds, and the figure 3 cut off signifieth so many Shillings. Now I add all these odd Remains together, viz. 1: ¾, 8 d. unto 10 l. 3 s. So the whole Sum is 10: 3: 9: ¾, the Resolution of the propounded Question, as by the following Operation.

〈 math 〉〈 math 〉

For the more clear demonstration of what hath been so far said of this Rule, I shall set down more Examples.

〈 math 〉〈 math 〉

In the first Example it is propounded 65937 l. at a Farthing per Pound, the ⅛ thereof is 8242 two Pences, and ¼ Farthing remaining, the ⅙ of 8242 two Pences is 1373 Shillings, and 4 two pences remain∣ing is 8 d. Next the 1/20 of 1373 Shillings is 68 Pounds, and 13 Shillings remaining; so taking all the Remains to this last Sum, it makes 68: 13: 8 ¼, and so much doth 65937 l. come to at a Farthing per Pound;

and in the same manner is the other Ex∣ample explained. The reason why I take the ⅛ and ⅙ in these Examples of a Far∣thing, and not 1/48, is because the ⅙ of ⅛ is equal to 1/48.

Proposition 2.

If the value of a Unite in the Sum of the propounded Question be 2 Farthings, then first take the ¼ and afterwards the ⅙; for although 2 Farthings be the 1/24 of a Shil∣ling, yet ¼ of 1/•• is equal thereunto, and done with more ease and less charge to the memory than to take the 1/24 of any Sum, having thus taken the ¼ to reduce them into two pences, if any thing remain they are so many half pence, then ⅙ reduceth them into Shillings as before.

Examples.

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

In the first of these Examples it is re∣quired to know how much 57632 l. at 2 quarters per Pound cometh unto, knowing 4 half pence do make 2 d. I therefore do take ¼ of 57632 which is 14408 two pen∣ces; then I take ⅙ of 14408 which is 2401 Shillings, and there remains 2 two pences which is 4 d. Then the 1/20 of 2401 Shil∣lings is 120 Pounds, and there doth remain 1 Shillings. So the Sum makes 120 l. 1 s. 4 d.

In the second Example the ¼ of 76597 is 19149 two pences, and there remains 1 half penny; the ⅙ of 19149 is 3191 Shillings, and there doth remain 3 two pences, which is 6 Pence, and the 1/20 of 3191 is 159 Pound, and 11 Shillings re∣maining. So the whole Sum doth amount 10 159 l. 11 s. 6 d. 2 quarters.

Proposition 3.

If the price of a Unite in Sum propound∣ed be 3 Farthings, then take the ¼ of Sum propounded which will be three pences, and if any thing remains it will be so many three farthings. Then the ¼ of the three pences is Shillings, and what doth remain is so many three pences, and the 1/20 of those Shillings is Pounds, as before.

Example.

〈 math 〉〈 math 〉

In the last of these Examples you find it required for to know how much 769327 l. at 3 Farthings per Pound cometh to? the ¼ of 769327 is 192331 three pences, and there remains three times three Farthings, which is 2 d. 1 qu. then the ¼ of 192331 is 48082 Shillings, and there remains 3 three pences, which is 9 d. Then the 1/20 of 48082 is 2404 Pounds and 2 Shillings remaining. So the whole amounts unto 2404: 2: 11: 1 qu.

Proposition 4.

If the price of an Unite of the Sum pro∣pounded to be resolved be an even part of a Shilling, and above the value of 3 Far∣things, divide the number given whose value is sought, by the Denominator of the Fraction representing the even part, the Quotient thereof will be Shillings, then take 〈◊〉〈◊〉 for Pounds as before.

Example. Let it be required to know how much 28376 l. at 1/•• cometh to? by the Table I find 1: ½ is 1/•• of a Shilling, therefore I divide the number 28376 by 8, or which is all one, I take the 〈◊〉〈◊〉 which is 35••7 Shillings, the 1/20 of 3547 Shillings

is 177 Pound and 7 Shillings remaining, as by the following Operation.

〈 math 〉〈 math 〉

Of every even part of a Shilling above the value of 3 Farthings, I have demon∣strated by these following Examples, which need little or no explanation, being so plain that at first sight they may be un∣derstood.

Examples.

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

Proposition 5.

If the price of a Unite of the propounded Sum be Pence, and not the even part of a Shilling, then you must divide the price given into even parts, as 7 d. is no even part of a Shilling, yet may be divided into even parts, as 4 d. and 3 d. make 7 d. and are both even parts of a Shilling, in which case work thus. First take ⅓ and then the 1/•• of the given number, and add the ⅓

and ¼ of the number given together, and you have the true number of Shillings, the 1/20 whereof is Pounds as before. Likewise 9 d. is ½ and ¼ or three times ¼, 11 d. is ½ and 1/•• and ⅙ or else it is 1/•• and ⅓ and ¼ of a Shilling, as by the following Operations doth more plainly appear, where you have an Example of every uneven part of a Shil∣ling, the price being any number of pence only.

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

Proposition 6.

When the price of a Unite of the given number be Pence and Farthings, and if it be an even part of a Shilling, work as in the fourth Proposition; but if it be uneven parts of a Shilling, as 1: ¼, 1: ¾, 2: 1/••, 2: ¾, or the like.

First find out some part of the price that is an equal part of a Shilling; then consider what part the Remainer of the price is of that even part, and divide that even part thereby, then take their Sum which is the true number of Shillings, the 1/20 whereof is Pounds, as before.

Example. It is required to know how much 28762 l. is worth at 1: ¼. First I take the 1/12 which is 2396 Shillings, and 〈 math 〉〈 math 〉 there remains 10 d. Now 2396 Shillings 10 d. being the value of the said number at 1 d. per Pound, and 1 Farthing being ¼ of a Penny, therefore the ¼ of the value at 1 d. per Pound 2396 Shillings and 10, which is 599 Shillings 2: ½ is the value of the said number at 1 Farthing per Pouud which added together gives 2996 Shillings one half penny, the 1/20 whereof 149: 16: 0: 2 qu. is the whole Sum required.

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

In the foregoing Examples you may perceive how the several Examples are wrought by dividing the given price into so many even parts as will make the com∣pleat price, although it may consist of 3, 4 or 5 Divisions, as in the 10. Example the price is 11 d. 3 qu. which is made up with the ½, and 1/••, and ¼ of the ⅓, and the ½ of the ¼, and the ½ of that ½, as 6, 4, 1, ½ and ¼, which all put together makes 11 ¾.

Proposition 7.

If the price of a Unit of the given num∣ber be 2 Shillings, cut off the figure toward the right hand of the given number, and double it, the Product will be Shillings, and the remaining figures towards the left hand are Pounds.

Example.

Let it be required to know how much 1283 Yards are worth at 2 Shillings per Yard, the figure toward the right hand which is 3 being cut off and doubled makes 6 Shillings, and the remaining figures 128 are Pounds. So * 1.58 the true worth of 1283 Yards at 2 Shillings per Yard cometh to 128 Pounds, 6 Shillings, as by the annexed Example may appear, being the answer and solution of the question, as was required.

By this Proposition we gather, that when the price of any number of Yards, Pounds, &c. is an even number of Shil∣lings, with the half of the even number,

multiply the given number, but double the first figure of the Product, and set it apart for Shillings, and the rest of the Product shall be Pounds.

Example.

Let it be required to know how much 5783 Ells of Linnen cometh unto at 14 s. per Ell, for the solu∣tion whereof, I take * 1.59 half of the price gi∣ven, which is 7, and therewith I do mul∣tiply the given num∣ber 5783; saying, 7 times 3 is 21; now ac∣cording to the Rule I double the first figure 1, which makes 2 Shillings, setting it apart, and I carry 2 to the next figure, saying, 7 times 8 is 56, and 2 as I carried is 58, set down 8 and carry 5, proceeding ac∣cording to the Instructions of the 4. Cha∣pter, and you will find the remaining part of the Product, to be 4048. So I find that 5783 Yards, Ells or Pounds at 14 is worth 4048 l. 2 s. as by the annexed Example doth appear.

Proposition 8.

If the price of a Unit of the given num∣ber be any odd number of Shillings, work the even number of Shillings by the last Example of the 7. Proposition, and for the odd Shillings take the 1/20 of the given num∣ber, and add these two together, and you will find your desire, as by these following Examples doth appear; where in the first towards your left hand it is required to know how much 7393 Yards cometh unto at 7 s. per Yard, for the even number of 〈 math 〉〈 math 〉 6 s. per Yard by the second Example of the 7. Proposition, 7393 Yards cometh to 2217 l. 18 s. then the 1/20 of 7393 is 369 l. 13 s. which added together make 2587 l. 11 s. the true value of 7393 Yards as was required, and in like manner is the other

Example operated for 3767 Yards at 9 s. per Yard.

Proposition 9.

If the price of a Unit of the given num∣ber be 5 s. then the Operation, although it be an odd number of Shillings, may by taking the 4 be done sooner, as by the fol∣lowing Examples doth more plainly ap∣pear.

〈 math 〉〈 math 〉

Proposition 10.

If the price of a Unite of the given num∣ber be Shill. and Pence, or Shill. Pence and Farthings, if the Shill. and Pence be an even part of a Pound, divide the number given by the Denominator of that Fraction repre∣senting the said even part, as by the follow∣ing

Example wherein it is required to know how much 576 Yards amount to at 3 s. 4 d. per Yard. First by the Table of the * 1.60 even parts of a Pound I find 3 s. 4 d. to be the ⅙ of a Pound, therefore I divide 576 by 6, which is the Denominator of the Fraction, or which is all one, I take the sixth part thereof which is 96, So I find that 576 Yards at 3 s. 4 d. cometh unto 96 l. as was required.

For further explanation sake I have set down these several following Examples.

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

Proposition 11.

When the price of a Unit of the number given is Shillings and Pence, but not the even part of a Pound, you must reduce them into even parts, as 8 s. 4 d. may be reduced into the even parts 6 s. 8 d. and 1 s. 8 d. which make 8 s. 4 d.

Likewise 7 s. 6 d. may be reduced into these even parts 5 s. and 2 s. 6 d. which make up 7 s. 6 d. add together the several Products of such reduced even parts, and their Sum is the answer to the question propounded, as by these following Exam∣ples may appear.

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

Proposition 12.

When the price of a Unite of the given number is Pence and Shillings, but cannot be reduced to even parts, then in such case multiply the given number by the number of Shillings contained in the price given, and after by the fourth or fifth Proposition hereof, then add the several Products found together, their Sum is the total in Shillings, as by this Example following.

Let it be required to know how much 5734 Yards cometh to at 9 s. 9 d. per Yard: according to the Rule I multiply the given number 5734 by 9, the Product gives me 51606 Shillings, therefore 9 d. being no even part of a Shilling, I by the fifth Pro∣position divide them into even parts 6 and 3, taking the ½ of the given number 2867,

and the ¼ which is 1433, these Products added together make 55906 Shillings and 6 Pence, the 1/20 thereof is 2795 Pounds, 6 Shillings and 6 Pence.

〈 math 〉〈 math 〉

Proposition 13.

When the price of a unit of any number given be Shill. Pence and Farthings, then tiply your given number by the number of Shillings contained in the given price, and your Product will be Shillings; then for the Pence and Farthings work by the sixth Proposition of this Chapter, as by the fol∣lowing Examples doth appear. In the first whereof it is required to know how much 1743 Yards come to at 7 s. 3 d. 3/••••, first I

multiply the given number 1743 by 7, the number of Shillings in the price given, the Product thereof is 12201 Shillings; then by the sixth Proposition I find the Product for 3 d. ¾, adding them together I find they do make 12745 Shillings 8 d. ¼.

〈 math 〉〈 math 〉

Proposition 14.

When the price of a Unite of any num∣ber given is Pounds, multiply the number given by the number of Pounds contained in the price, and the Product is the num∣ber of pounds required, as by these two following Examples you may discern.

〈 math 〉〈 math 〉

Proposition 15.

When the price of any given number is Pounds and Shillings, work for the Pounds as by the last Proposition, and for the Shil∣lings as by the 7. and 8. then add the seve∣ral Products together, and the Sum is the value sought as was required, as by these following Examples.

