divide by 3, but if to New-York currency, divide by 4, and the true answer is obtained.* 2. If pounds only are given, place a cipher at the right hand, and divide as before. The answer will be dollars. 3. If shillings, pence and farthings only are given, the answer may be obtained by dividing the farthings by 4, and putting the quotient at the right hand of the pence; then dividing this number by 12, and placing the quotient at the right hand of the shillings, and dividing these by 20; and then divide the quotient as before directed. EXAMPLE.-Reduce 1 s. 6d. to cents. 1216 201.50 3)75 (25 cents, Ans. 4. The following Rules may be convenient. 66 86 66 66 SECTION XII. COMPOUND MULTIPLICATION. Compound Multiplication teaches how to find the amount of a given sum of different denominations, multiplied by a simple number. *Ten thirds of a dollar are equal in value to one £.; 100 thirds of a dime, 1000 thirds of a cent, and 10,000 thirds of a mill, are all of the same value. Ten fourths of a dollar make a pound New-York currency; 100 fourths of a dime, &c. It is commonly the most simple method to consider the multiplication of each denomination of the compound number by the simple factor, as a distinct question; and the several products thus obtained will be the whole amount sought. What is the weight of 5 casks of raisins, each weighing 2 cwt. 3 qrs. and 25 lbs.? qrs. lbs. cwt. 2 3 25 5 5 5 10 15 125 It is evident that the number of cwt. in one cask repeated as many times as there are casks, will give the number of cwt. in all the casks; the same is true of the quarters, and of the pounds. Hence 10 cwt. 15 qrs. and 125 lbs. expresses the true weight of the 5 casks. But 125 lbs. is equal to 4 qrs. and 13 lbs. ; then we may reckon the 4 qrs. with the other quarters and merely write 13 lbs. under the column of pounds. Then 15 qrs.- +4=19, which are equal to 4 cwt. and 3 qrs.; then 10 cwt.+4=14 cwt. We have then 14 cwt. 3 qrs. and 13 lbs., as seen in the following example 2 3 25 14 3 13 RULE. : 1. The multiplier may be written under the lowest denomination. Multiply each denomination successively beginning with the lowest, setting down the excess, and carrying from each denomination to the higher, as in Compound Addition. 2. When the multiplier is a composite number, the multiplication may be made by the component parts, as in the following example: What is the weight of 36 bales of cotton, each weighing 3 cwt. 2 qrs. and 17 lbs.? cut. qrs. lbs. 6 21 3 18 6 131 1 24 Here we first multiply by 6, which gives the weight of 6 bales, and this product by 6, which evidently gives the weight of 36 bales, 6 times 6-weight of 36 bales. What will 47 yards of cloth cost, at 1 £. 8 s. 9 d. per yard? In this case the multiplier 47 is not a composite number; that is, it cannot be produced by the multiplication of any other numbers; but we may multiply the price of one yard by 6, and that product by 8, which will give the cost of 48 yards; then if we subtract the price of one yard from this product, the cost of 47 yards will remain. 9 Cost of 45 yds. 67 11 3 Ans. yards as before. Hence, Or we may multiply the price of one yard by 9, and this product by 5, which will give the price of 45 yards; then multiplying the price of one yard by 2, will give the price of 2 yards, which added to the cost of 45, will give the cost of 47 2. When the multiplier cannot be produced by the multiplication of small numbers, two such numbers as come nearest to it may be taken; and then the value of the odd parts may be found and added to the amount; or if too large, subtracted from it. What is the cost of 3475 yards of cloth, at 2 £. 3 s. 5 d. per yard? Here we first multiply by 10, which of course gives the price of 10 yards; the price of 10 yards multiplied by 10 gives the price of 100 yards, which, multiplied by 10, gives the price of 1000 yards; and this multiplied by 3 gives the price of 3000 yards. The price of 100 yards multiplied by 4 gives the price of 400 yards; the price of 10 yards multiplied by 7 gives the price of 70 yards, and the price of 1 yard multiplied by 5, gives the price of 5 yards. Add these several amounts to the price of 3000 yards and we have 7543 £. 12 s. 11 d., the price of 3475 yards. Hence 3. When the multiplier is larger than 12, multiply the given price or quantity by 10, which gives the amount of 10, this product multiplied by 10 gives the amount of 100, &c. Multiply the given price or quantity by the units, and the price or amount of 10 by the tens in the multiplier; and so on as to the 100's. The several products added together will be the amount sought. What is the cost of 9 cwt. of tea at 7 s. 6 d. per pound? In this example we first multiply by 7, which gives the price of 7 lbs. ; this product multiplied by 8 gives the price of 56 lbs. cwt. This multiplied by 2 gives the price of 1 cwt. which multiplied by 9 gives the price of 9 cwt. The same result may be obtained by multiplying first by 7, then by 4, which gives the price of a quarter; and this, multiplied by 4, gives the price of 1 cwt. SECTION XIII. COMPOUND DIVISION. Compound Division teaches the method of finding how many times a given simple number is contained in any sum or quantity of different denominations. A compound number may be divided by a simple number by regarding each denomination of the former as a distinct dividend. If 3 £. 13 s. and 8 d. be divided equally between 2 men, how much will each receive? If the pounds, shillings, and pence in this question could be separately divided by 2 without a remainder, the division would be perfectly easy. But 3 £. cannot be divided by 2 without a remainder, neither can the 13 s. But if we take one pound from the 3, 2 will remain, which may be equally divided between two men, by giving them I each. We may now call the 1 £. 20 s. and put it with the 13 s., making 33 s., which cannot be divided by 2 without a remainder of 1, which we may call 12 d. and reckon it with the 8 d. making 20 d.; this number can be divided by 2 without a remainder, so that the whole is divided into two equal parts. £. S. d. 2)2 32 20 1 16 10 But this change of the numbers may be made more conveniently as we proceed in the operation. From the preceding example and illustration, we deduce the following RULE. 1. Write the numbers as in Simple Division, and divide the several denominations of the dividend successively by the divisor. 2. When there is a remainder after dividing any denomination, reduce it to the next lower and add it to the number in that denomination in the given sum. Divide these, and if there is a remainder, reduce it to the next lower denomination and proceed as before till the work is finished. If the divisor can be separated into its component parts, divide by each of them separately. EXAMPLE. If 27 lambs cost 10 £. 2s. 6d. what will 1 cost? £. S. d. 9) 10 2 6 3) I 2 6 If Here we divide first by 9 and that quotient by 3, because 3 times 9 are 27. the same money had been paid for 9 lambs, they would have cost 1 £. 2s. 6d. each; but since there are 3 times 9 lambs, it is evident, that one will cost only a third part of 1£. 2s. 6 d. to 7 s. 6 d. 7 6 Divide 12 £. 16 s. 6 d. by 57. |