17. If 7 were to be the denominator to the following quantity, To reduce compound fractions to simple fractions. 1. What is of } ? OPERATION. This question may be analyzed by saying, ×}=} Ans. If of an orange be divided into 4 equal parts, one of those parts is of the orange; and, if of be, it is evident that of will be seven times as much. And 7 times is. If, therefore, of be, of will be 3 times as much; and 3 times is 1. RULE. Change mixed numbers and whole numbers, if there be any, to improper fractions; then multiply all the numerators together for a new numerator, and all the denominators together for a new denominator; the fraction should then be reduced to its lowest terms. If there be numbers in the numerator similar to those in the denominator, they may be cancelled in the operation. Ans. 1582TT. 756 TT• of 157 of 57% of 100? Ans. 5087800 = 575812. 5. What is of § of 4 of 21? of 4 of 7 of 1? 7 11.17 23 7 XX 17 23 X: X 7 RULE 2. When there are any two numbers, one in the numerators, and the other in the denominators, which may be divided by a number without a remainder, the quotients arising from such division may be used in the operation of the question, instead of the original numbers. The quotients also may be cancelled, as other numbers. 1. Reduce of of off to its lowest terms. be 2 and 1. We write the 2 above the 14, and 1 below the 7. We also find a 21 among the numerators, and a 27 among the denominators, which may be divided by 3, and that their quotients will be 7 and 9. We write the 7 above the 21, and 9 below the 27. We again find a 5 among the numerators, and a 25 among the denominators, which may be divided by 5, and that their quotients will be 1 and 5. We write the 1 over the 5, and the 5 below the 25. We then multiply the 4, 2, 7, and 1 together for a numerator = 56, and the 1, 9, 5, and 11 for a denominator =495. The answer will therefore be 5. 3. What is the value of 4 of 2% of 13 of 14 of $ 34 ? NOTE. The above rule will apply, when the product of several numbers is to be divided by the product of other numbers. 4. What is the continued product of 8, 4, 9, 2, 12, 16, and 5, divided by the continued product of 40, 6, 6, 3, 8, 4, and 20 ? The product of 4 and 9 in the upper line is equal to the product of 6 and 6 in the lower, therefore they are cancelled; and the product of 2 and 12 in the upper line is equal to the product of 3 and 8 in the lower line; also the product of 16 and 5 in the upper line is equal to the product of 4 and 20 in the lower line; these are all cancelled. We also find, that the 8 in the upper line and the 40 in the lower line may be divided by 8, and their quotients will be 1 We write the 1 above the 8, and the 5 below the 40. usual process, we now find our answer is t. and 5. By the 5. What is the continued product of 12, 13, 14, 15, 16, 18, 20, 21, and 24, divided by the continued product of 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11? 3 2 3 2 2 2 7 2 12×13×14×15×16×18×20×21×24 26208 2×3×4×5×6×7×8×9×10×11 1 1 1 1 1 1 1 1 1 CASE VII. -2382 11 [Ans. To find the least common multiple of two or more numbers. 1. What is the least common multiple of 4, 6, 8, 16, and 20? OPERATION. 2)4 6 8 16 20 2)2 3 4 8 10 2)1 3 2 4 5 1 3 1 2 5 In this operation, we have divided the given numbers by that number which would divide most of the given numbers without a remainder, that is, by one of the prime factors, and have 2×2×2×3×2×5 = 240 Ans. continued the division until no number would divide two of them. We have then multiplied all the divisors and the last quotients together and found their product to be 240, which is a common multiple of the given numbers, 4, 6, 8, 16, and 20. It is not only a common multiple, but it is the least common multiple. To prove this, we assume the following admitted propositions, which are self-evident : 1. Every number not prime itself is the product of two or more primes or factors, and is resolvable into its original primes by division. 2. The least number which contains all the prime factors of two or more numbers is the least common multiple of those quantities. The factors of each number in the question are as follows: 2, 2, and 2=2×2×2=8 16 66 20 66 2, 2, 2, and 2 = 2 × 2 × 2 × 2 = 16 2, 2, and 5 =2×2×5 = 20 Now 240 is the least number which contains all the factors common to each of these numbers, 2 × 2 × 2 × 2 × 3 ×5= |