MULTIPLICATION & DIVISION OF COMPOUND NUMBERS. ¶ 41. 1. A man bought 2 yards of cloth, at 1 s. 6 d. per yard; what was the cost? 2. If 2 yards of cloth cost 3 shillings, what is that per yard? 3. A man has three pieces of cloth, each measuring 10 yds. 3 qrs.; how many yards in the whole? 4. If 3 equal pieces of cloth contain 32 yds. 1 qr., much does each piece contain? how 5. A man has five bottles, each containing 2 gal. 1 qt. 1 pt.; how much wine do they all contain? 6. A man has 11 gal. 3 qts. 1 pt. of wine, which he would divide equally into five bottles; how much must he put into each bottle? 7. How many shillings are 3 times 8 d.? 9 d.? 3 X 10 d.? 4 X 7 d. ? - 7 X 6 d.? 8. How much is one third of 2 shillings? of 2 s. 4 d. ? of 14 d.? 3 x 9 d.? 10 X of 2 s. of 3 s of 21 d. ? 10. If 6 yards of cloth cost 7. 14 s. 4 d., what is the price per yard? Here, as the numbers are large, it will be most convenient to write them down before multiplying and dividing. 1 OPERATION. 6)7 14 4 2 cost of 6 yards. 1 5 8 3 price of 1 yard. Proceeding after the man 5 8 3 price of 1 yard. 6 number of yards. Ans. 7 14 4 2 cost of 6 yards. 6 times 3 qrs. are 18 qrs.ner of short division, 6 is con4 d. and 2 qrs. over; we set tained in 7£. 1 time, and 1£. down the 2 qrs.; then, 6 times over; we write down the 8 d. are 48 d., and 4 to carry quotient, and reduce the re makes 52 d. = 4 s. and 4 d. mainder (1.) to shillings, over, which we write down; (20 s.,) which, with the given again, 6 times 5 s. are 30 s. shillings, (14 s.,) make 34 s.; and 4 to carry makes 34 s. 16 in 34 s. goes 5 times, and 4 & 1£. and 14 s. over; 6 times over; 4 s. reduced to pence 1. are 6., and 1 to carry = 48 d., which, with the makes 7£., which we write given pence, (4 d.,) make 52 down; and it is plain, that the d. ; 6 in 52 d. goes 8 times, and united products arising from 4 d. over 4 d. 16 qrs., the several denominations is which, with the given qrs. the real product arising from (2) 18 qrs. ; 6 in 18 qrs. goes the whole compound number. 3.times; and it is plain, that the united quotients arising from the several denominations, is the real quotient arising from the whole compound 11. Multiply 3£. 4 s. 6 d. by 7. 13. What will be the cost of 5 pairs of shoes at 10 s. 6 d. a pair? 15. In 5 barrels of wheat, each containing 2 bu. 3 pks. 6 qts., how many bushels ? number. 14. At 2£. 12 s. 6 d. for 5 pairs of shoes, what is that a pair? 16. If 14 bu. 2 pks. 6 qts. of wheat be equally divided into 5 barrels, how many bushels will each contain ? 17. How many yards of 18. If 9 coats contain 39 cloth will be required for 9 yds. 3 qrs. 3 na., what does 1 coats, allowing 4 yds. 1 qr. coat contain? 3 na. to each? 19. In 7 bottles of wine, 20. If 5 gal. I gill of wine each containing 2 qts. 1 pt. 3 be divided equally into 7 bot tles, how much will each contain? gills, how many gallons? 21. What will be the 22. If 8 silver cups weigh weight of 8 silver cups, each 3 lb, 9 oz. 1 pwt. 16 grs., what weighing 5 oz. 12 pwt. 17 is the weight of each? grs.? 24. If 119 cwt. 1 qr. of su 23. How much sugar in 12 hogsheads, each containing gar be divided into 12 hogscwt. 3 qrs. 21 lb. ? heads, how much will each hogshead contain? 25. In 15 loads of hay, each weighing 1 T. 3 cwt. 2 qrs., how many tons? 26. If 15 teams be loaded with 17 T. 12 cwt. 2 qrs. of hay, how much is that to each team? When the multiplier, or divisor, exceeds 12, the operations of multiplying and dividing are not so easy, unless they be composite numbers; in that case, we may make use of the component parts, or factors, as was done in simple numbers. Thus 15, in the example] 15 being a composite numabove, is a composite number ber, and 3 and 5 its compoproduced by the multiplica- nent parts, or factors, we may tion of 3 and 5, (3 × 5 divide 17 T. 12 cwt. 2 qrs. by 15.) We may, therefore, one of these component parts, multiply 1 T. 3 cwt. 2 qrs. by or factors, and the quotient one of those component parts, thence arising by the other, or factors, and that product by which will give the true the other, which will give the answer, already taught, true answer, as has been al- (T 20.) ready taught, (T 11.) OPERATION. T. cit. qr. 1 3 2 One factor, as OPERATION. 