for the quotient, without remainder. Lastly, we collect the several quotients, and have 6 s. 11 d. for the cost per yd., as was required. We may also divide by the factors of 30, thus : 5) 10 9 41 £0 6s. 11 d. 2. If 185 yds. cost £604 6 s. 8 d., what is the cost per yd. ? Answer, £3 5 s. 4 d. 3. If 284 yds. cost £700 4 s. 9 d., what is the cost per yd. ? Answer, £2 9 s. 32 d. 4. If 316 t. cost £4034 5 s. 4 d., what is the cost per ton? Answer, £12 15 8. 4 d. 5. If 297 A. cost £4039 16 s. 41 d., what is the cost per acre ? Answer, £13 12 s. 01 d. 6. If 921 cwt. cost £9169 14 s. 14 d., what is the cost per cwt. ? Answer, £9 19 s. 11 d. 7. Divide 78 s. m. 49 A. 1 R. 28 P. 10 s. yds. Of s. f. by 385. Answer, 129 A. 3 R. 6 P. 15 s. yds. 54 s. f. For further practice, the student may reverse the 6 examples which immediately follow the first example of the preceding article ; in each of which he may divide the whole cost by the number of yards, which will give the cost per yard. 327. When the dividend and divisor represent quantities of the same kind, we must observe whether the quotient should or should not also be of the same kind. If it should, we operate as in the preceding article. For example, suppose that, with £15, we have gained £27 10 s. 3 d., and that we would find the gain on each pound. It is plain that this gain should be the fifteenth part of the whole gain, and therefore of the same name with the divisor and dividend : consequently, we divide £27 10 s. 33 d. by 15, and have £1 16 s. 8 d. for the required gain. 328. But, if the nature of the question requires that the quotient should be of a different kind, we reduce the divisor and dividend to the lowest denomination of the dividend ; after which, considering the units of the dividend as being of the same kind with those of the quotient that we seek, we divide as in Art. 326. For example, if we would know how much land can be bought for £483 5 s. 7 d., at the rate of £5 per acre, it is evident, from the nature of the question, that the quotient must be acres and parts of an acre, and that the number of these must be determined by the number of times that £5 is contained in £483 5 s. 73 d. Now, as dissimilar units cannot have a ratio to each other, in order to express the ratio of the dividend to the divisor both numbers must be rendered homogeneous. But this can only be done by reducing both numbers to the lowest denomination of the dividend; or (324) the dividend to a fraction of its principal unit. We therefore reduce the numbers, thus : £483 5 s. 72 d. 463950 qrs.; £5 = 4800 qrs. The abstract ratio of these numbers expresses the required number of acres; that is, 443852 A. = 96 A. 2 R. 25 P., is the number of acres and parts of an acre required by the question. Or thus: 5 s. 7{d.= £: then 4833= £15,465, and 15465 = 5= 3963 A. = 96 A. 2 R. 25 P., as before. E.camples. 1. How much corn, at 8 s. per bushel, can be bought for £26 10 s.? Answer, 66 bush. 1 pk. 2. How much cloth, at 4 s. per yd., can we buy for £20 16 s. 4 d.? Answer, 104 yds. O qr. 11 n. 3. At £4 per acre, how much land can be bought for £314 19 s. 6 d.? Answer, 78 A. Ž R. 39 P. 4. Having planted 25 bushels of potatoes, and harvested 437 bush. 2 pks., what is the rate of increase ? Answer, 17 bush. 2 pks. for each bushel, or 17} for 1. Division of a Compound Number by a Compound Number: 329. When the divisor and dividend are both compound numbers, but of different kinds, we reduce the divisor to a fraction of the principal unit, and divide the dividend by this fraction in its lowest terms; that is, (212,) we multiply the dividend by the denominator of this fraction, and divide the product by the numerator. Examples. 1. If 15 cwt. 2 qrs. 8 lbs. of sugar cost £39 1 g. 2 d., what is it per cwt. ? 