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

Proposition 16.

When the given price of a Unite of the number given is Pounds, Shillings, Pence and Farthings. First, find out the whole value of the given number in Shillings, Pence and Farthings, by the Instructions of the 13. Proposition of this Chapter, as if there had been no Pounds in the given price, then for the Pounds work as by the directions of the 15. Proposition, then add these two Products thus found together, and their Product or Sum shall be the whole value of the given number at the price pro∣pounded.

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

Proposition 17.

If the given number whose value is re∣quired be of several Denominations, as Hundreds, Quarters and Pounds; for the Hundreds, follow the directions of the last Proposition, and for 2 quarters take half the given price, for 1 quarter take ¼, for 14 pound take ⅛, for 7 pound take 1/16 of

the given, then add all these Products to∣gether, and you have the full value requi∣red, as by these following Examples doth appear.

〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

CHAP. XIX.

The Rule of Fellowship without Time.

THe Rule of Fellowship without Time, is when several Merchants do put in each man his several and particular Sum of Mony, without any respect at all to any Time, but to have such a proportion of the gain as he hath in the Stock.

Example.

A, B, C and D make a Stock of 20 Pound, in which A hath 5 Pound, which is ¼,

	l.
A	5
B	10
C	3
D	2
	20

B hath 10 Pound which is ½, C hath 3 l. which is 3/20, and D hath 2 Pound which

2/20. By this 20 Pound they do gain 7 l. which makes the Stock 27 Pound; now that person that hath ¼ in the first Stock, must also have 1/•• of the profit, which is 1 l. 15 s. and he that hath ½ of the first Stock must have ½ of the Profit which is 3 l. 10 s. and he that hath and he that hath 3/20 of the first Stock must likewise have 3/20 of the profit, which is 1 l. 1 s. and he that hath 2/20 of the first Stock must have 2/20 of the profit, which is 14 s. as you see.

		Stock.	Profit.
A	1/••	5	1 15
B	½	10	3 10
C	3/20	3	1 01
D	2/20	2	14
		20	7

Now to find out each mans several share in the profit or loss, you must collect the several Sums paid into Stock together, and by addition find their Sum, then by the Rule say, As the Sum of the several part∣ners shares is to the profit or loss, so is each mans particular share in Stock to his particular share in the profit or loss, and

as many shares as the principal Stock doth contain, so many Operations will there be of the Rule of Three, the fourth propor∣tional of each Operation being the respe∣ctive share in the profit or loss, belonging to that share of the Stock, which is the third number in the same Operation.

Example 1.

Four Partners, viz. A, B, C and D make up a Stock, in which Stock they have each different and unequal shares. A puts in 160 l. B put in 80 Pound, C puts in 144 Pound, and D put in 96 l. now there. was gained by the whole Stock 126 l the question is how much of the 126 l. each Partner to is have according to his share in the Stock.

	l.	s.	d.
A	160
B	80
C	144
D	96
	480

According to the former directions, I add the several shares together, and they do make 480 Pounds, now because there are four several shares, there will be four Operations by the Rule of Three, which are as follow.

If 480 l. give 126 l. what will 160 l? Ans. 42 l.

If 480 l. give 126 l. what will 80 l? Ans. 21 l.

If 480 l. give 126 l. what will 144 l? Ans. 37 l. 16 s.

If 480 l. give 126 l. what will 96 l? Ans. 25 l. 04 s.

And these fourth proportionals found are due to the several partners over and above their respective shares, as to A 42 l. to B 21 l. to C 37: 16, and to D 25 l. 4 s. as you see following, which being added altogether make 126 l. as before, equal to the Sum that was gained.

A	42
B	21
C	37: 16
D	25: 04
	126

The adding together of these fourth proportionals found is the proof of the single Rule of Fellowship, for if they make up the same Sum as was gained by the whole Stock, or lost, then your several Operations have been right worked, if there be any difference you may certainly conclude you have committed some errour.

Example 2.

Two Partners join together and are re∣solved to make up a Stock, viz. A and B. The first Partner A doth resolve to have 4/7 of the profit, and layeth in 1368 Pound. It is required to know how much B must pay into Stock to gain the remaining 3/7 of the profit.

By the Rule of Three, say, If 4/7 of the gain which the first man A taketh, did pay into Stock 1368/1, how much must 3/7 of the gain lay into Stock, which is the second man B profit, which by multiplying and dividing you will find to be 1026l. and so much must B lay into Stock to gain 3/7 of the profit, if A gaineth 4/7 by putting in 1368l.

Example 3.

Six Partners, viz. A, B, C, D, E and F build a Ship between them, amounting to the Sum of 2376 Pounds: the several Partners put in as followeth.

	l.	s.
A put in	222	15
B	271	05
C	519	15
D	297	00
E	816	15
F	148	10
	2376

These Partners let their Ship out to Freight for a certain Voyage for 1250 l. It is required to know how much of the said Freight is due to the several Partners. Therefore by the Rule of Three, as 2376l. the whole value of the Ship is to 1250 l. the Freight, so is the several respective shares put in by the said Partners, to their respective shares in 1250l. the Freight, which by multiplying and dividing I find to be as followeth.

Answer.

	l.	s.	d.
A	117	03	9
B	195	06	3
C	273	08	9
D	156	05	0
E	429	13	9
F	78	02	6
	1250

If there happen to be a Loss, the several Partners must bear a proportional share in the Loss, as they did in the Gain, which is to be found out after the same manner. For as the general Stock of the Partners is to the general Loss, so is the particular share in the Stock to the particular share in the Loss.

Example 4.

Three Merchants adventure to Sea 794 Pound. Viz. A hath 148: 17: 6, B hath 198: 10: 0, and C hath 446: 12: 6. These Merchants have lost out of their

general Stock 213 l. 10 s. I require to know each mans share in the loss.

As 794 l. is to 213 l. 10 s. so is 148 l. 17 s. 6 d. to 40 l. 00 s. 7 d. ½

As 794 l. is to 213 l. 10 s. so is 198 l. 10 s. 0 d. to 53 l. 7 s. 6 d.

As 794 l. is to 213 l. 10 s. so is 446 l. 12 s. 6 d. to 120 l. 1 s. 10 d. ½.

Answer.

A hath lost	40	00	7 ½
B	53	07	6
C	120	01	10 ½
	213	10

Here in this Example A hath lost 40 l. 7 d. ½, B hath lost 53: 7: 6, and C hath lost 120: 1: 10: ½; each bearing such a proportional share in the Loss as they have in the general Stock, and their respective shares in the Loss maketh up the just Sum of 213 l. 10 s. the general Loss as before.

CHAP. XX.

The Rule of Fellowship with Time.

THE Rule of Fellowship with Time, is when each Partners share in the Stock hath a Relation to the Time that it hath remained in Stock. In this Rule you must multiply each Partners particular share by the Time it hath continued in Stock; these several Products you may call for distinction sake their new layings in. Then by the Rule of Three say, as the Sum of their new layings in is to the profit made, so is each mans particular new laying in to his particular share of profit.

Example 1.

Two Partners join Stocks together, viz. A laid in the first of April 475 Pounds, and B laid in the first of July 525 Pounds, and by the last of December they had gained 360 Pound; now I desire to know each

mans proportion of the profit answerable to his share in Stock and Time.

A hath for 9 Months	475
B hath for 6 Months	525
	1000

A share multiplied by his Time is	4275
B share multiplied by his Time is	3150
	7425

As	is to	so is	to
7425	360	4275	207	5	5	¼: 6075/7425
7425	360	3150	152	14	6	½: 1350/7425

A share for 475 l. 9 mon. is	207: 5: 5: ¼ 6075/7425
B share for 525 l. 6 mon. is	152: 14: 6: ½ 1••25/7445
Total Sum is	360

Example 2.

Two Partners are resolved to make a Stock together, viz. A putteth in 380 l. the first of May, and B cannot put any thing in till the first of August next. I

demand how much B ought to put into receive ½ of the profit at the years end.

Multiply A share 380 by 8 Months, the time his Mony abideth in Company, the Product thereof is 3040 l. for his new lay∣ing in. Now because B putteth in nothing until the first of August, his Mony will lie in but 5 Months. Divide 3040 by 5, and thereof will come 608 l. and so much must B lay in to receive ½ of the profit, as was required.

Example 3.

Two Partners joined together in a Stock of 750 Pound, by which Stock they gained 240 Pound, which put to the Stock maketh 990 l. A his Mony was in 12 Months, and B his Mony was in 8 Months; when they divided the Mony, A received 671 l. 8 s. 6 d. ¾ and 3/7, and B received 318 l. 11 s. 5 d. of. 4/••. I require to know how much each mans Stock was at first? Then by the Rule of Three, if 12 Months give 750, 8 Months will give 500; there the Stock of A first put in was 500, which subtracted from 750, there remains the Stock of B, as was required.

Example 4.

Three Graziers hire a piece of Ground for 37: 10. A put into the said Ground 60 head of Oxen 15 days, B put into the said piece of Ground 48 head of Oxen 21 days, and C put in 75 head of Oxen 12 days: I desire to know how much each Grazier must pay.

Oxen. days.

A	60	15	900
B	48	21	1008
C	75	12	900
			2808

If	give	what will
	l.	s.		l.	s.	d.
2808	37	10	900	12	0	4	½	1296/2808
2808	37	10	1008	13	9	2	¾	••216/2808
2808	37	10	900	12	0	4	½	1••96/2808

	l.	s.	d.
A is to pay	12	0	4	½	1296/2808
B is to pay	13	9	2	¾	••••16/28088
C is to pay	12	0	4	½	1296/2808
	37	10	0	0

Which three several Sums being added together make 37 l. 10 s. which is the Sum the three Graziers are to pay among them∣selves.

Example 5.

Three Merchants entred in a Trade of Company, as A, B, C, and put in several Sums of Mony for a Stock. A put in the first of January 250 l. and the first of July following he taketh out again 40 l. B. lay∣eth in the first of April 425 l. and the first of September following he puts in 75 l. C laid in the first of June 360 l. and the first of September following he took out again 60 l. at the years end they found they had gained 275 l. I require to know how much each Partner shall have of the said 275 l. gained.

Answer. First multiply the mony that A laid in which was 250 l. by 12 Months, the time it continued in Stock, the Product thereof is 3000 l. from thence subtract 6 times 40 l. which is 240, for the mony which he took out again, and there doth remain 2760, which is the just laying in of A. Then multiply 425 l. the mony B

paid, by 9 Months his time of continuance, the Product thereof is 3825, whereto add 4 times 75 l. which is 300, for the mony he laid in more the first of September, the Sum whereof is 4125, the second man B his layings in. Then for the mony that C paid which was 360 l. multiply by 7 Months his time, the Product thereof is 2520, from whence subtract 4 times 60 l. for the mony C took out the first of September which is 240, and there doth remain 2280 the third man C layings in.

A his new layings in	2760
B	4125
C	2280
	9165

Then according to the Instructions of the first Example of this Chapter, add all the new layings in together, the Sum whereof is 9165, then by the Rule of Three the several Operations are as follow.

As	is to	so is	to
l.	l.	l.	l.	s.	d.	f.
9165	275	2760	82	16	3	2	4170/9165
9165	275	4125	123	15	5	1	5535/9165
9165	275	2280	68	08	2	3	8625/9165

A is to have	82	16	3	2	4170/9165
B	123	15	5	1	5535/9165
C	68	08	2	3	8625/9165
	275

Which shares aforesaid being added all together make up the sum of the whole gain 275.

The Proof of this Rule is not different from that of Fellowship without Time laid down in the last Chapter.

CHAP. XXI.

Of Interest Simple and Compound, Purchases and Valuation of An∣nuities.

INterest is of two Kinds, Simple and Compound.

Simple Interest is when it is required so much per Cent. for so many years.

Compound Interest is as much as to say, Interest upon Interest, that is, you add the

Interest every year to the Principal, and the next year you accompt Interest for their Sum being added both together.

Of Simple Interest.

Let it be required to know, if the In∣terest of 100 l. for 12 Months be 6 l. what shall the Interest of 375 l. be for the same time.