3) 17 12 2 3 one of the factors. The other factor, 5) 5 17 2 3 10 2 17 12 5 the other factor. 2 the answer. 27. What will 24 barrels of flour cost, at 2£. 12 s. 4 d. a barrel ? Ans. 1 3 2 28. Bought 24 barrels of flour for 62 £. 16 s.; how much was that per barrel ? 30. If 1 cwt. of sugar cost 3 £. 7 s. 8 d., what is that per 29. What will 112 lb. of sugar cost, at 74 d. per lb. ? Note. 8, 7, and 2, are fac-lb.? tors of 112. 32. Bought 84 pipes of 31. How much brandy in 84 pipes, each containing 112 brandy, containing 9468 gal. 1 qt. 1 pt.; how much in a pipe? gal. 2 qts. 1 pt. 3 g.? 33. What will 139 yards of cloth cost, at 3 £. 6 s. 5 d. per yard? 34. Bought 139 yards of cloth for 461 £. 11 s. 11 d.; what was that per yard? 139 is not a composite num- When the divisor is such a ber. We may, however, de- number as cannot be produced compose this number thus, by the multiplication of small 139100+30 + 9. numbers, the better way is to We may now multiply the divide after the manner of 1 price of 1 yard by 10, which long division, setting down will give the price of 10 yards, the work of dividing and reand this product again by 10, ducing in manner as fol which will give the price of lows: 100 yards. We may then multiply the price of 10 yards by 3, which will give the price of 30 yards, and the price of 1 yard by 9, which will give the price of 9 yards, and these three products, added together, will evidently give the price of 139 yards; thus: 33 4 332 1 99 12 29 17 5 price of 1 yd. 10 2 price of 10 yds. 10 8 price of 100 yds. £. d. 139) 461 li 11 (3£. 417 44 11 891 (6.8. 57 12 695 (5 d. 695 The divisor, 139, is contained in 461 £. 3 times, (3£.,) and a remainder of 44., which must now be reduced to shillings, multiplying it by 20, and bringing 461 11 11 price of 139 yds. in the given shillings, (11 s.,) Note. In multiplying the making 891 s., in which the price of 10 yards (33£. 4 s. divisor is contained 6 times, 2 d.) by 3, to get the price of (6 s.,) and a remainder of 30 yards, and in multiplying 57 s., which must be reduced the price of 1 yard (3. 6 s. to pence, multiplying it by 12, 5 d.) by 9, to get the price of and bringing in the given 9 yards, the multipliers, 3 and pence, (11 d.,) together mak9, need not be written down, ing 695 d., in which the dibut may be carried in the visor is contained 5 times, mind. (5 d.,) and no remainder. The several quotients, 3, 6, 5 d., evidently make the answer. The processes in the foregoing examples may now be pre sented in the form of a RULE for the Multiplication of RULE for the Division of Com Compound Numbers. pound Numbers. 1. When the divisor does I. When the multiplier does not exceed 12, multiply suc-not exceed 12, in the manner cessively the numbers of each of short division, find how denomination, beginning with many times it is contained in the least, as in multiplication the highest denomination, unof simple numbers, and carry der which write the quotient, as in addition of compound and, if there be a remainder, numbers, setting down the reduce it to the next less dewhole product of the highest nomination, adding thereto the denomination. number given, if any, of that so continue to do II. If the divisor exceed 12, II. If the multiplier exceed 12, and be a composite num- and be a composite, we may diber, we may multiply first by vide first by one of the comone of the component parts, ponent parts, that quotient by that product by another, and another, and so on, if the comso on, if the component parts ponent parts be more than be more than two; the last two; the last quotient will be product will be the product re- the quotient required. quired. III. When the multiplier III. When the divisor exexceeds 12, and is not a com-ceeds 12, and is not a composite, multiply first by 10, posite number, divide after the and this product by 10, which manner of long division, setwill give the product for 100; ting down the work of diand if the hundreds in the mul-viding and reducing. tiplier be more than one, multiply the product of 100 by the number of hundreds; for the tens, multiply the product of 10 by the number of tens; for the units, multiply the multiplicand; and these several products will be the product required. |