To transform into twelfths the terms must be multiplied by 4. Therefore, when the fractions proposed have not the same denominator, reduce them to a common denominator, and then proceed as before. 130. We have seen that fractions cannot be added together without being of the same kind, or of the same denominator. Let us now determine a method which will enable us to reduce fractions to a common denominator. The common denominator of two or more fractions is such a number as will contain the several proposed denominators, without remainders; thus, in our last example, 12 is a common denominator of 3, 4, 6; so is 24, 36, &c., any multiple of 12; but 12 is the least, and it will be preferable to find that one, because we shall thereby shorten the operations. 131. What is the least common denominator of §, §, 18, 11, 48? First, let it be understood that to resolve a number into its prime factors is to divide it by 2, and the quotient by 2, and this second quotient again by 2. When the quotient can no longer be divided by 2, we must try by 3, then by 5, by 7, &c., until the quotient is a prime number. The several divisors and the last quotient are the prime factors of the numbers. Let the above denominators be resolved into their prime factors. 6=7×3 Now, we observe that the two of the first line, the three twos of the second, and the two of the third, are contained in the three twos of the fourth line, for which reason they are struck out; the three of the first line, the two threes of the third, and the three of the fourth are contained in the two threes of the fifth line, then they are also struck out. The product of the remaining factors will evidently be the least common denominator, or 2x2x2x3x3x5=360. 132. In practice, the operation is carried on in this manner: The several divisors, and the last quotient 5, are the factors found above. But, by inspection, we easily see that 6 and 8 are contained in 24; and, therefore, may be neglected, and the process is thus abridged : and the least common denominator is 2×3×3×4x5=360, as before. 133. To transform the proposed fractions into others, having 360 for their denominator, it is merely necessary to multiply the terms of every fraction by the quotient resulting from dividing 360 by its denominator. Now, 6 is contained in 360, 60 times; therefore, § 60 × 5 388. 60 × 6 45 X 5 45 × 8 20 × 7 20 × 18 Add together 51, 173, 73, 85, 123, 1611, and 13§. Here we notice that 2 and 4 are contained in 8, 3 in 6, and 5 in 20; thus, we have only to consider 6, 20, and 8 to determine the least common denominator :— Therefore, 2×2×3×5x2=120. Transforming all the fractions to this denomination, and adding together the numerators, we find 43, or 423. Put down in the column of the fractions, and the 4 is added with the wholes, and we find the sum to be 8241. 134. EXERCISES. 1. What is the sum of,, and ? §,, and ? 99 2. Add together 31, 47, 5, 11, and 13. 3. Find the sum of 34+7+12433+1413+19517. 4. A draper has 4 pieces of black cloth, the first contains 32 yards; the second, 25 yards; the third, 44 yards; and the fourth, 36 yards. How many yards are there altogether. 5. A workman performs, in 5 days, a work which it takes another 8 days to do. What part of the work would both together perform in 1 day? 6. In an army, the cavalry equal, and the artillery of the infantry. What part of the infantry are the cavalry and artillery together? 7. A weaving machine makes, the first day, cloth; the second day,; the third day, of a piece of ; and on the fourth day, How much has been done during these four days? 8. Two persons, intending to meet, start at the same time, from two different towns; the first could perform the distance in 8 days, and the second in 9 days. What part of the distance will they have achieved in 2 days? 9. Water is admitted into a reservoir through four apertures; the first would fill it in 24 hours, the second in 36 hours, the third in 30 hours, and the fourth in 40 hours. What part of the reservoir is filled up in 1 hour? Also in 3 hours? 10. A requires 5 yards of cloth for a coat, 23 yards for a pair of trowsers, 14 yards for a waistcoat, and yard for a pair of gaiters. How much does he want altogether? SUBTRACTION OF FRACTIONS. 135. From a piece of cloth, yard long, a tailor makes use of 3. How much remains? We are led by the question to subtract from , viz., to take away the numerator of the lesser from the numerator of the greater, then -=}. = Also, fa or }; }‡—&=&; H−1}=1 or 3. From 13% Remainder 64 or 63 From 1941 Take 176% Remainder 18 or 181 136. A performs of a work whilst B does of it. How much does one perform more than the other? Evidently, the lesser work must be taken away from the greater; but it happens sometimes that there is some difficulty in ascertaining at once the greater of two fractions, which have different denominators; the difficulty is easily removed by reducing them to the same denominator, and then the lesser is subtracted from the greater as before. Now, we find that =11, and 3=114; therefore, 13-3=117-111==the difference between the work performed by A and by B. 144 In like manner: 1--11--ά; 1-=-18-11. 24 24 137. Let it be required to subtract 5 from 11. 1st method: 111—54=23—16-69-32-37=6}. Find the difference of 17 and 14%. 1st method: 178-144-139-194-973-832=4y=238. 2nd method 178-144-1733-1438=238; or this method may be put under the following form, as in addition : 173 Since cannot be subtracted from, we add 1 whole to }} and §8+}}=77, from which is taken away, the remainder is ; now, add 1 to 14, (because 1 was added to the top quantity,) and we say 15 from 17 leaves 2. 138. EXERCISES. 1. Required, the difference of and; of and; of 18 and of 3 and 4. 2. Subtract 5 from 83; 3. Find the value of -411. 4. Find the value of 1-3÷43-34+148-94-1+1. 5. of an oil barrel is emptied. What part remains? 6. 12 7. The first day A performed of his work; the second, . How much has he to finish? of a work is achieved. How much remains to be done? 8. From a piece of linen, 30% yards long, 5 yards+73+10% were sold. How much is there left? 9. A jar weighs 5 lbs., when filled with water it weighs 367 lbs. The weight of the water is required. 10. of a pole stands in the ground and in the water. How much is there above the water? 11. A basin would be filled in 7 hours by a tap, and emptied in 10 hours by a pipe. If both tap and pipe be opened at the same time, when the basin is empty, how much water would it contain at the end of 1 hour? 12. The sum of two numbers is 73; one is 24. The other is required. |