Example 1.

For the effecting this Operation by the Rule of Practice, work thus. It is here requi∣red to know the Interest of 375 l. at 6 per Cent. I therefore multiply 6l. by 3, the Product thereof is 18l. which * 1.61 is the Interest of 300 l. but there is 75l. more to know the Interest of, whichis three quarters of a hundred, that is, of 100, the half whereof is 50 l. the quarter 25 l. which make up 75 l. therefore under 18 l. I set down 3 l. the half of 6 l. and 1 l. 10 s. the quarter of 6 l. and add all these toge∣ther their Sum is 22 l. 10 s. the Interest of 375 l. for 12 Months as was required.

Example 2.

I desire to know if the Interest of 100 l. for 12 Months be 6 l. what shall the Interest of 450 l. be for 9 Months.

This question may be answered by two Operations of the Rule, as followeth.

If	require	what shall	answer
100 l.	6 l.	450 l.	27 l.
12 mon.	27 l.	9 mon.	20: 05

But by the first Example of the Double Rule of Three, you do this and such like questions at one Operation. Thus

Say, if 100 l. for 12 Months gains 6 l. what shall 450 l. gain for 9 Months?

Set down the several numbers given in order as follow with their denominations over head.

The Operation is thus, you have here five numbers given to find out a sixth pro∣portional, wherein 100 l. is the first Term, 12 the second, 6 the third, 450 and 9 the fifth Term.

Therefore multiply the two first Terms, viz. 100 and 12 together, the Product

thereof is 1200, then multiply the three last Terms, viz. 60, 450 and 9 into one 〈 math 〉〈 math 〉 another, and the Product thereof will be 24300, which is the Dividend to be divi∣ded by the Divisor 1200; now because the Divisor and Dividend have each two Cyphers, therefere I cut off the said Cy∣phers, and there remains 12 for the Di∣visor, and 243 for the Dividend, which being divided the Quotient is 20, which is so many Pounds, but there remains 3, which being the quarter of 12 I set down in the Quotient 5 s. So I find the Interest of 450 l. for 9 Months is 20 l. 5 s. at the Rate of 6 l. per 100 l. as before.

But to avoid Fractions in this manner of working, and to find out the true and

exact Interest of any Sum of Mony pro∣pounded, not only in Months but in days, observe the following Directions.

Reduce your Interest, whether it be 6 l. 7 l. or 8 l. into Pence as you see.

〈 math 〉〈 math 〉

I likewise convert the given numbers of Time into Days, and because the Year doth contain 365, and 6 Hours, it being better to allow more than less, I therefore do reckon 366 Days.

Example 3.

Let it be required to know the Interest of 360 l. for 40 days, at 6 per Cent. per. An. set down the given numbers as you see;

then multiply the first and second number together, viz. 100 and 366, the Product thereof is 36600, then multiply the three 〈 math 〉〈 math 〉

last numbers together, and their Product will be 20736000; make the former num∣ber 36600 your Divisor, and 20736000 your Dividend; but because both the Di∣visor and Dividend do contain each two Cyphers, cut off two Cyphers from both Divisor and Dividend, then 366 will be your Divisor, and 207360 your Dividend, and the Quotient of such Division will give 566 Pence 204/366, which by Reduction doth make 47 s. 2 d. 204/366, and so much is the exact Interest of 360 Pounds for 40 days at 6 l. per Cent.

When the given Sum whose Interest you desire to know, doth consist of several De∣nominations, work according to the fol∣lowing directions.

Example 4.

I desire to know the Interest of 1236 l. 16 s. 8 d. for 12 Months or a whole Year.

By the Rule of Three.

If 100 l. give 6 l. what will 1236 l. 16 s. 8 d. give.

Or thus, set down the given number whose Interest you do require 1236: 16: 8, and multiply it by 6, as you see, setting

down the Product of each Denomination multiplied by 6, as you see.

〈 math 〉〈 math 〉

Take away the two figures towards the right hand, or else make a separation be∣tween them, and then it will be 74 l. and 21 hundred parts now to find out the value of this 21 hundred parts, you must multiply it by 20, and by taking two figures from the Product it makes, 4 Shillings 20 parts, multiply this 20 parts by 12, doing as before, you have 2 d. and 40 parts. Lastly multiply this 40 parts by 4, and your Pro∣duct will be 1 Farthing, 60/100 parts of a Farthing. So the exact Interest of 1236: 16: 8 for a Year is 74 l. 4 s. 2 d. 1 f. 60/100.

〈 math 〉〈 math 〉

By this following Table you may with more expedition and ease find out the true and exact interest of any Sum of mony given, wherein take notice that the Penny is divided into 100 parts, of which 25/100 is the value of 1 Farthing, 50/100 the value of of a half penny, and 75/100 the value of three farthings.

A Table of Interest at 6 l. per Cent.

	1 Day.	2 Days.	4 Days.
	s.	d.	C.	s.	d.	C.	s.	d.	C.
1
5			1			2			4
10			2			4			7
15			3			6			11
1			4			8			15
2			8			16			32
3			12			24			48
4			15			31			62
5			19			39			78
6			23			47			94
7			27			55		1	10
8			31			63		1	26
9			35			71		1	42
10			39			78		1	56
20			79		1	58		3	12
30		1	18		2	36		4	68
40		1	58		3	15		6	24
50		1	97		3	94		7	80
60		2	36		4	72		9	36
70		2	76		5	51		10	92
80		3	15		6	31	1	0	48
90		3	55		7	10	1	2	04
100		3	94		7	89	1	3	60
200		7	89	1	3	78	2	7	20
300		11	83	1	11	67	3	10	80
400	1	03	78	2	7	56	5	2	40
500	1	07	72	3	3	45	6	6	00

A Table of Interest at 6 per Cent.

	8 Days.	10 Days.	20 Days.
	s.	d.	C.	l. s.	d.	C.	l. s.	d.	C.
1			1			1			3
5			8			10			19
10			15			19			39
15			23			29			59
1			30			39			78
2			64			78		1	57
3			96		1	17		2	36
4		1	24		1	57		3	15
5		1	56		1	97		3	94
6		1	88		2	36		4	73
7		2	20		2	76		5	52
8		2	52		3	14		6	31
9		2	84		3	55		7	10
10		3	12		3	94		7	89
20		6	24		7	89	1	3	78
30		9	36		11	83	1	11	67
40	1	0	48	1	3	77	2	7	56
50	1	3	60	1	7	72	3	3	45
60	1	6	72	1	11	66	3	11	34
70	1	9	84	2	3	61	4	7	23
80	2	0	96	2	7	55	5	3	12
90	2	4	8	2	11	50	5	11	01
100	2	7	20	3	3	45	6	6	90
200	5	2	40	6	6	90	13	1	80
300	7	9	60	9	10	36	19	8	71
400	10	4	80	13	1	81	26	3	61
500	13	0	00	16	5	26	32	10	52

A Table of Interest at 6 per Cent.

	1 Day.	2 Days.	4 Days.
	s.	d.	C.	s.	d.	C.	s.	d.	C.
600	1	11	67	3	11	34	7	9	60
700	2	3	61	4	7	23	9	1	20
800	2	7	56	5	3	12	10	4	80
900	2	11	50	5	11	1	11	8	40
1000	3	3	45	6	6	90	13	0	00
2000	6	6	90	13	1	80	26	0	00
3000	9	10	36	19	8	70	39	0	00
4000	13	1	80	26	3	60	52	0	00
5000	16	5	25	32	10	50	65	0	00
6000	19	8	71	39	5	40	78	0	00
7000	23	0	16	46	0	30	91	0	00
8000	26	3	61	52	7	20	104	0	00
9000	29	7	06	59	2	10	117	0	00
10000	32	10	51	65	9	00	130	0	00
	1 Month.	2 Months.	3 Months.
1			6			12			18
5			30			60			90
10			60		1	20		1	80
15			90		1	80		2	70
1		1	20		2	40		3	60
2		2	40		4	80		7	20
3		3	60		7	20		10	80
4		4	80		9	60	1	2	40
5		6	00	1	0	00	1	6	00
6		7	20	1	2	40	1	9	60
7		8	40	1	4	80	2	1	20
8		9	60	1	7	20	2	4	80
9		10	80	1	9	60	2	8	40

A Table of Interest at 6 per Cent.

	8 Days.	10 Days.	20 Days.
	s.	d.	C.	l. s.	d.	C.	l. s.	d.	C.
600	15	7	20	19	8	71	39	5	42
700	18	2	40	23	0	16	46	0	32
800	20	9	60	26	3	62	52	7	23
900	23	4	80	29	7	07	56	2	13
1000	26	0	00	32	10	52	65	9	04
2000	52	0	00	65	9	04	131	6	09
3000	78	0	00	98	7	57	197	3	13
4000	104	0	00	131	6	09	263	0	17
5000	130	0	00	164	4	61	328	9	22
6000	156	0	00	197	3	13	394	6	26
7000	182	0	00	230	1	65	460	3	31
8000	208	0	00	263	0	17	526	0	35
9000	234	0	00	295	10	69	591	9	39
10000	260	0	00	328	9	21	657	6	43
	6 Months.	9 Months.	12 Months.
	s.	d.	C.	s.	d.	C.	s.	d.	C.
1			36			54			72
5		1	80		2	70		3	60
10		3	60		5	40		7	20
15		5	40		8	10		10	80
1		7	20		10	80	1	2	40
2	1	2	40	1	9	60	2	4	80
3	1	9	60	2	8	40	3	7	20
4	2	4	80	3	7	20	4	9	60
5	3	0	00	4	6	00	6	0	00
6	3	7	20	5	4	80	7	2	40
7	4	2	40	6	3	60	8	4	80
8	4	9	60	7	2	40	9	7	20
9	5	4	80	8	1	20	10	9	60

A Table of Interest at 6 per Cent.

	1 Month.	2 Month.	3 Months.
	l.	s.	d.	l.	s.	d.	l.	s.	d.
10		1	0		2	0		3	0
20		2	0		4	0		6	0
30		3	0		6	0		9	0
40		4	0		8	0		12	0
50		5	0		10	0		15	0
60		6	0		12	0		18	0
70		7	0		14	0	1	1	0
80		8	0		16	0	1	4	0
90		9	0		18	0	1	7	0
100	0	10	0	1	0	0	1	10	0
200	1	0	0	2	0	0	3	0	0
300	1	10	0	3	0	0	4	10	0
400	2	0	0	4	0	0	6	0	0
500	2	10	0	5	0	0	7	10	0
600	3	0	0	6	0	0	9	0	0
700	3	10	0	7	0	0	10	10	0
800	4	0	0	8	0	0	12	0	0
900	4	10	0	9	0	0	13	10	0
1000	5	0	0	10	0	0	15	0	0
2000	10	0	0	20	0	0	30	0	0
3000	15	0	0	30	0	0	45	0	0
4000	20	0	0	40	0	0	60	0	0
5000	25	0	0	50	0	0	75	0	0
6000	30	0	0	60	0	0	90	0	0
7000	35	0	0	70	0	0	105	0	0
8000	40	0	0	80	0	0	120	0	0
9000	45	0	0	90	0	0	135	0	0
10000	50	0	0	100	0	0	150	0	0

A Table of Interest at 6 per Cent.

	6 Months.	9 Months.	12 Months.
	l.	s.	d.	l.	s.	d.	l.	s.	d.
10		6	0		9	0		12	0
20		12	0		18	0	1	4	0
30		18	0	1	7	0	1	16	0
40	1	4	0	1	16	0	2	8	0
50	1	10	0	2	5	0	3	0	0
60	1	16	0	2	14	0	3	12	0
70	2	2	0	3	3	0	4	4	0
80	2	8	0	3	12	0	4	16	0
90	2	14	0	4	1	0	5	08	0
100	3	0	0	4	10	0	6	0	0
200	6	0	0	9	0	0	12	0	0
300	9	0	0	13	10	0	18	0	0
400	12	0	0	18	0	0	24	0	0
500	15	0	0	22	10	0	30	0	0
600	18	0	0	27	0	0	36	0	0
700	21	0	0	31	10	0	42	0	0
800	24	0	0	36	0	0	48	0	0
900	27	0	0	40	10	0	54	0	0
1000	30	0	0	45	0	0	60	0	0
2000	60	0	0	90	0	0	120	0	0
3000	90	0	0	135	0	0	180	0	0
4000	120	0	0	180	0	0	240	0	0
5000	150	0	0	225	0	0	300	0	0
6000	180	0	0	270	0	0	360	0	0
7000	210	0	0	315	0	0	420	0	0
8000	240	0	0	360	0	0	480	0	0
9000	270	0	0	405	0	0	540	0	0
10000	300	0	0	450	0	0	600	0	0

The Ʋse of the former Table.

