of pence contained in 250l.; which product, again multiplied by 4, the number of farthings in one penny, the product gives the number of farthings contained in 250.; or the answer. Rule. In reduction afcending, divide the given number by the number of units of that denomination which make one of the next greater; and divide that quotient by the number of units of the fame denomination which make one unit of the next higher, and proceed in this manner till the whole is finished. Thus, as in the foregoing example I reduced 250%, inta 240000 farthings; fo inverfely, here I fay, in 240000 farthings, how many pounds? In this example, I first divide the given number of farthings by 4, the number of units of that denomination which are contained in an unit of the next higher denomination; or the number of farthings contained in one penny, and the quotient gives the number of pence contained in 240000 farthings; which quotient is again divided by 12, the number of pence contained in one fhilling, and that quotient gives the number of fhillings contained in the given fum of farthings; and lastly, thefe fhillings are again divided by 20, the number of fhillings in one pound, and the quotient is 250l. for the answer. Thus it may be feen, that reduction afcending and defcending prove each other. For if the fum be performed by reduction defcending, it must be proved by reduction afcending, as in the two foregoing examples; and if it be in reduction afcending, it must be proved by reduction descending. In reduction descending, when the fum confifts of feveral denominations, the number in each denomination, after the firft, is to be added to the denomination to which it belongs; as in the following example: Example. 239 10 4780 ΙΟ 6 Example 3. In 2391. 10. 6d. how many pence?-In this example, after having reduced the pounds into thillings by multiplying by 20, I add to the product the to fhillings which stand in the place of fhillings; and having reduced the fhillings into pence by multiplying by 12, I add to that product the 6 pence which stand in the place of pence; and the answer to the work is thus found to be 57,486 pence. 4790 12 57480 57486 24 10 3 20 480 10 490 Example 4. In 24 tons, 10 hundreds, and 3 quarters, how many pounds?-This is performed as the foregoing, but having refpect to the table of avoirdupois weight. I therefore multiply the tons by 20 to reduce them to hundred weights, as cc hundred is 1 ton, and to the product I add the 10cwt. I then multiply by 4 to bring the hundreds into quarters, to which I add the 3 quarters: and, laftly, mul tiply the quarters by 28, the number of pounds in a quarter: the product is 54964 for the 15704 answer. 1960 3 1963 3926 54964 Examples in Reduction Afcending. Example 5. In 24,649,721 minutes, how many days, hours, and minutes? Example 6. In 47,398 grains of troy weight, how many ounces, pennyweights, and grains ? Example 7. In 29,472,986 fquare perches of land, how many acres, roods, and perches? Example 5. 6,0)2464972,1 2) 410828,41 12) 205414 17117,10 Example 6. 2)47398 12)23699 2,0) 197,4 22 98 14 Example 7. 4,0)2947298,6 4) 736824,26 184206 VOL. I. 20 In Y In the first of thefe examples I divide the minutes by 60, to bring them into hours, by cutting off the o from the 6, and dividing by 6 only, as taught in Divifion, and the quotient is 41082$ hours, and there remains 4 minutes, which, placed before the cut off from the dividend, makes 41 minutes for a remainder. This quotient I again divide by 2, and that quotient by 12, which is equal to dividing by 24, the number of hours in 1 day, and the quotient is 17117 days, and there remains 10; but to find the true remainder, I multiply the 10, the last remainder, by 2, the first divifor of the hours, and the product is 20 for the true remainder: thus the answer to the question is, 17117 days, 20 hours, 41 minutes. In the fecond example, I divide the grains by 2 and 12, to bring them into pennyweights, as 24 grains make 1 pennyweight; the quotient is 1974 pennyweights, and 22 grains remain. The pennyweights I divide by 20, to bring them into ounces, and the quotient is 98 ounces, and I pennyweight zemains, to which I bring down the 4 I cut off from the dividend, and the laft remainder is 14 pennyweights: thus the anfwer is 98 ounces, 14 pennyweights, and 22 grains. In the laft example, I divide the perches by 40, as 40 perches make 1 rood, and the quotient is 736824 roods, and 26 perches; the roods I divide by 4, to bring them into acres; and the answer is found to be 184,206 acres, and 26 perches. Reduction, both afcending and defcending, may be performed by one divifor or multiplier: thus, to bring farthings into pounds, the pounds may be divided by 960, the number of farthings in a pound, and the quotient will give the number of pounds, and the remainder (if any) muft be refolved into the inferior denominations. And to reduce hundredweights into fingle pounds, they may be multiplied by 112, the pounds in an hundred-weight, and the product is the answer. But the method before laid down is the more regular, and at the fame time the more expeditious way of performing this rule. Examples Examples of both Kinds for Practice. Example 8. If 420 pieces of cloth contain 8420 ells Flemish, it is required to know how many ells English they contain? --Anf. 5052 ells English. Example 9. In 220 puncheons of rum how many hogfheads?-Anf. 293 hogfheads, and 21 gallons remain. Example 10. A filversmith hath 1000 ounces of filver to be made into fpoons, falts, and tankards; each spoon to weigh 2 oz. 12 pwt. each fait 3 oz. and each tankard 30 oz. and to make an equal quantity of each, it is defired to know how many he can make of each ?-Anf. 28, and 64 pwt. remain. In the first of these examples the 8420 ells Flemish are reduced into quarters of a yard, by multiplying by 3 (as there are 3 quarters in an ell Flemish), and then brought into English ells, by dividing these quarters by 5, the number of quarters in an ell English. The 9th example is wrought in the fame manner: viz. by reducing the 220 puncheons into gallons, by multiplying by 84, and bringing these gallons into hogfheads, and by dividing them by 7 and 9, which is equal to 63, the gallons in a hogfhead. This method of reduction always takes place when the less denomination Y 2 denomination is not contained any certain number of times exactly in the greater. In the Icth example I reduce the weight of 1 fpoon, 1 falt, and I tankard, into pennyweights, by multiplying by 20, and then add them together; and by the total 712, I divide the 1000 ounces of filver, which is also reduced into pwts. and it quotes 28 of each, and 64 pwts. remain. By reduction we are enabled to reduce the coin of one country into that of another, without having recourse to the rule of three, or exchange. Example 11. What is the value, in Englifh coin, of 350 ducats, at 45. 2d. per ducat ? Example 12. In 2461. 18s. 6d. Flemish money, how much English, the course of exchange being 30s. 6d. per pound fterling? Example 13. How much money English is there in 4420 pieces of eight, the course of exchange being at 49fd. sterling? Example 12. |