141. When the price is an aliquot part of a shilling. RULE. Having placed the price, and its corresponding aliquot part, as before directed, divide by the aliquot part, and the quotient will be shillings, which reduce to pounds by dividing it by 20. If there be a remainder after the first division, it is shillings; reduce it to pence, which divide by the aliquot part, and the quotient is pence P. 7. What is the worth of 826 lemons, at 3d, each ? P Here if we consider the articles as worth a shilling each, they will cost in the whole as many shillings as there are articles given; wherefore if they cost an aliquot part of a shilling each, the whole will evidently amount to the same part of so many shillings: thus, in Example 7, 826 lemons at 1s. each will cost 826 shillings; but at 3d. each (since 3d. is of a shilling) they will amount to of 826 shillings: and the like in other cases. 142. When the price is an aliquot part of a pound. RULE I. Having placed the price and the aliquot part as before, divide by the aliquot part, and the quotient will be pounds. II. The remainder (if any) must be reduced to shillings, and divided by the aliquot part for shillings; if there be a second remainder reduce it to pence, and divide for pence; if a third, reduce it to farthings, and divide for farthings". 17. What must be given for 135 pullets, at 2s. 6d. each? a The observations in the two foregoing notes are equally applicable to this rule; thus, in Example 17, 135 pullets at 17. each will cost 1357.; but since the price (2s. 6d.) is of a pound, it is evident that the whole will amount to of 1351. and the like may be shewn of the other Examples. 143. When the price is not an aliquot part. RULE I. Divide the price into sums which are aliquot parts of the whole, (viz. of a shilling, or a pound, as the case may be,) or of which one is an aliquot part of the whole, and the rest aliquot parts either of the whole, of this, or successively of one another. II. Divide by these several aliquot parts, and the quotients being added together, the sum will be the answer in shillings, if the aliquot parts are of a shilling; and in pounds, if they are aliquot parts of a pound'. This rule will be evident from an explanation of the 29th Example, where if 371oz. had cost 1s. each, the whole would have cost 371 shillings; but at 3d. each (since 3d. is of a shilling) they will cost of 371 shillings, or 92s. 9d.; and at d. each (since d. is of 3d.) they will cost of their value at 3d., that is of 92s. 9d., or 15s. 5d.; wherefore, if the value at 3d. each be added to the value at d. each, the sum will evidently be the value at 3d. each: and the like may be shewn in every other case, 29. What is the value of 371 ounces of tobacco, at 3d. per S. 20 108 2 Answer 5l. 8s. 2d. Explanation. Here 3d. not being an aliquot part of a shilling, I divide it into two sums, 3d. and, of which 3d. is of a shilling, and d. of 3d.; I divide the given number 371 by 4, and the quotient is 92s. 9d.; this, which is the value at 3d., I divide by 6, because d. is of 3d.; and the quotient is 15s. 5d.; I add the two quotients together, and reduce the 108 shillings into pounds, by dividing by 20. In the second operation, I divide the given price 3d. differently; I take 2d., which is, and 1d., which is, both aliquot parts of a shilling; I therefore divide the top line by both, add the quotients together, and divide by 20, which gives the answer the same as in the first operation, 30. What sum will 123 sawyers earn, at 17s. 6d. each '? OPERATION. Explanation. I find that 17s. 6d. will conveniently resolve into aliquot parts, viz. 10s. of a pound, 5s. = of 10s., and 2s. 6d. = of 5s. I therefore divide successively by these, and the sum of the quotients is the answer, 31. What sum will pay for 215 cheeses, at 17s. 10d. each? Explanation. When there are several aliquot parts, we are not obliged to take each part out of the next preceding one; we may take it out of any of the preceding parts, as may be most convenient. In this example 2s. is (not of 5s. but) of 10s.; I therefore divide 1077. 10s. (and not 532. 15s.) by it. In like manner 10d. is (not of 2s, but) of 5s.; I therefore divide 537. 15s. (and not 217. 10s.) by 6. 6 To prove this example, multiply 371 by 3d., and divide by 12 and 20. To prove operations of this kind, multiply the given price by the number of particulars (Art. 105.); in the present instance it will be 17s. 6d. X 123. |