Let it be required to find the Interest of 975 l. for 12 Months or a Year. First, I find the Interest of 900 l. for 12 Months to be 54 l. which I set down, then I find the Interest of 70 l. for the same time to be 4 l. 4 s. which I set under the 54 l. next I find the Interest of 5 l. for the like time to be 6 s. which I likewise set down, as you see.

	l.		l.	s.
The In∣terest of	900	for 12 mon. is	54	00	00
70	04	04	00
5	00	06	00
			58	10	00

Add these several Sums together, and their total Sum will be 58 l. 10 s. which is the exact Interest of 975 l. for 12 months at 6 l. par Cent. as was required.

Of Compound Interest.

Compound Interest is when Interest upon Interest is reckoned on any Sum of Mony.

Example.

Let it be required to know the Interest of 200 l. for 3 years, reckoning Interest upon Interest, to effect which by the Rule of Three, observe these following pro∣portions.

As	is to	so is	to
100	006	200	012	1. year.
100	006	212	024	2. year.
100	006	224: 14: 4: 3: 03	8: 47: 0: 2	3. ye.

As by the following Example doth appear.

In this Operation I say at first, if 100 l. give 106 l. Interest and Principal for one year, what shall 200 l? and by Multipli∣cation I reduce 200 l. into the lowest De∣nomination, which gives me 192000 far∣things, I multiply these farthings by my se∣cond Term 106, the Product is 20352000, and because I am to divide by 100, I cut off the two Cyphers towards the right hand, which is all one and the same, and there doth remain 203520 farthings, which reduced

〈 math 〉〈 math 〉

makes 212 l. for the Principal and Interest of 200l. the first year, the Proportion now lies. If 100 l. give 106 l. what will 212 l. give? and by the foregoing Operation I find 212 l. to contain 203520 Farthings, which I again multiply by 106, the Pro∣duct thereof is 21573120, cutting off two figures as before, there remains 215731 Farthings, which reduced make 224 l. 14s. 4 d. 3 f. the Principal and Interest upon Interest of 200 Pound for two years. Lastly, I multiply 215731 by 106, the the Product is 228674 (86, cutting off two figures as before there remains 228674 Farthings, which reduced gives 238 l. 4 s. ½ for the Principal and Interest upon In∣terest of 200l. for three years. And in like manner must you work if time be longer, so that if you multiply 228674 Farthings by 106, and from the Product thereof cut off two figures towards your right hand, the Remainer gives you the number of far∣things contained in the Principle and In∣terest upon Interest of 200 l. for 4 years.

But because questions of Interest upon Interest require an Operation for each year, I have therefore inserted the following Table.

A Table shewing the Principal and Interest upon Interest for one Pound for any Term under 31 Years.

Years.	l.	s.	d.	f.
1	1	1	2	1
2	1	2	5	2
3	1	3	9	3
4	1	5	3	0
5	1	6	9	0
6	1	8	4	2
7	1	10	0	3
8	1	11	10	2
9	1	13	9	2
10	1	15	9	3
11	1	17	11	2
12	2	0	3	0
13	2	2	7	3
14	2	5	2	2
15	2	7	11	1
16	2	10	9	2
17	2	13	10	1
18	2	17	1	0
19	3	0	6	0
20	3	4	1	3
21	3	7	11	3
22	3	12	0	2
23	3	16	4	1
24	4	0	11	1
25	4	5	9	1
26	4	10	11	0
27	4	16	4	1
28	5	2	1	2
29	5	8	3	0
30	5	14	8	3
31	6	1	7	1

The first Column of this Table is the Column of years from 1 to 31, and over against each number of years doth stand the Sum that 20 Shillings or one Pound doth come to at Interest upon Interest for the same time.

Suppose it were required to know by this Table the increase of 50 l. for 12 years at Interest upon Interest. Look in the Table over against 12 years and you will find 2 l. 0 s. 3 d. which is the increase of 1 Pound for 12 years, which multiplied by 50, as you see, viz.

	l.	s.	d.
	2	0	3
			50
50 Times 3 d. is		12	6
50 Times 2 l. is	100
Total is	100	12	6

So that 100 l. 12 s. 6 d. is the increase of 50 l. for 12 years, allowing Interest upon Interest.

Of Purchases and Annuties.

If you would purchase any Sum of Mony to be paid you any number of years to come, you must first know what it is worth in ready mony, allowing Interest upon Innerest. To effect this say, As 106l. is to 100 l. so 100 l. to the present worth of 100 l. to be paid at 12 Months or a year. Then for the second year say, as 106 l. is to 100 l. so is the present worth of one year to the present worth for two years, and in like manner is the present worth of any Sum of Mony given found out for any number of years to come, as by the follow∣ing Example doth more plainly appear.

Example.

Let it be required to know how much 50 l. to be paid at two years end is worth in ready mony, by the Rule of Three according to the preceding Instructions say,

〈 math 〉〈 math 〉

Here you see 50 l. to be paid at one years end is worth in ready money 47 l. 169 parts, or 3 s. 4 d. Then for the second year make this last Sum found out 47: 169, your third number, saying,

〈 math 〉〈 math 〉

By this last Operation the fourth num∣ber found out is 44: 499, by which you see that 50 l. to be paid at the end and expiration of two years is worth no more than 44 Pound 10 Shillings in ready mony, and in like manner if you would know the present worth of 50 l. at the end and

ration of three years, make 44: 499 your third number; saying,

As 106 l. is to 100 l. so is 44: 499 to 41: 980.

But to avoid this trouble I have inserted the following Table shewing the present worth of one Pound in ready mony, to be paid any number of years under 31, and if you know the present worth of one Pound, by Multiplication you may know the present worth of any other Sum of Mony.

Annexed to this Table is another Table, shewing the true value of 1 l. Annuity to continue 31 Years in ready mony, whereby you may easily know the true value of any Purchase to continue the same Term of years.

A necessary Table shewing the present worth of 1 l. to be paid any number of years under 31, at 6 l. per Cent.

Years.	l.	s.	d.
1	18	10	2	94339
2	17	9	2	88999
3	16	9	2	83962
4	15	10	0	79209
5	14	11	1	74726
6	14	1	1	70496
7	13	3	2	66506
8	12	6	2	62741
9	11	10	0	59190
10	11	2	0	55839
11	10	6	2	52678
12	9	11	1	49697
13	9	4	2	46884
14	8	10	0	44230
15	8	4	1	41726
16	7	10	2	39365
17	7	5	0	37136
18	7	0	0	35034
19	6	7	1	33051
20	6	2	3	31181
21	5	10	2	29415
22	5	6	2	27750
23	5	2	3	26180
24	4	11	1	24698
25	4	7	2	23300
26	4	4	3	21981
27	4	1	3	20737
28	3	11	0	19563
29	3	9	1	18453
30	3	5	3	17411

A Table shewing the true value of 1 l. Annuity to continue any number of years under 31, in ready Mony at 6 l. per Cent.

Years.	l.	s.	d.	q.	Decimals.
1		18	10	2	0	94340
2	1	16	08	0	1	83339
3	2	13	05	2	2	67301
4	3	09	03	2	3	46510
5	4	04	03	0	4	21236
6	4	18	04	1	4	91732
7	5	11	07	3	5	58238
8	6	04	02	1	6	20979
9	6	16	00	1	6	80169
10	7	07	02	1	7	36008
11	7	17	08	3	7	88687
12	8	07	08	0	8	38384
13	8	17	00	2	8	85268
14	9	05	10	3	9	29498
15	9	14	03	0	9	71224
16	10	02	01	2	10	10589
17	10	09	06	2	10	47725
18	10	16	06	2	10	82760
19	11	03	02	0	11	15811
20	11	09	04	3	11	46992
21	11	15	03	1	11	76405
22	12	00	10	0	12	04158
23	12	06	00	3	12	30337
24	12	11	00	0	12	55035
25	12	15	07	2	12	78335
26	13	00	00	2	13	00316
27	13	04	02	2	13	21053
28	13	08	01	2	13	40616
29	13	11	10	3	13	59071
30	13	15	04	2	13	76482

The Ʋse of the foregoing Tables.

It is required to know the present value in ready Mony of 36 l. Annuity or Lease that is to continue 21 years.

By the former Table you find 1 l. An∣nuity or Lease to continue 21 years is worth 11 l. 15 s. 3 d. 1 f. therefore the answer to your question must be 36 times 11 l. 15 s. 3 d. 1 f.

	l.	s.	d.	f.
Therefore	11	15	3	1
Multiplied by				36
36 times 1 Farthing is			9	0
36 times 3 Pence is		9	0	0
36 times 15 Shillings is	27	0	0	0
36 times 11 Pounds is	396	0	0	0
Total Sum is	423	9	9	0

By Decimals thus, look for the Decimal value of one Pound Annuity or Lease for 21 Years, which you will by the same Table find to be 11: 76405, which mul∣tiply by 36 Pound, the Product will be

423 Pound, 50580, which in value is ten Shillings, being three Pence more than the other.

〈 math 〉〈 math 〉

Therefore the Purchase of 36 Pound for 21 years is worth 423 l. 9 s. 9 d. as was required.

CHAP. XXII.

Of Rebate or Discount.

REbate or Discount is when mony is due at a certain time, as at 2, 3, 6 or 12 Months, and the person to whom the mony is payable may sometimes have such occasion for his mony as to desire it before it be due; the person that is to pay the mony hath no reason to pay his mony before it be due, without some considera∣tion

of Rebate or Discount for so much time at 6, 7 or 8 per Cent. as they do agree.

Observe this as a general Rule, that upon Rebatements there must be no more mony paid down than would augment it self unto the whole Sum due, if it were put forth to Interest for so much time as it is paid or received before hand.

Example 1.

Let it be required to know the Rebate of 250 l. for 12 Months at 6 per Cent. Then by the Rule of Three say,

As 106 is to 100, so is 250 l. 〈 math 〉〈 math 〉

	l.	s.	d.
Mony to be laid down	235	16	11	¾
Mony rebated or discounted	14	03	00	¼
Total	250

By which I gather that the Interest of 235: 16: 11: ¾ for 12 Months at 6 l. per Cent. is 14: 3: 0: ¼, which makes up the whole debt 250 Pound, as before.

Example 2.

I demand how much I must rebate or discount for 165 Pounds for 9 Months at 8 Pound per Cent. simple or bare In∣terest.

Say by the Rule of Three. 〈 math 〉〈 math 〉

	l.	s.	d.
Total Sum due is	165	00	00
To be paid	155	13	02
At 8 l. per Cent. for 9 months is rebated	9	6	10

Example 3.

I demand the Discount or Rebate of 548 l. for 6 Months at 7 per Cent. simple Interest.

Say by the Rule of Three. 〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

Total Sum due is	548
Paid down	529	09	04
Rebated for 6 Months at 6 l. per Cent.	18	10	08

Quest. Let it be required to know the Rebate of 500 l. for 9 mon. at 6 l. per Cent.

Example. To resolve this Question, you must find out the Interest of 500 l. for 9 Months, the Interest of 100 l. for 9 Months or three quarters of a year is 4 l. 10 s. therefore the Interest of 500 l. for the same time must of necessity be 22 l. 10 s. now I say if 522 l. 10 s. give 500 l. what will 500 l. give.

〈 math 〉〈 math 〉

Here you find the answer to the que∣stion last stated, by the Quotient which doth give you 114832 Pence, which is reduced as followeth.

