From the above, the student will see that the method pur sued may be greatly varied, according to the caprice of the operator; also, that the best methods to practise, practice alone will suggest. 1 8.. 1 0 2... 1 It is usual in business to take the nearest cent, and to reject fractions of a cent; thus, in this last example, we should call the amount 32 dollars and 94 cents. The student, however, should not, in any case, be satisfied till he has obtained the exact answer. 334. When the quantity is compound, and the value of its principal unit also compound, we find, as in the first 12 examples of the preceding article, the value of the given number of principal units, and then, as in the last 8 examples, take parts of the value of the principal unit for the lower denomi nations. To find the value of 11 cwt. 1 qr. 14 lb., at £3 15 s. 6 d. per cwt., we proceed thus: We first multiply £3 15 s. 6d. by 11; that is, having multiplied 3 by 11, we take parts of 11, considered as pounds, for the 15 s. 6d. Having thus taken 11 times £3 15 s. 6 d., the value of 1 cwt., we have the value of 11 cwt. Then, to have the value of 1 qr., we take of £3 15 s. 6 d., which gives 18 s. 10 d. Lastly, to have the value of 14 lb., or qr., we take of 18 s. 10 d., the value of 1 qr. Having added the several products, we have £42 18 s. 93 d. for the value of the whole, as required. Or thus: We here multiply £3 15 s. 6 d. by 11, as in art. 321; we then take parts for 1 qr. 14 lb., and having added, we have £42 18 s. 9 d. for the whole value, as before. Proof.-By Fractions. 15 s. 6 d. 15 s. 31 s. Again, 1 qr. 14 lb. = 11⁄2 qr. = = 1 cwt. Then, 91 × 151 = 40 = £, and £331 £45. qr.cwt., and 113 cwt. £42 18 s. 93 d. 13741 320 By Decimals. 15 s. 6 d. = = 15,5 s. £,775; therefore, £3 15 s. 6 d. £3,775. Again, 1 qr. 14 lb. = 1,5 qr.,375 cwt. ; therefore 11 cwt. 1 qr. 14 lb. 11,375 cwt. Then, 11,375 × 3,775= £12,940625, and reducing the decimal, we have £42 18 s. 93 d. = This method often requires a great many figures, and is there fore very tedious; the scholar may, however, sometimes use it as proof. Examples. 1. 4 cwt. 3qr. 14lb. at £2 10s. 6d. per cwt. £12 6s. 21 d. 73 2. 7 0 19 "" 3 14 81 66 26.15 7 1 43 52 3 103 66 211 7 41 6. What is the cost of 297 T. 13 cwt. 2qr. 7 lb., at £13 12 s. 0 d. per ton? of 11 square yards for £1? Ans. £4049 0s. 10128 d. 7. How much work will £74 2 s. 9 d. pay for, at the rate Ans. 815 sq. yds. 48 sq. f. 2 qr. 31 lb. of potatoes, what 10 s. 2 d.? 8. If we pay £1 for 5 cwt. quantity should we have for £63 Ans. 17 T. 11 cwt. 1 qr. 3 lb. 63 oz. 9. What is the cost of 91 acres at £15 10 s. 1 d. per acre? Ans. £1411 1 s. 4 d. 10. How much land has the man who pays £15 10 s. 1 d. tax, at the rate of £1 for every 91 Acres? Ans. 1411 A. 0 R. 11 P. Calculation by Complements. 335. As a general discussion of this subject would perhaps occupy more pages than are contained in this entire work, we shall here chiefly consider the advantages it affords in Practice calculations. n When a quantity is multiplied by any proper fraction,, it is diminished by such part of itself as is expressed by the com plement of that fraction, that is, by m m n ths of itself. Thus: n=1, that is, when the numerator is only one unit less than the denominator, the above property affords great advantage in calculation, seeing that we have only 1 to subtract—th. For example, if we would multiply 3848 by m = , as the complement of is, we have only to subtract from 3848 its eighth part, thus: 3848 - 481 3367, which is easier than the following: 3848 8) 26936 Again, if we would find the cost of 3848 yds. at 17 s. 6d. per yd., as 17 s. 6d. is £7, we subtract from 3848 its eighth, as above, and have £3367 for the required cost. This is much easier than the following: What is the cost of 97 yds. at 831 cts. per yd.? Then as 83 is § of a Here 97 yds. at $1, would be $97. dollar, and its comp., we subtract from 97 its sixth part, and have $80,831 for the required cost. This is certainly more expeditious than the following: cts. 831 12X8+1=97 1000. 8 8000 831 $80,831 |