To illustrate this, we may proceed with the above example, thus:

Assuming 27 s. for the multiplicand, we multiply by 3 s., which are similar units, (332,) and take parts of 1 s. for 93d.,

thus:

From the above, the student will see that the method pur sued may be greatly varied, according to the caprice of the ope rator; also, that the best methods to practise, practice alone will suggest.

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. 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. 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.

15 s. 6 d.

31

=

13 qr.

Then, 91 X 151

409

40.

£3, and £31-£157. qr.cwt., and 113 cwt.

13741

320

By Decimals.

£42 18 s. 93 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. 93 d.

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.

66 3 14

26 15

71

29 14 21

6. What is the cost of 297 T. 13 cwt. 2qr. 7 lb., at £13

12 s. 0 d. per ton?

Ans. £4049 0s. 10158 d. 2 s. 9 d. pay for, at the rate Ans. 815 sq. yds. 413 sq. f. 2 qr. 31 lb. of potatoes, what 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. 11⁄2d. 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?

7. How much work will £74 of 11 square yards for £1?

8. If we pay £1 for 5 cwt. quantity should we have for £63

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

m

When a quantity is multiplied by any proper fraction, it is diminished by such part of itself as is expressed by the complement of that fraction, that is, by

m - n

m

ths of itself. Thus:

336. When m—n=1, that is, when the numerator is only one unit less than the denominator, the above property