〈 math 〉〈 math 〉

At last we have found that if 522 l. 10 s. give 500 l. then 500 l. will give 478 l. 9 s. 4 d. which subtracted from 500 l. giveth 21 l. 10 s. 8 d. for the Rebate of 500 l. for 9 Months at 6 l. per Cent.

This following Table will be of very great use for the ready casting up of the Rebate of any Sum of Mony for any time propounded.

A Table of Rebate at 6 per Cent.

	1 Month.	2 Month.	3 Months.
	l.	s.	d.	l.	s.	d.	l.	s.	d.
1			1			2			4
2			2			5			7
3			3			7			11
4			5			10		1	2
5			6		1	0		1	6
6			7		1	2		1	9
7			8		1	5		2	2
8			9		1	7		2	4
9			11		1	9		2	8
10		1	0		2	0		2	11
20		2	0		3	11		5	11
30		3	0		5	11		8	10
40		4	0		7	11		11	10
50		5	0		9	11		14	9
60		6	0		11	11		17	9
70		7	0		13	10	1	0	8
80		8	0		15	10	1	3	8
90		8	11		17	10	1	6	7
100		9	11		19	10	1	9	7
200		19	10	1	19	7	2	19	1
300	1	9	11	2	19	5	4	8	8
400	1	19	12	3	19	2	5	18	3
500	2	9	9	4	19	0	7	7	9
600	2	19	8	5	18	10	8	17	4
700	3	9	8	6	18	7	10	6	11
800	3	19	7	7	18	5	11	16	5
900	4	9	7	8	18	2	13	6	0
1000	4	19	6	9	18	0	14	15	7

A Table of Rebate at 6 l. per Cent.

	4 Months.	5 Months.	6 Months.
	l.	s.	d.	l.	s.	d.	l.	s.	d.
1			5			6			7
2			9		1	0		1	2
3		1	2		1	6		1	9
4		1	7		1	11		2	4
5		1	11		2	5		2	11
6		2	4		2	11		3	6
7		2	9		3	5		4	1
8		3	2		3	11		4	8
9		3	6		4	5		5	3
10		3	11		4	11		5	10
20		7	10		9	9		11	8
30		11	9		14	8		17	6
40		15	8		19	6	1	3	4
50		19	7	1	4	5	1	9	2
60	1	3	6	1	9	3	1	14	11
70	1	7	5	1	14	0	2	0	9
80	1	11	4	1	19	0	2	6	7
90	1	15	4	2	3	11	2	12	5
100	1	19	3	2	8	9	2	18	3
200	3	18	4	4	17	7	5	16	6
300	5	17	8	7	6	4	8	14	9
400	7	16	11	9	15	1	11	13	0
500	9	16	1	12	3	11	14	11	3
600	11	15	3	14	12	8	17	9	6
700	13	14	6	17	1	6	20	7	9
800	15	13	9	19	10	3	23	6	0
900	17	12	11	21	19	0	26	4	3
1000	19	12	2	24	7	10	29	2	6

A Table of Rebate at 6 per Cent.

	7 Months.	8 Months.	9 Months.
	l.	s.	d.	l.	s.	d.	l.	s.	d.
1			8			9			10
2		1	4		1	6		1	8
3		2	0		2	4		2	7
4		2	8		3	1		3	5
5		3	4		3	10		4	4
6		4	1		4	7		5	2
7		4	9		5	5		6	0
8		5	5		6	1		6	11
9		6	1		6	11		7	9
10		6	9		7	8		8	7
20		13	6		15	5		17	3
30	1	0	3	1	3	1	1	5	10
40	1	7	1	1	10	9	1	14	5
50	1	13	10	1	18	6	2	3	1
60	2	0	7	2	6	2	2	11	8
70	2	7	4	2	13	10	3	0	3
80	2	14	1	3	1	6	3	8	11
90	3	0	10	3	9	3	3	17	6
100	3	7	8	3	16	11	4	6	1
200	6	15	3	7	13	10	8	12	3
300	10	2	11	11	10	9	12	18	4
400	13	10	6	15	7	8	17	4	6
500	16	18	2	19	4	7	21	10	7
600	20	5	10	23	1	6	25	16	9
700	23	13	6	26	18	6	30	2	10
800	27	7	1	30	15	5	34	9	0
900	30	8	8	34	12	4	38	15	1
1000	33	16	4	38	9	3	43	1	2

A Table of Rebate at 6 l. per Cent.

	10 Months.	11 Months.	12 Months.
	l.	s.	d.	l.	s.	d.	l.	s.	d.
1			11		1	0		1	2
2		1	11		2	1		2	3
3		2	10		3	2		3	5
4		3	9		4	2		4	6
5		4	9		5	3		5	8
6		5	8		6	3		6	9
7		6	8		7	4		7	11
8		7	7		8	4		9	1
9		8	7		9	5		10	2
10		9	6		10	5		11	4
20		19	1	1	0	10	1	2	8
30	1	8	7	1	11	3	1	14	0
40	1	18	1	2	1	8	2	5	3
50	2	7	7	2	12	2	2	16	7
60	2	17	2	3	2	7	3	7	11
70	3	6	8	3	13	0	3	19	3
80	3	16	2	4	3	5	4	10	6
90	4	5	9	4	13	10	5	1	10
100	4	15	3	5	4	3	5	13	2
200	9	10	6	10	8	6	11	6	5
300	14	5	9	15	12	10	16	19	7
400	19	0	11	20	17	1	22	12	10
500	23	16	2	26	1	4	28	6	0
600	28	11	5	31	5	7	33	19	3
700	33	6	8	36	9	10	39	12	5
800	38	1	11	41	14	1	45	5	8
900	42	17	2	46	18	5	50	18	10
1000	47	12	5	52	2	8	56	12	1

The Explanation of the foregoing Tables.

Let it be required to find the Rebate of 348 l. for five Months at 6 l. per Cent.

In the foregoing Table I look in the Column of five Months and in the same

l.	l.	s.	d.
300	7	06	04
40	0	19	06
8	0	3	11
348	8	9	9

Column over against 300 l. I find 7: 6: 4, and against 40 l. I find 19: 6, and against 8l. I find 3s. 11d. which Sums I set down one under another, as in the Example; and by Addition I find the Rebate of 348 l. to be 8 l. 9 s. 9 d. And in like manner you must do for any greater or lesser Sum.

CHAP. XXIII.

Equation of Payments.

THIS Rule teacheth how to bring the several Times appointed for the Paiment of a Sum of Mony due unto one Time for payment of the whole Sum. As admit, a Merchant doth owe 900 l. to be paid at three several payments, viz. 300 l. at three Months, 300 l. more at six Months, and 300 l, more at nine Months. But it is concluded by the Debtor and Cre∣ditor to pay the Mony all together at one Time. It is required to know the Time the whole Sum ought to be paid in, that neither Debtor nor Creditor may be losers.

The Rule.

Multiply each Payment by his own Time, add the several Products together, and divide the Sum thereof by the whole Debt, the Quotient sheweth you the exact Time for payment of the whole Sum.

300 multiplied by 3 Months is	900
300 multiplied by 6 Months is	1800
300 multiplied by 9 Months is	2700
	5400

The several Products added together make 5400, which divided by 900 l. the whole Debt, the Quotient gives 6 Months for the exact Payment of the whole Sum at one Payment.

The Proof.

First find out what the Interest of the Mony doth come to, that should have been paid before the time found out for payment of the whole Sum. Likewise find out what the Interest of the Mony comes to, that should have been paid after the said time found out for payment of the whole Sum; and if the Interest of the Mony paid before be equal to the Interest of the Mony paid after, then the time found out is the exact time for paiment of the whole Sum.

Example.

In the foregoing Example where 300 l. is to be paid at 3 Months, 300 l. more at 6 Months, and 300 l. more at 9 Months, it is found out by the said Operation that 6 Months is the exact time for payment of the whole Sum of 900 l. To prove this, there was 300 l. to be paid at 3 Months, but is not paid unto 6 Months, which is 3 Months longer, and the Interest of 300l. for 3 Months at 6 l. per Cent. is 4 l. 10 s.

The 300 l. paid at 6 Months is the time equated or found out.

The other 300 l. to be paid at 9 Months is to be now paid at 6 Months, which is 3 Months before his time of Payment, the Interest whereof for 3 Months is 4 l. 10 s. as before, therefore 6 Months is the exact time for payment of the whole Sum of 900 l.

A Merchant oweth 750 l. to be paid at 4 several payments; 200 l. at 4 Months; 250 l. more at 8 Months, 150 l. more to be paid at 10 Months, and 150 l. more at 12 Months. I require to know what time the whole 750 l. ought to be paid in,

that neither Debtor nor Creditor may be losers?

l.	l.
200 multiplied by 4 Months is	800
250 multiplied by 8 Months is	2000
150 multiplied by 10 Months is	1500
150 multiplied by 12 Months is	1800
	6100

〈 math 〉〈 math 〉

The several Sums of payment being multiplied by their respective time, the Sum of their Products is 6100, which divided by 750, the Quotient gives eight Months and 2/15 for the time of payment altogether.

There is owing to a Merchant 480 l. to be paid 120 l. ready mony, 120 l. at four Months, 120 l. more at eight Months, and 120 l. more at 12 Months. I demand the exact time for payment of the whole Sum 480 l. so that neither Debtor nor

Creditor may have any advantage nor be damnified.

l.	l.
120 multiplied by 0 Months is	000
120 multiplied by 4 Months is	480
120 multiplied by 8 Months is	960
120 multiplied by 12 Months is	1440
	2880

〈 math 〉〈 math 〉

The several Products added together make 2880, which divided by the whole Sum 480 l. the Quotient giveth 6 Months, as was required.

CHAP. XXIV.

Of Tare, Tret and Cloff.

TAre, Tret and Cloff are allowances of Merchandize sold by weight.

Tare is the allowance for the weight of the Cask, Chest, Bag, &c. wherein any Commodity is put.

Tret is an allowance of 4 pound weight, upon every subtle 100 pound weight; that is to say, to allow 104 pound for every 100 pound.

Cloff is an allowance of 2 pound upon every draught that doth exceed 336 pound, or 300 gross weight.

Quest. 1. If one subtle Hundred cost 3 pound, what shall 2704 pound cost?

By the Rule of Three say, If 104 l. cost 13 l. what shall 2704 l. cost?

〈 math 〉〈 math 〉

By the Operation I find 2704 pounds weight to be worth 78 pound Sterl.

Quest. 2. If 100 l. be worth 13 s. 4 d. what will 2496 l. be worth, allowing 4 l. for every 100 for Tret?

Divide the given number 2496 by 26, because the 4 l. Tret is the 26. part of 104, and the Quotient will be 96, which is to be abated for the Tret out of 2496, and there will remain 2400 l. nett at 13 s. 4 d. per Hundred. Then by the Rule of Three say, If 100 cost 13: 4, what will 2400 l. cost? the answer will be 1600 l. the an∣swer to the question, as was required.

CHAP. XXV.

The Rule of Barter.

BArter teacheth how to sell Wares for Wares, or change one Merchandize for another.

As for Example.

There are two Merchants, viz. A and B, A hath 24 C. weight of Tobacco at 4 d. ½ per pound, and B hath Sugar at 7 d. per pound. I demand how much Sugar must B deliver to A for the Tobacco?

By the Rule of Three say, If 1 l. of Tobacco cost 4 d. ½, what will 24 C. 〈 math 〉〈 math 〉

〈 math 〉〈 math 〉

By the foregoing Operation you find that 24 C. will cost 50 l. 8 s. Then say again,

If 7d. buy 1 pound of Sugar, how many pound will 50l. 8s.

By multiplying and dividing you will find that 50 l. 8 s. will buy 1728 pound of Sugar at 7d. per pound, therefore B must deliver 1728 pound of Sugar to A at 7 d. per pound for 24 C. of Tobacco at 4 d. ½ as was required.

〈 math 〉〈 math 〉

Two Merchants barter one with the other: A hath 128 Yards of Broad Cloth at 9 s. 6 d. and B hath Holland Cloth at 3 s. 4 d. per Ell. I demand how much

Holland Cloth B must deliver to A for his 128 Yards of Broad Cloth?

