d. d. ..... ..... To illustrate this, we may proceed with the above example, thus : 18. 3 d. is te of £1 27 1 13 9 ....3 8. 9 d. 3 d. is of 1 s. 3 d. 1 13 9 3 qrs. is of 3 d.... Ø Ø Ø cost, at 3 d. (barred.) 0 i 81 cost, at....... O s. 08 d. £5 2 111 cost, at.... 3 s. 9 d. Or thus : 2 s. 0 d. is it of £1 27 1 s. 8 d. is 1 of £1 2 14 0 cost, at .2 8. O d. 1 s. is 1 of 2 S....... 2 5 0 cost, at .1 s. 8 d. 11 d. is of 1 8..... i Ø cost, at 1 s. (barred) 1 d. is of ij d..... 0 3 41 cost, at ...O s. 14 d. 0 0 6 cost, at ..O s. 01 d. £5 2 11; cost, at .... ..3 s. 9 d. Or thus : 2 s. 6 d. is of £1...... 27 1 s. 3 d. is 1 of 2 s. 6 d. 3 7 6 cost, at.........2 s. 6 d. 3 d. is } of 1 s. 3 d..... 1 13 9 cost, at.........1 s. 3 d. 3 qrs. is 1 of 3 d........ 0 1 81 cost, at.........0 8.0 d. £5 2 111 cost, at.........3 8. 9fd. Again, without barring, thus : 3 s. 4 d. is l of £1... 27 0 1 11 £5 2 111 Assuming 27 s. for the multiplicand, we multiply by 3 s., which are similar units, (332,) and take parts of 1 8. for 9d., £ d. 8. d. thus : 8. 4 6 d. is S....... 27 3 s. 9 d. £4 ls. 6 9 1 81 £5 2 111 1 7 3 s. 9 d. 0 1 87 £5 2 111 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. PRACTICE TABLES. 8. 50.. 1. Aliquot parts of a Dollar. Aliquot parts of a Shilling. d. 6.. 331 4. 3 1..., 8. d. ...... 1 3......... 5 0........ 1 0......... 4 0......... 08.......... 3 4. 06. 2 6.. so 2 03.. 1 8......... 0 2........ 120 io Ö.......... .... 0 4.... 0..... ......... Aliquot parts of a Cwt. cwt. 14 Aliquot parts of a Ton. T. •14 1 ..16 . 1 17 yds. 61,50 « Examples. d. 1. 11 yds. at 18 s. 8 d. per yd. 10 5.4 2. 12 yds. 2 6 1 10 0 3. 15 yds. 1 4 1 0.0 4. 16 yds. 1 3 1 0 0 5. 18 yds. 6 8 6 0 0 6. 1 9 1 9 9 7. 19 yds. 4 6 4 5 6 8. 27 yds. 11 91 15 18 41 17 19 53 17 6 28 17 6 85 19 11 261 0 0 56 15,00 24,161 75,80 35,197 17. 19 3 14 89,433 18. 17 2 0 76,561 20 52,361; 32,93145 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 denominations. To find the value of 11 cwt. 1 qr. 14 lb., at £3 15 s. 6 d. per cwt., we proceed thus : 90 87,50 49,75. " s. 15 d. 6 15 1qr. is cwt. 3 value of 11 cwt. at £3 prod. for 15 s., or 3 times 2 15 5 s. ; i. e. value of 11 cwt. 6 d. is to of 5 s. 2 at 15 s. 0 5 6 value of 11 cwt. at 6 d. 14 lb. is qr. 0 18 10 value of 1 qr. 0 9 54 value of 14 lb. £42 18 93 We first multiply £315 s. 6 d. by 11; that is, having multiplied 3 by 11, we take parts of 11, considered as pounds, for the 15 s. 6 d. 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. 103 d. Lastly, to have the value of 14 lb., or 3 qr., we take 1 of 18 s. 101 d., the value of 1 qr. Having added the several products, we have £42 18 s. 9 d. for the value of the whole, as required. Or thus : 1 qr. is 1 3 15 ៩ 6 11 cwt. lqr. 14 lb. 41 10 6 14 lb. is 1 18 101 9 51 £42 18 93 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. =151 s. = 31s. = £j), and £336= £160. Again, 1 qr. 14 lb.=1: qr. iqr.= cwt., and 113 cwt. =fcwt. Then, x 40 = £42 18 s. 98 d. 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 X 3,775 = £42,940625, and reducing the decimal, we have £42 18 s. 9 d. £ 13741 17 This method often requires a great many figures, and is therefore 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. 2. 7 0 19 3 14 81 26 15 7 3. 13 2 7 2 3 93 29 14 21 4. 9 1 12 5 11 63 52 3 10 5. 29 3 7 1 41 211 7 41 6. What is the cost of 297 T. 13 cwt. 2qr. 7 lb., at £13 12 s. 01 d. per ton ? Ans. £40490 s. 10758 d. 7. How much work will £74 2 s. 9 d. pay for, at the rate of 11 square yards for £1 ? Ans. 815 sq yds. 448 sq. f. 8. If we pay £1 for 5 cwt. 2 qr. 34 lb. of potatoes, what quantity should we have for £63 10 s. 2 d. ? Ans. 17 T. 11 cwt. 1 qr. 3 lb. 6 oz. 9. What is the cost of 91 acres at £15 10 s. 13 d. per acre ? Ans. £1411 1 s. 41 d. 10. How much land has the man who pays £15 10 s. 11 d. tax, at the rate of £1 for every 91 Acres ? Ans. 1411 A. O 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. When a quantity is multiplied by any proper fraction, mit is diminished by such part of itself as is expressed by the complement of that fraction, that is, by ths of itself. Thus: nQ Q ( mn) MQ — mot ng QX Also, Q т - п m n m m m m n = 36 m If we put Q=36, and 7, we shall have 36 X = of 36 =36-8=28. 336. When m =1, that is, when the numerator is only one unit less than the denominator, the above property n |