15 cwt. 2 qrs. 8 lbs. = 154 cwt. = - 149 cwt. 8. d. 39 1 2 7 109) 273 8 2 (£2 10 s. 2 d. 218 20 1090 18 12 109) 218 (2 d. 218 *** Having divided the whole cost, £39 1 s. 2 d. by 109, which is the value of the given quantity in cwts., we have, for the value of 1 cwt., £2 10 s. 2 d. 2. If 13 yds. 2 qrs. of cloth cost £2 5 s. 6 d., what is it per yd ? Answer, 3 s. 4, d. 3. If 14 cwt. 3 qrs. 9 lbs. of sugar cost £40 9 s. 6 d., what is it per cwt. ? Answer, £2 14 s. 7 d. + 4. If 15 t. 16 cwt. of hay cost £71 2 s., what is it per ton? Answer, £4 10 s. For more examples the student is referred to examples 8, 9, 10, 11, 12, 13, in Art. 325; in each of which the whole cost, divided by the fractional equivalent of the quantity in cwts., will give the price per cwt. 330. When the divisor and dividend are of the same kind, and the nature of the question requires that the quotient shall, in kind, agree with them, we operate as in the preceding article. But, if the question requires that the quotient shall be of a different kind, we reduce the dividend and divisor to the lowest denomination mentioned in either, and divide as before. Or, we reduce each of the given compound numbers to a fraction of the principal unit, divide as in fractions, and reduce the quotient as usual. Examples. 1. How much sugar can be bought for £39 1 s. 2 d., at £2 10 s. 2 d. per cwt ? Here the divisor and dividend are of the same kind, but the d, question demands a quotient of a different kind; we therefore say, £39 18. 2 d. 9374 d. Also £ 2 10 s. 2 d. 602 d. Then 93074 is the abstract ratio of the price of the whole quantity sought to the price of a unit of that quantity: but the quantities evidently have the same relation to each other that their prices have: wherefore, as the denominator 602 represents one unit of the quantity sought, or 1 cwt., and the numerator 9374, the quantity itself in units of the same order as those of the denominator, it follows that 337 cwt. = 15 cwt. 2 qrs. 8 lbs. is the quantity sought. Or thus : £39 1 s. 2 d.=39770=498; and £ 2 10 g. 2 d.= 2133=2f2b = 126: then, 4687 X 13=-487=15 cwt. 2 qrs. 8 lbs., as before. 2. At £3 5 s. 4 d. per yd., how many yards can you buy for £604 6 8. 8 d. ? Answer, 185. 3. At £2 9 s. 39 d. a yard, how many yards can be bought for £700 4 s. 9 d. ? , 4. At £12 15 s. 4 d. a ton, how many tons can we buy for £4034 5 8. 4 d. ? Answer, 316. 5. At 13 12 s. 01 d. a ton, how many tons can be bought for £4039 16 s. 41 d.? Answer, 297. 6. At £9 19 s. 11 d. a ton, how many tons can be bought for £9169 14 s. 11? Answer, 921. 7. At £4 10 s. a ton, how much hay can be bought for £71 2 s.? Answer, 15 t. 16 cwt. 8. At £3 15 s. 6 d. per cwt., how much sugar can be bought for £42 18 s. 9 d.? Answer, 11 cwt. 1 qr. 14 lbs. 9. At £2 10 s. 6 d. per cwt., how much can be bought for £12 6 s. 21 d.? Answer, 4 cwt. 3 qrs. 14 lbs. 10. At £3 14 s. 81 d. per cwt., how much can be bought for £26 15 s. 71 d.? Answer, 7 cwt. Oqr. 19 lbs. 11. At £2 3 s. 9 d. per cwt., how much can be bought for £29 14 s. 21 d. ? Answer, 13 cwt. 2 qrs. 7 lbs. 12. At £5 11 s. 6 d., per cwt., how much can be bought for £52 3 s. 10 d. ? Answer, 9 cwt. 1 qr. 12 lbs. SECTION XVII. PRACTICE AND CALCULATIONS BY COMPLEMENTS. Practice. 331. PRACTICE, so called from its continual application in the purchase and sale of goods, is an easy and expeditious method of multiplying compound numbers. It consists in resolving the lower denominations of a compound number into convenient aliquot parts or measures of the principal unit, and of each other; and then ascertaining the value of each aliquot or part, by taking a corresponding part of the value of the unit or of the value of that part which the aliquot measures. 332. When one of the given numbers' is simple, either of them may be taken for multiplicand or multiplier. For example, if we would find the cost of 27 yds. of cloth, at 3 s. 9 d. per yd., it is evident that this cost is 27 times 3 s. 9 d. But 27 units of whatever kind, at 3 8. 94 d. per unit, will give the same result. The number 27, therefore, or multiplier, may be referred to the kind of units that we seek, and thus become the multiplicand; that is, it may be considered as £27, 27 s., or 27 units of any kind to which we can reduce 3 s. 9 d. For it is evident (59) that both the given numbers must be referred to the same kind of unit, before each can thus take the place of the other. Assuming £27 for the multiplicand, we take parts of £1 for the multiplier 3 s. 9 d., thus : £ 2 s. is £ 15 of £1 27 is 1 of 2 s. 2 14 s. 0 d. cost of 27 units, at 2 0 per unit. 6 d. is I of 1 s. 1 7 0. at 10 3 d. is I of 6 d. 0 13 6. at 0 6 3 qrs.is of 3 d. 0 6 9. at 0 3 £5 23. 11 d. total, or cost at 3 91 333. When any of the lower denominations is too small a part of the preceding aliquot to give a convenient divisor, we introduce a convenient aliquot, at pleasure, and bar the product which it gives, as not belonging to the question : after which we proceed as usual. 8. d. 1 s. 0.... |