By the Rule of Three say, If 1 Yard of Broad Cloth cost 9: 6 what cost 128 Yards?

〈 math 〉〈 math 〉

If 3 s. 4 d. buy one Ell of Holland Cloth, what will 60 l. 16 s. 〈 math 〉〈 math 〉

Two Merchants barter, A hath 432 Yards of Canvas at 14d. per Yard, B giveth A in barter for those 432 Yards of Canvas at 14 d. 24 pieces of Cloth. I desire to know how much A giveth for one piece of Cloth.

By the Rule of Three. If 1 yard cost 14 d. what will 432 yards? 〈 math 〉〈 math 〉

If 24 pieces cost 25 l. 4 s. what cost 1? 〈 math 〉〈 math 〉

Here I find by the Operation A giveth 21 s. a piece for the 24 pieces of Cloth, as was required.

Two Merchants, viz. A and B do barter one with another, A hath Nutmegs at 5 s. per pound to sell for ready money, but in barter he will sell for 6 s. per pound, and B hath Cinnamon at 4 s. 6 d. per pound, to sell for ready mony. I desire to know how much B must sell his Cinnamon a pound for in Barter that he may not be a loser.

By the Rule of Three. If 5 s. give 6 s. what will 4 s. 6 d? An∣swer 5 s. 4 d. ⅔.

Two Merchants, viz. A and B do barter one with another. A hath Currans at 35 s. per Hundred, ready money, which he barters to B at 42 s. per Hundred: B hath Raisins of the Sun, which cost him 28 Shillings per Hundred, but he barters them at 33 Shillings 3 Pence per Hundred, I desire to know how much in 100 l. each man gaineth?

By the Rule of Three If 35 s. gain 7 s. what will 100 l. gain? Answer 20 l.

If 28 s. gain 33 s. 3 d. what will 100 l. gain? Answer 18 l. 15 s.

By multiplying and dividing you will find,

A gaineth 20 l. per Cent.

B gaineth 18 l. 15 s. per Cent.

CHAP. XXVI.

Of Loss and Gain.

THis Rule sheweth how to find out the Loss or Gains upon any parcel of Goods sold, and it likewise sheweth how to find what price you must sell your Goods for to gain so much per Cent.

Quest. 1. Suppose I buy broad Cloth at 7 s. 6 d. per Yard, and I sell the same Cloth at 8 s. 9 d. per Yard, I desire to know how much is gained in 100 l.

First see whether there be Gain or Loss in one Yard, and how much it is.

	s.	d.
Sold for per Yard	8	9
Given per Yard	7	6
Gained per Yard	1	3

By the Rule of Three say, If 7 s. 6 d. gain 1 s. 3 d. what will 100 l. gain? Answer 16l. 13s. 4d. Multiply and divide, and you will find 16 l. 13 s. 4 d. is gained per 100l. the answer to the question, as was required.

Quest. 2. If one Yard of Broad Cloth be worth 9s. 6d. for how much must 36 Yards be sold for to gain 20 per Cent.

By the Rule of Three say, If 100 l. give 120 l. what will 9 s. 6 d. give? Answer 11 s. 4d. 8/10. Multiply and divide, and you will find 11s. 4d. 8/10. Then by the same Rule again say,

If 1 Yard give 11: 4: 8/10, what will 36 Yards? 20: 10: 4: 8/8. Multiply and divide,

and you will find that you must sell 36 Yards of that Broad Cloth for 20: 10: 4: 8/10 to gain 20l. per Cent.

Quest. 3. If a piece of Broad Cloth con∣taining 34 Yards cost 18 l. 14 s. and I sell it for 10 s. 8 d. per Yard, I demand whe∣ther I gain or lose, and how much per Cent.

By the Rule of Three say, If 1 Yard give 10 s. 8 d. what will 34 Yards? Answ. 18 l. 2 s. 8 d. which is less than the price it cost by 11 s. 4 d. therefore you have lost, now to find how much you you have lost per Cent. say,

If 18 l. 14 s. lose 11 s. 4 d. what will 100 l. lose? 3 l. 0 s. 7 d. ¼.

By multiplying and dividing you will find you lose 3 l. 0s. 7d. ¼ in the 100 l. the answer to the question, as was required.

Quest. 4. If one Yard of Broad Cloth cost me 13 s. 4 d. and I sell 24 Yards of the same Cloth for 14l. 17 s. 6d. I desire to know whether I gain or lose, and how much in the 100 l. of mony.

By the Rule of Three say If 1 Yard cost 13 s. 4 d. what will 24 Yards? Answer 16 l. 0 s. 0 d.

By multiplying and dividing I find 24 Yards at 13 s. 4d. will cost 16 l. and I sold the 24 Yards for 14 l. 17 s. 6d. So by Sub∣traction I find I have lost upon the 24 Yards 1 l. 2 s. 6 d. Now to know how much per Cent. say,

If 16 l. lose 1 l. 2. s. 6 d. what will 100l? Answer 7 l. 0 s. 7 d. ½.

By multiplying and dividing you will find you have lost 7l. 0s. 7d. ½ per Cent. the answer to the question as was required.

Quest. 5. If I give for 32 Yards of Cloth after the rate of 7 s. 6 d. and for 18 Yards 12 s. 6 d. and I sell one Yard with another for 10 s. per Yard. I desire to know whe∣ther I gain or lose, and how much per Cent.

By the Rule of Three say,

If 1 Yard cost 7 s. 6 d. what cost 32 Yards? Answer	12	0	0
If 1 Yard cost 12: 6, what cost 18 Yards? Answer	11	5	0
50	23	5	0

So I find the two pieces contain 50 Yards and cost me 23 l. 5 s. 0 d. but I sell them round for 10 s. per Yard, and by the Rule of Three say,

If 1 Yard sell for 10 s. what will 50 Yards sell for? Answer 25 l.

By multiplying and dividing you will find it to amount unto 25 l. so that I have gained 1 l. 15 s. Now to know how much per Cent. I have gained,

By the Rule of Three say, If 23 l. 5 s. gain 1 l. 15 s. what will 100 l. gain? Answer 7 l. 10 s. 6 d. ¼.

Multiply and divide, and you will find you have gained 7: 10: 6: ¼ in 100 l. which is the answer to the question, as was re∣quired.

〈1 page duplicate〉〈1 page duplicate〉

〈1 page duplicate〉〈1 page duplicate〉

Quest. 6. If I buy 100 of Oranges at 2 a Penny, and another at 3 a penny, and sell them 5 for two pence, I require to know how much is lost or gained?

By the Rule of Three say,

If 2	Or. cost 1 d. what cost 100,	Ans. 4:	2
If 3	Or. cost 1 d. what cost 100,	Ans. 2:	9⅓
5	200	6:	11⅓

If 5 Oranges cost 2 d. what cost 200? Answer 80 d. or 6 s. 8 d.

So I find by multiplying and dividing the several Operations, that there is 3 d. ⅓ of a penny lost, being the answer to the question, as was required.

CHAP. XXVII.

Allegation Medial.

TEacheth how to find out a Common or Medium Rate or Price in a mix∣ture of several Rates and Prizes together, which must be performed by reducing the several prices into one Denomination.

Multiply the particulars of the several mixtures by his respective price, and di∣vide their Sum by the whole quantity, the Quotient is the Medium or Common Price required.

Quest. 1. A Cooper hath Wine of several prices. Viz. 36 Gallons at 2 s. 40 Gallons at 2 s. 6 d. 60 Gallons at 3 s. and 120 Gal∣lons at 3 s. 4 d. He intends to put all these Wines together, but would know the com∣mon price.

In the following Work you have 36 Gal∣lons multiplied by 24 pence, which is 2 s. the Product is 864 Pence. 40 Gallons by 30 Pence, or 2 s. 6 d. the Product thereof

is 1200 Pence. 60 Gallons by 36 Pence or 3 s. the Product thereof is 2160 Pence. And 120 Gallons by 40 pence or 3 s. 4 d. the Product thereof is 4800; the Sum of these several Products is 9024, which if you divide by 250 Gallons the Sum of the seve∣ral quantities, the Quotient will be 36 d. 24/250 or 3 s. 0 d. 24/250, which is the Medium or common Price, as was required.

〈 math 〉〈 math 〉

Quest. 2. A Mealman hath 230 Bushels of Meal at 9 d. 368 Bushels at 14 d. and

485 Bushels at 16 d. He is intended to mix all these together, and desires to know the common price.

〈 math 〉〈 math 〉

The common price here found out is 13 d. ¾, as was required, and in like man∣ner for all other questions of this nature.

CHAP. XXVIII.

Alligation Alternate.

ALligation Alternate teacheth how to mix divers sorts and prices together, in such manner that you may make them of what price or quantity you please.

Quest. 1. A Goldsmith hath four sorts of Gold, the first sort is worth 36 Crowns the pound weight, the second sort is worth 38 Crowns the pound, the third sort is worth 42 Crowns the pound, and the fourth sort is worth 45 Crowns the pound: and with these it is required to make a mixture, worth 41 Crowns the pound. I desire to know how much of each sort he must take.

First, set down the numbers of which the Alligation is to be made, one under the other, viz. 36, 38, 42 and 45, in such order as you see in the following Example, and the common price unto which you would reduce them, to the left hand, which

in this Example is 41, then observe which of the four numbers before set down are greater than the common number 41, and which are less, and with a Semicircle link a greater and a lesser number together, for two greater or two smaller numbers than the common number may not be coupled together. 〈 math 〉〈 math 〉

After these numbers are according to the former directions linked or coupled together, next observe how much each of the lesser numbers differeth from the com∣mon number 41, and set that difference against his respective number to which he is coupled; as here 36 differeth 5 from the common number 41, which I do not set down against 36 but against 45, the num∣ber to which he is coupled; so likewise 38 doth differ 3 from the common number 41, which I do not set against 38 but against 42 the number to which he is coupled: likewise you must observe how much each

of the greater numbers doth exceed the common number 41, as 42 doth exceed the common number 41 but 1, which I set against the number 38, to which he is cou∣pled, and 45 doth exceed the common number 41 but 4, which I set against the number 36 to which he is coupled. The Sum of these differences which is 13 sheweth the number of pounds the composition or mixture must weigh, and the several diffe∣rences do shew how many pound of each sort you must take. As 4 sheweth you must take 4 pound of 36 Crowns fine, 1 pound of 38 Crowns fine, 3 pound of 42 Crowns, and 5 pound of 45 Crowns fine; each pound of this mixture containing 13 l. is worth 40 Crowns, as was desired.

The Proof.

		Crowns.
4 l. of 36	Crowns fine is	144
1 l. of 38	38
3 l. of 42	126
5 l. of 45	225
		533

So likewise 13 l. of 41 Crowns 533

But suppose it were required to make a mixture of a certain weight as well as a certain fineness. As, let it be required to make a mixture of 8 pound weight of the several former finenesses, and to the former common price of 41 Crowns, how much must be taken of each sort?

Then by the Rule of Three say, As the Sum of the several differences which is 13 is to the weight propounded 8 pound, so is each particular difference to the weight de∣sired, as followeth.

As 13 is to 8, so is 4 to	26/13.
As 13 is to 8, so is 1 to	8/13.
As 13 is to 8, so is 3 to	11/••••.
As 13 is to 8, so is 5 to	3/••••.
	8 l.

So there must be 2 l. 6/13 of the fineness of 36 Crowns, 8/13 parts of a pound of 38 Crowns, 1 pound 11/13 of 42 Crowns, and 3 pound 1/13 of 45 Crowns, which added all together make 8.

To prove the Work.

2 6/13 of 36	Crowns is	88 8/13
8/13 of 38	23 5/13
1 11/13 of 42	77 7/13
3 1/13 of 45	138 6/13
		328

8 pound of 41 Crowns is 328

Therefore you may certainly conclude you have committed no error in your work.

