54, which is produced by multiplying 6 and 9, contains all these factors, and one of them, viz. 3, repeated. The reason why 3 is repeated is because it is a factor of both 6 and 9. By reason of this repetition, a number is produced 3 times as large as is necessary for the common multiple. When the least common multiple of two or more numbers to be found, if two or more of them hav a common factor, it may be left out of all but one, because it will be sufficient that it enters once into the product. These factors will enter once into the product, and only once, if all the numbers which have common factors be divided by those factors; and then the undivided numbers, and these quotients be multiplied together, and the product mul tiplied by the common factors. If any of the quotients be found to have a common factor with either of the numbers, or with each other, they may be divided by that also. Reduce 3, 4, %, and 4, to the least common denominator. The least common denominator will be the least common multiple of 4, 9, 6, and 5. Divide 4 and 6 by 2, the quotients are 2 and 3. Then divide 3 and 9 by 3, the quotients are 1 and 3. Then multiply ng these quotients, and the undivided number 5, we have 2 × 1 × 3 × 5 = 30. Then multiplying 30 by the two common factors 2 and 3, we have 30 × 2 × 3 = 180, which is to be the common denominator. Now to find how many 180ths each fraction is, take 2, 5, , and of 180. Or observe the factors of which 180 was made up in the multiplication above. Thus 2 × 1 × 3 X 5 × 2 × 3 = 180. Then multiply the numerator of each fraction by the numbers by which the factors of its denominator were multiplied. The factors 2 and 2 of the denominator of the first fraction, were multiplied by 1, 3, 3, and 5. The factors 3 and 3, of the second, were multiplied by 2, 1, 5, and 2. The factors 2 and 3, of the third, were multiplied by 2, 1, 3, 5; and 5, the denominator of the fourth, was multiplied by 2, 2, 1, 3, and 3. XXIII. If a horse will eat of a lushel of oats in a day, how long will 12 bushels last him? In this question it is required to find how many times of a bushel is contained in 12 bushels. In 12 there are 36, therefore 12 bushels wil! last 36 days. At & of a dollar a bushel, how many bushels of corn may be bought for 15 dollars? First find how many bushels might be bought at of a dollar a bushel. It is evident, that each dollar would buy 5 bushels; therefore 15 dollars would buy 15 times 5, that is, 75 bushels. But since it is instead of of a dollar a bushel, it will buy only as much, that is, 25 bushels. This question is to find how many times of a dollar, are contained in 15 dollars. It is evident, that 15 must be reduced to 5ths, and then divided by 3. 15 5 75 (3 25 bushels. The above question is on the same principle as the fol lowing. How much corn, at 5 shillings a bushel, may be bought for 23 dollars? The dollars in this example must be reduced to shillings, before we can find how many times 5 shillings are contained in them; that is, they must be reduced to 6ths, before we can find how many times & are contained in them. 23 6 138 (5 Ans. 273 bushels. 23138 and & are contained 273 times in 138. If 7 yds. of cloth will make 1 suit of clothes, how many suits will 48 yards make? If the question was given in yards and quarters, it is evident both numbers must be reduced to quarters. In this instance then, they must be reduced to 8ths. 739 and 48384 384 (59 354 30 630 suits. Ans. In the three last examples, the purpose is to find how many times a fraction is contained in a whole number. This is dividing a whole number by a fraction, for which we find the following rule: Reduce the dividend to the same denomination as the divisor, and then divide by the numerator of the fraction. Note. If the divisor is a mixed number, it must be re duced to an improper fraction. N. B. The above rule amounts to this; multiply the dividend by the denominator of the divisor, and then divide it by the numerator. At of a dollar a bushel, how many bushels of potatoes may be bought for of a dollar. as many times as 1 is contained in 3. is contained in Ans. 3 bushels. 0 If of a ton of hay will keep a horse 1 month, how many horses will of a ton keep the same time? 9 are contained in as many times as 3 are contained Ans. 3 horses. 10 At of a dollar a pound, how many pounds of figs may be bought for 2 of a dollar? 5ths and 4ths are different denominations; before one can be divided by the other, they must be reduced to the same denomination; that is, reduced to a common denominator. and = 16. are contained in 15 as many times as 4 are contained in 15. Ans. 3 lb. 15 2 At 7 dolls. a yard, how many yards of cloth may be bought for 575 dollars? 5 73=38 and 57% = 461. 5ths and 8ths are different de nominations; they must, therefore, be reduced to a common denominator. 3885 3 10 in 9. 46 304 and 412305. 2305 (304 2128 7177 yards. 304 177 From the above examples we deduce the following rule, for dividing one fraction by another : If the fractions are of the same denomination, divide the numerator of the dividend by the numerator of the divisor. If the fractions are of different denominations, they must first be reduced to a common denominator. If either or both of the numbers are mixed numbers, they must first be reduced to improper fractions. Note. As the common denominator itself is not used in the operation, it is not necessary actually to find it, but only to multiply the numerators by the proper numbers to reduce them. By examining the above examples, it will be found that this purpose is effected, by multiplying the numerator of the dividend by the denominator of the divisor, and the denominator of the dividend by the numerator of the divisor. Thus in the third example; multiplying the numerator of by 5 and the denominator by 1, it becomes 5, which reduced is 33 pounds as before. XXIV. A owned of a ticket, which drew a prize. A's share of the money was 567 dollars. What was the whole prize? of a number make the whole number. Therefore the whole prize was 5 times A's share. 567 Ans. 2835 dollars. A man bought of a ton of iron for 13 dollars; what was it a ton? make the whole, therefore the whole ton cost 7 times 135. A man bought was it a ton? 12 are 5 times as must cost of 40. 96 dollars. A man bought was it a ton? 13 7 Ans. 95 dolls. of a ton of iron for 40 dollars; what much as. If cost 40 dollars, 2 12 of 40 is 8, and 8 is of 96. Ans. of a ton of hay for 17 dollars; what are 3 times as much as . Since cost 17 dollars, must cost of 17, and must cost of 17. 17 (3 or multiplying first by 17 5 5/3 5 Ans. 23 dolls. 85 (3 281 dolls. If 4% firkins of butter cost 33 dollars, what is that a fir kin? 4. First we must find what costs. is part of 22, therefore will cost of 33 dollars, and § will cost 22 of 33 dollars. 33 5 The six last examples are evidently of the same kind. In all of them a part or several parts of a number were given to find the whole number. They are exactly the reverse of the examples in Art. XVI. If we examine them still farther, we shall find them to be division. In the last example, if 4 firkins instead of 42 had been given, it would evidently be division; as it is, the principle is the same. It is therefore dividing a whole number by a fraction; the general rule is, to find the value of one part, and then of the whole. To find the value of one part, divide the dividend by the nume→ rator of the divisor; and then to find the whole number, multiply the part by the denominator. Or, according to the two last examples, multiply the dividend by the denominator of the divisor, and divide by the numerator. N. B. This last rule is the same as that in Art. XXIII. This also shows this operation to be division. Note. If the divisor is a mixed number, reduce it to an improper fraction. If of a yard of cloth cost of a dollar, what will a yard cost? It is evident that the whole yard will cost 5 times, which 2 dollars. 2 = is If of a yard of cloth cost of a dollar, what is that a yard? If cost, 4,7 times or vard. must cost of; of is; being 11 dollars must be the price of a = |