¶The second Chapter treateth of the rule of three compounde, which are foure in nomber.
THere belongeth to the fyrst & seconde partes of the rule of thre compound alwaies fyue numbers: whereof (in the first
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THere belongeth to the fyrst & seconde partes of the rule of thre compound alwaies fyue numbers: whereof (in the first
part of the rule of three compo••nde the seconde nomber and the 〈◊〉〈◊〉, are alwayes of one semblaunce, and ••ke denomination: whose rule is thus, multiply ye first nōber by the seconde, & that shalbe your diuisor: then mul∣tiplie the other three nōbers the oneby the other to be your diuidende. Exā∣ple, of this first part: if 100. crowns in 12. monthes, do gaue 15 .li. what will 60. crownes gaine. in 8. monthes? Answere, first multiplie 100. crownes by 12. monthes, & therof cōmeth 1200. for your diuisor: then multiply 15 .li. by 60. crownes, & by 8. monthes & you shal haue 7200. diuide 7200. by 1200, & therof cōmeth 6 .li. so many li. wyl 6. crownes gaine in 8. mōthes: this que∣stiō may be done by ye double rule of 3. yt is to say by ye rule of 3. at 2. times, but yet this rule of 3 cōpoūd is more brief. 〈 math 〉〈 math 〉
2. In the seconde part of the rule of third compound, the 3. nomber is like vnto the fift, wherof the rule is thus: multiplie the 3. nomber by the 4. the product shalbe your diuisor: then mul∣tiply the first nomber by ye seconde, & the product therof by the fift, yt whiche nomber shalbe your diuidend, or nō∣ber yt is to be diuided: as by example,
When 60. crownes in 8 monthes do gaine 6 .li. in howe many monthes wil: 100. crownes gain. 15 .li. Aunswer Multiply the thirde nomber 6. by the fourth nōber 100, & ther of cōmeth 600 then multiplye the first nōber 60. by the secōd nōber 8. & by the fift nōber 15 thereof will come 7200. then diuide 7200. by 600. & ye quotiēt wilbe 12: in so many monthes will 100. crownes gaine 15 .li. This question may like∣wise be done by the double rule of 3. 〈 math 〉〈 math 〉
3. In the thirde part of the rule of 3. compound, there may be 5. nombers or more: & in this rule the first nōber & the last alwayes dissemblaunt the one to thother: & the questiō is from the last nōber vnto the first, wherof ye rule is thus: multiply that nomber which you woulde know by those nō∣bers which do giue the value, & diuide the product of the same, by ye multipli∣cation of the nōbers which are alrea∣dy valued, as by exāple. If 4. deniers Parisis, be worth 5. deniers arnols, & 10. deniers tournois, be world 12. de∣niers of Sauoy, I demaūd how many deniers Parisis are 8. deniers of Sa∣uoy worth▪ Aunswere: Multiply 8. de∣niers of Sauoy (which is the nomber yt you would know) by 4. deniers pa∣risis, & by 10 deniers t〈…〉〈…〉is whiche are ye nōbers that giue ye value, & they make 320: then multiplie 5. denters tournois, by 12 deniers of sauoy (which are the nombers alredye valued) and they make 60: lastly diuide 320. by 60
and you shall finde 5. deniers ⅓ parisis, so muche are the denies of Sauoye worth. 〈 math 〉〈 math 〉
4. In the fourthe parte of the rule of three compounde: the first nomber and the last are always semblant and of one denomination, and the questiō of this rule, is alwayes from the last nōber to the last sauing one. Where∣of there is a rule which is thus. You must multiplye that nomber whiche you woulde knowe, by the nombers that are alreadye valued, and diuide the product of the same, by the multi∣plication which commeth of the nom∣bers that giue the value, as by exāple
If 4: deniers Parisis, bee worth 5. Deniers Lournois, and 10. Deniers Lournois, be worthe 12. Deniers of Sauoy, I demaunde how many De∣niers
of Sauye. are 15. Deniers Pa∣risis worth. Aunswere: Multiplye 15. Deniers Parisis that you woulde knowe, by 5. Deniers Lournois, & by 12. Deniers of Sauoye, which are the nombers alreadye valued, and they make 900 Diuide the same by 4 times 10. which are the nombers that doe giue the value, and you shal finde 22. Deniers ½ of Sauoye, so much are the 15. Deniers Parisis worth. 〈 math 〉〈 math 〉