Quest. 2. A Vintner hath several sorts of Wine, viz. he hath Wine of 14 d. the Gallon, Wine of 19d. the Gallon, Wine of 23d. the Gallon, and he hath Wine of 28d. the Gallon, now he doth intend to make a mixture of 48 Gallons; now it is required to know how many Gallons of each price he must take to make one Gallon with another worth 21 d.

	d.	Gal.	s.	d.
21	14	2 at 14	2	4
19	7 at 19	11	1
23	7 at 23	13	5
28	2 at 28	4	8
		18 at 21	1: 11	6

Say by the Rule of Three,

As 18 to 48, so 2 to	5 Gallons 6/18
As 18 to 48, so 7 to	18 Gallons 1/18
As 18 to 48, so 7 to	18 Gallons 12/18
As 18 to 48, so 2 to	5 Gallons 6/18
	48

The Proof is as followeth.

		d.
5 Gallons 6/18 of 14	Pence is	74 ••••/18
18 Gallons 12/18 of 19	354 12/18
18 Gallons 12/18 of 23	429 6/18
5 Gallons 6/18 of 28	149 6/18
		1008 d.

48 Gallons of 21 Pence is 1008 d.

Therefore you may conclude your Ope∣ration is without error.

Quest. 3. A Drugster hath three sorts of Druggs, the first sort at 3 s. the pound, the second at 7 s. and the third at 15 s. per pound. Of these sorts he is to make 3 par∣cels containing each 24 pound, one parcel to be reckoned 11 s. per pound, the other 12s. and the other 13s. how much must be taken of each sort to make up each parcel?

〈 math 〉〈 math 〉

The common prices propounded are 11 Shillings, 12 Shillings, and 13 Shillings; likewise the prices given of which the mixture is to be made are 3 Shillings, 7 Shillings and 15 Shillings, and because two of the three given prices are less than the common price, I couple the two lesser with the greater as you see, and the differences of the 2 lesser I place against the greater, and the differences of the greater against the lesser; then proceed as before, saying,

As 20 the whole difference is to 24 the whole quantity of each parcel, so is each particular difference to its respective quan∣tity enquired after, as followeth.

1.	As 20 to 24, so 4 to 4 ⅘
As 20 to 24, so 4 to 4 ⅘
As 20 to 24, so 12 to 14 ⅖
	24

2.	As 20 to 24, so 3 to 3 ⅗
As 20 to 24, so 3 to 3 ⅗
As 20 to 24, so 14 to 16 ⅘
	24

3.	As 20 to 24, so 2 to 2 ⅖
As 20 to 24, so 2 to 2 ⅖
As 20 to 24, so 16 to 19 ⅕
	24

Here you see in the first parcel that is to be 11 s. per pound, to make a mixture of 24 l. you must take 4 pound ⅘ of 3 s. per pound, 4 l. ⅘ of 7 s. per pound, and 14 l. ⅖ of 15 s. per pound; and in the second parcel of 12 s. the common price round, you must take 3 l. ⅗ of 3 s. per pound, 3 l. ⅗ of 7 s. per pound, and 16 l. ⅘ of 15 s. per pound; and in the third parcel of 13 s. round, you must take 2 l. ⅖ of 3 s. and 2 l. ⅖ of 7 s. and 19 l. ⅕ of 15 s. per pound, and if you add up each parcel found out they will each amount unto 24 l.

Quest. 4. A Goldsmith hath five sorts of Gold, one is 15 carrects fine, 17 car∣rects fine, 20 carrects fine, 23 carrects fine, and 26 carrects fine. All these five sorts he doth intend to mix with such an

allay, as that a mixture of 124 ounces should be but 12 carrects fine, I desire to know how much of either sort of Gold must be taken to make up 124 l.

To answer this and all other questions of this nature, set down their several and particular finenesses one under another, as you see in the following Example, and set down the common fineness of the Gold which is demanded, viz. 12 carrects to∣wards the left hand, also set down a Cypher under the particulars for an alloy, set the alloys difference, which is 12, against all the other Sums, as in the Example you may see is done, and set the difference of all the other Sums against the alloy, as 3, 5, 8, 11 and 14. Then by the Rule of Three say, As 101 the whole difference is to 124 the whole quantity, so is each particular dif∣ference to each particular quantity sought. 〈 math 〉〈 math 〉

	oz.
As 101 to 124, so 12 to	14	74/101
As 101 to 124, so 12 to	14	74/101
As 101 to 124, so 12 to	14	74/101
As 101 to 124, so 12 to	14	74/101
As 101 to 124, so 12 to	14	74/101
As 101 to 124, so 41 to	50	34/101
	124

CHAP. XXIX.

Of Exchange.

EXchange is the way of remitting mony from one place or Country to another, by paying mony in one place, and receiving the like value in another, with allowance sometimes of profit and sometimes of loss.

It will be of little or no consequence to give you the names of the particular Coyns of several Countries, because with us in England they sometimes rise and sometimes fall in value, according as the Exchange runs, I therefore do only intend to lay down some necessary questions as follow.

Quest. 1. A Merchant payeth at London, or any other Sea Port Town of England 256 l. Sterling, to receive the value thereof at Amsterdam Exchange at 33 s. 9 d. Flemish, for 20 s. Sterling I desire to know how much Flemish mony I must receive at Am∣sterdam.

By the Rule of Three say, If 1 l. Sterl. give 33 s. 9 d. Fl. what will 256 l? 〈 math 〉〈 math 〉

Quest. 2. A Merchant at Amsterdam doth deliver 432 l. Flemish, to receive the value thereof at London Exchange, 33 s. 9 d. for 20 s. Sterling, I desire to know how much Sterling mony I must receive at London?

By the Rule of Three say, If 33 s. 9 d. Fl. give 1 l. Ster. what will 432 l. Fl. 〈 math 〉〈 math 〉

Multiply and divide and you will find he must receive 256 l. Sterling. These two questions do prove one another, as you may plainly see.

Quest. 3. A Merchant at London delivers 374 l. 2 s. Sterling to be received in Paris, the like value Exchange at 58d. per French Crown. I demand how many French Crowns must I receive in Paris?

By the Rule of Three say, If 58 d. give 1 Fr. Cr. what wil 374 l. 2 s? 〈 math 〉〈 math 〉

By multiplying and dividing you will find you must receive at Paris 1548 French Crowns.

Quest. 4. A Merchant at Paris doth de∣liver 2870 French Crowns, the like value

to be received in London, Exchange at 57d. ½ per Crown. I desire to know how much Sterling mony I must receive at London?

By the Rule of Three say, If 1 Fr. Cr. give 57 d. ½ how much will 2870 〈 math 〉〈 math 〉

Quest. 5. A Merchant at London doth de∣liver for Franckford 140 l. Sterl. Exchange for Frankford at 46d. Sterl. the Florin of 66 Kreutrars. The Question is how many Florins must he receive at 60 Kreutrars at Frankford?

By the Rule of Three say. If 46 d. give me 66 Kreutr. what will 140 l? 〈 math 〉〈 math 〉

Quest. 6. An Merchant in Antwerp takes up mony by Exchange, at the rate of 33 s. 4 d. Flemish, to pay at London 20 s. Sterl. but when the Bill of Exchange came due, he was forced to take up mony in London to pay the said Bill, to pay at Antwerp for every 20s. Sterling 33 s. 10 d. Flemish. I desire to know whether I do win or lose, and how much in the 100 l. of mony?

By the Rule of Three say. If 33s. 10d. give 33s. 4d. what will 100l? Answer 98 l. 10s. 5d. which subtract from 100l. gives 1l. 9s. 7d. for the remainer, and so much is lost in 100 l. of mony.

Quest. 7. A Merchant at Cales doth re∣ceive a Bill of Exchange from London for 1500 Duckets which is the value of 365 l. 12 s. 6 d. Sterling delivered at London. I desire to know what the Ducket is worth?

If 1500 Duckets cost 365l. 12s. 6d. what cost 1 Ducket? Answer 4 l. 10 s. ½.

The Exchange between London and France is much different from the Flemish Exchange, because in France it is delivered in French Crowns, which is worth 50 Sous Tournois the piece.

For in France they do account by De∣niers Tournois, accounting 12 Deniers 1 Sous Tournois, and 20 Sous Tournois 1 Livre or Frank, which is also 1 Pound Tournois; the French Crown in Com∣merce is current for 51 Sous Tournois, but in Exchange it will pass for no more than 50 Sous Tournois, which is 2 l. 10 Sous Tournois the Crown.

Quest. 8. A Merchant in London doth deliver 360 l. 2 s. 4 d. Sterling Exchange at 5 s. 8 d. the French Crown, and to re∣ceive at Paris 50 Sous Tournois for every French Crown. I desire to know how many Livres I shall receive at Paris for the 360 Pound paid in London?

By the Rule of Three say, If 5 s. 8 d. give 2 Livres 10 Sous, what will 360 l. give? Multiply and divide and you will find 3177 Livres 10 Sous Tour∣nois, and so much must I receive in Paris for my 360 Pound.

Quest. 9. A Merchant at Paris doth de∣liver by Exchange 3784 Livres Tournois, at 50 Sous Tournois the French Crown, and for each French Crown to receive here

5 s. 7 d. Sterling in London. I desire to know how much Sterling mony I shall receive for my 3784 Livres or Pounds Tournois?

By the Rule of Three say, If 2l. 10 Sous Tournois give 5s. 7d. what will 3784? Or, If 5 l. Tournois give 10s. 2 d. Sterling, what will 3784?

Multiply and divide either of these pro∣portions, and you will find the fourth pro∣portional to be 422l. 10s. 11d. and so much you must receive at London for your 3784 Livres paid or delivered in Exchange at the rate before mentioned.

CHAP. XXX.

Of Gauging.

IF you would know how many Gallons any Vessel contains in Wine or Ale mea∣sure, you must first take the Diameter of the head of the said Vessel in Inches, then the Diameter at the Bung withinside in

Inches, then in the respective following Table look for the Diameter of the Bung in Inches before taken, which you will find in the respective Table in the Column in∣tituled Inches, and the number that you find stand against it set down twice, then in the same Columns intituled Inches look for the Diameter of the head Inches, and the number standing against it set down once under the number twice set down, and add them all together, multiply their Sum by the length of the Vessel in Inches, from the Product cut off 3 figures towards the right hand, and those remaining figures towards the left hand are the number of Gallons required, and the 3 figures cut off are parts of a Gallon either of Ale or Wine measure, according to the Table from whence you take your numbers.

Example 1.

There is a Cask or Vessel whose Diameter at the Bung is 26 Inches, the Diameter at the head is 17 Inches, and the length of the said Vessel is 41 Inches. I desire to know how many Gallons of Wine he doth con∣tain? In the Wine Table over against 26

Inches the Diameter at the Bung, I find the number 0: 766, which according to instru∣ctions I set down twice; likewise against the number 17 I find 328, which I set down: the Sum of all these three is 1: 860, which multiplied by 41 the length, the Product is 76: 260, from whence I cut off the three figures towards the right hand, and there will remain 76, which sheweth the Vessel doth contain 76 Gallons 260/1000.

Diameter at Bung	26	0	766
The same		0	766
Diameter at Head	17	0	328
		1	860
			41
		1	860
		74	40
Content in Gallons		76	260/1000

Example 2.

There is a Cask or Vessel whose Diameter at Bung is 22 Inches, his Diameter at Head 16 Inches, and his Length 36 Inches. I do

desire to know how many Gallons of Ale measure it doth contain? out of the re∣spective Table for Ale measure I find the Diameter at the Bung 22 Inches, and against it this number 450, which I set down twice, I also find out the Diameter at head 16 Inches in the same Table, and against it I find the number 238, which I set down as you see. Add all these together and their Sum will be 1138, which multiply by 36, the length of the Cask or Vessel, the Pro∣duct will be 40: 968, and cutting off three figures towards the right hand, there re∣mains 40, which sheweth the Cask doth contain of Ale measure 40 Gallons 968/1000.

Diameter at Bung	22	0	450
The same again		0	450
Diameter at Head	16	0	238
		1	138
			36
		6	828
		34	14
Content in Gallons		40	968/1000

A Table shewing how to find the Content of any Cask or Vessels in Gallons, &c.

Inc	Ale	Inc	measure	Inc	Wine	Inc	measure.
1	0	001	31	0	892	1	0	001	31	1	089
2	0	004	32	0	951	2	0	005	32	1	160
3	0	008	33	1	011	3	0	010	33	1	234
4	0	015	34	1	073	4	0	016	34	1	310
5	0	023	35	1	137	5	0	028	35	1	388
6	0	032	36	1	203	6	0	041	36	1	469
7	0	045	37	1	271	7	0	056	37	1	551
8	0	059	38	1	341	8	0	072	38	1	636
9	0	075	39	1	412	9	0	092	39	1	724
10	0	093	40	1	485	10	0	113	40	1	813
11	0	112	41	1	561	11	0	137	41	1	904
12	0	134	42	1	638	12	0	163	42	2	000
13	0	157	43	1	717	13	0	192	43	2	096
14	0	182	44	1	797	14	0	221	44	2	194
15	0	208	45	1	880	15	0	255	45	2	295
16	0	238	46	1	965	16	0	296	46	2	398
17	0	264	47	2	051	17	0	328	47	2	504
18	0	301	48	2	139	18	0	367	48	2	611
19	0	335	49	2	219	19	0	409	49	2	721
20	0	371	50	2	321	20	0	453	50	2	833
21	0	409	51	2	415	21	0	500	51	2	948
22	0	450	52	2	510	22	0	548	52	3	065
23	0	491	53	2	608	23	0	600	53	3	184
24	0	535	54	2	707	24	0	653	54	3	305
25	0	580	55	2	808	25	0	708	55	3	428
26	0	628	56	2	911	26	0	766	56	3	554
27	0	677	57	3	016	27	0	826	57	3	682
28	0	728	58	3	123	28	0	888	58	3	813
29	0	781	59	3	232	29	0	933	59	3	945
30	0	836	60	3	341	30	1	020	60	4	080

CHAP. XXXI.

The Extraction of the Square Root.

IT is necessary to insert the Table of the Squares of all the nine Digits, as fol∣loweth.

Roots.	Squares.
1	1
2	4
3	9
4	16
5	25
6	36
7	49
8	64
9	81

Example.

It is required to extract the square Root of 22429696, set a mark or prick with your Pen, under the first figure towards your right hand, and so under every second figure as you see, so will your given num∣ber

be divided into four parts or points, whereby you must note that the Root will admit of four figures, for the square Root of the given number doth always admit of as many figures as the given number doth admit Points. Then proceed to the Ex∣traction thus, beginning at the left hand the first Point being 22, by the former Table of Squares, I find 16 to be the great∣est Square number contained therein, whose Square Root is 4, therefore I set down 4 in the Quotient, and the Square thereof 16 I set underneath 22, and I subtract one from the other and there remains 6, which I set underneath; to this remainer 6 I draw down the next Point 42, which doth make the number 642; next I double the Quo∣tient 4 which is 8, which I set down one figure short of the right hand as you see; then do I seek how many times 8 will go in 64, which is 7 times, set down 7 under that vacant figure towards your right hand 2, and likewise set down 7 in the Quotient, and multiply 87 by 7, whose Product 609 set under the line, and subtract it from the number made of the last remainer, and the Point last brought down 642, and the re∣mainer of this subtraction will be 33, to

which remainer draw down the next point 96, and it makes the number 3396; next double the Quotient 47, it makes 94, which set under the number 3396, that it may stand one figure short towards the right hand, as you see in the following Example; then ask how many times 94 is contained in the number or figures standing over them 339? answer is three times, there∣fore I set down 3 to the right hand of 94, also I set down 3 in the Quotient, then do I multiply 943 by 3 the last figure in the Quotient, and the Product of the multi∣plication 2829 I set underneath the line, and subtract it from the number made of the last remainer, and the point last brought down which is 3396, and of this subtra∣ction there doth remain 567; lastly to this remainer I draw down the next point 96, and it makes the number 56796, now double the Quotient 473, which makes 946, and set it under the number 56796, that it stand one figure short towards your right hand as you see, and ask how many times 946 is contained in the figures stand∣ing over him 5679? answer is 6 times, set∣ting down 6 to the right hand of 946, and also in the Quotient, and multiply 9466 by

6 the last figure set in the Quotient, and the Product of such multiplication I set down under the line being 56796, which I subtract from the number composed of the last remainer, and last point brought down which is 56796, and there will remain 0, and having no more points to draw down, I therefore conclude I have finished my Ex∣traction, wherefore the square Root of the given number 22429696 is 4736, as was required, as by the following Example doth appear.

〈 math 〉〈 math 〉

The Proof of the Extraction of the Square Root, is done by multiplying the Root found in it self, and the Product is the number given whose Square Root is required.

Example.

Let it be required to prove this Example where the Root found is 4736; therefore I multiply 4736 by it self, that is to say, by 4736, and the Product is 22429696 the number given.

〈 math 〉〈 math 〉

CHAP. XXXII.

The Extraction of the Cube Root.

I Shall first insert a Table containing the Cubes of all the nine Digits, which I would advise the learner to be compleatly able to remember without book.

Roots.	Cubes.
1	1
2	8
3	27
4	64
5	125
6	216
7	343
8	512
9	729

Example.

Let it be required to extract the Cube Root of the given number 42144192.

First point the given number as you see, setting a prick under the first figure to∣wards your right hand, and so under every third figure as you see, every prick except the last towards the left hand containing three figures, and that containeth the two figures 42: therefore in the foregoing Table, find out the greatest Cube number in 42, which I find to be 27, whose Cube Root is 3, which place in the Quotient, and subtract the Cube number 27 from 42, and there will remain 15, which place under neath; to this remainer bring down the next point 144, so have you the number 15144, which call your Resolv. draw a line under; square the Quotient 3 which is 9, multiply the said Spuare 9 by 300, and the Product thereof 2700, place under the Resolvend, and call it the Treble Square. Again, multiply the Quotient 3 by 30, which makes 90, place 90 under the Treble Square, and call it the Treble Quotient, add the Treble Square and Treble Quotient to∣gether into one Sum, and it makes 2790, which call your Divisor, then examine how often you can take your Divisor in your Resolvend, which is four times, set 4 in the Quotient, then multiply the Treble

Square 2700 by the 4 last placed in the Quotient, and subscribe the Product un∣derneath your Divisor, square the last said figure in the Quotient 4, which is 16, thereby multiply the Treble Quotient 90, and the Product thereof 1440 place under the last Product 10800, cube the figure 4 last placed in the Quotient, which by the foregoing Table is 64, and place it under the preceding Products, then add these three Products into one Sum, and subtract the Sum thereof 12304 from your Resolv. and there will remain 2840, to this remain∣der draw down the next point 192, and the number 2840192 is a new Resolvend, under which draw a line, then square the Quotient 34 which is 1156, which Square multiply by 300, and the Product 346800 set under the Resolvend, and call it the Treble Square as before; then multiply the Quotient 34 by 30, and the Product thereof 1020, set under the Treble Square, and call it the Treble Quotient, and the Sum of the Treble Square and Treble Quotient is 347820, which call your Divisor, and examine how often you can take your last Divisor in your last Resolvend, which will be 8 times, therefore set 8 in the Quotient,

multiply your last Treble Square by the last figure in your Quotient 8, the Product thereof 2774400 subscribe under your last Divisor, square the last figure in your Quo∣tient 8, which is 64, by which multiply the last Treble Quotient, the Product 65280 set under the last Product, then cube the last figure of your Quotient 8, which by the foregoing Table you will find to be 512, which place under the pre∣ceding Products, and add all these three Products into one Sum, which makes 2840192, which if you subtract from your last Resolvend there will remain 0, and in such manner must you proceed by reiterating the work by squaring the Quotient, until all the figures are brought down.

But in this Example the work is finish∣ed, for there is no more figure to draw down, and 0 remains; so the Cube Root of 42144192 is 348, as was required.

〈 math 〉〈 math 〉

To prove the Extraction of the Cube Root.

Multiply the Cube Root found by it self, then multiply the Product of the multipli∣cation by the Root found, and the Product of this second multiplication is the given number whose Cube Root is required.

Example.

〈 math 〉〈 math 〉

It is required to prove this Example of the Extraction of the Cube Root, wherein the Cube Root found is 348, which mul∣tiply by 348, and the Product of the mul∣tiplication will be 121104, which multi∣plied again by 348 the Root found produ∣ceth the given number to be extracted 42144192, whereby it is concluded the Operation is right.

FINIS.

Notes

• * 1.1

  	C.	qr.	l.
  Bought	45	3	16
  Sold	28	1	24
  Remains	17	1	20
• * 1.2

  l.	s.	d.	f.
  7836	14	7	1
  4524	17	9	3
  3311	16	9	2
  7836	14	7	1
• * 1.3

  〈 math 〉〈 math 〉
• * 1.4

  〈 math 〉〈 math 〉
• * 1.5

  〈 math 〉〈 math 〉
• * 1.6

  〈 math 〉〈 math 〉
• * 1.7

  〈 math 〉〈 math 〉
• * 1.8

  〈 math 〉〈 math 〉
• * 1.9

  〈 math 〉〈 math 〉
• * 1.10

  〈 math 〉〈 math 〉
• * 1.11

  〈 math 〉〈 math 〉
• * 1.12

  〈 math 〉〈 math 〉
• * 1.13

  〈 math 〉〈 math 〉
• * 1.14

  〈 math 〉〈 math 〉
• * 1.15

  〈 math 〉〈 math 〉
• * 1.16

  〈 math 〉〈 math 〉
• * 1.17

  〈 math 〉〈 math 〉
• * 1.18

  〈 math 〉〈 math 〉
• * 1.19

  〈 math 〉〈 math 〉
• * 1.20

  〈 math 〉〈 math 〉
• * 1.21

  3698	1
  7396	2
  11094	3
  14792	4
  18490	5
  22188	6
  25886	7
  29584	8
  33282	9
• * 1.22

  〈 math 〉〈 math 〉
• * 1.23

  〈 math 〉〈 math 〉
• * 1.24

  〈 math 〉〈 math 〉
• * 1.25

  〈 math 〉〈 math 〉
• * 1.26

  〈 math 〉〈 math 〉
• * 1.27

  〈 math 〉〈 math 〉
• * 1.28

  〈 math 〉〈 math 〉
• * 1.29

  〈 math 〉〈 math 〉
• * 1.30

  〈 math 〉〈 math 〉
• * 1.31

  〈 math 〉〈 math 〉
• * 1.32

  〈 math 〉〈 math 〉
• * 1.33

  〈 math 〉〈 math 〉
• * 1.34

  〈 math 〉〈 math 〉
• * 1.35

  〈 math 〉〈 math 〉
• * 1.36

  〈 math 〉〈 math 〉
• * 1.37

  2152	Crowns.
  538	Angels.
  1076	Pounds.
• * 1.38

  〈 math 〉〈 math 〉
• * 1.39

  〈 math 〉〈 math 〉
• * 1.40

  〈 math 〉〈 math 〉
• * 1.41

  〈 math 〉〈 math 〉
• * 1.42

  〈 math 〉〈 math 〉
• * 1.43

  〈 math 〉〈 math 〉
• * 1.44

  〈 math 〉〈 math 〉
• * 1.45

  〈 math 〉〈 math 〉
• * 1.46

  〈 math 〉〈 math 〉
• * 1.47

  〈 math 〉〈 math 〉
• * 1.48

  〈 math 〉〈 math 〉
• * 1.49

  〈 math 〉〈 math 〉
• * 1.50

  〈 math 〉〈 math 〉
• * 1.51

  〈 math 〉〈 math 〉
• * 1.52

  〈 math 〉〈 math 〉
• * 1.53

  〈 math 〉〈 math 〉
• * 1.54

  〈 math 〉〈 math 〉
• * 1.55

  〈 math 〉〈 math 〉
• * 1.56

  〈 math 〉〈 math 〉
• * 1.57

  1	times 12 is	12
  2	24
  3	36
  4	48
  5	60
  6	72
  7	84
  8	96
  9	108
• * 1.58

  〈 math 〉〈 math 〉
• * 1.59

  〈 math 〉〈 math 〉
• * 1.60

  〈 math 〉〈 math 〉
• * 1.61

  〈 math 〉〈 math 〉