876 6. Change to a mixed number. 13. Change 125 to an improper fraction. V. To change or reduce compound fractions to simple fractions. MENTAL OPERATIONS. 1. What part of an orange is a 2. What part of an apple is a 3. What part of a bushel is a 4. What part of a quart is a of a half? FOR THE SLATE. Ans. . This question may be analyzed by saying, if of an apple be divided into 5 equal parts, that of an apple; and, if of be of will be 7 times as much. 7 be, ofwill be 4 one of these parts is of We therefore induce the following RULE. Change mixed numbers and whole numbers, if there be any, to improper fractions; then multiply all the numera tors together for a new numerator, and all the denominators together for a new denominator; the fraction should then be reduced to its lowest terms. 7. What is of 1 of 7? OPERATION. 7××7=45 Ans. 8. What is 9. Change of of & of ? Ans... Ans. 2. of 1 of 2 of 2% of 7 to a simple fraction. 2720. NOTE 1. If there be numbers in the numerators and denominators, that be alike, an equal number of the same value may be cancelled. In performing this question, we perceive that there is a 4 and 5 and 7 among the numerators, and also the same numbers among the denominators; these we cancel before we commence the operation. 11. Required the value of of of 1 of 13 of 52. 12. Reduce of off of § of to a simple fraction. NOTE 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. 14. Reduce of of 7 to a simple fraction. In performing this question, we find that the 15 among the numerators and the 9 among the denominators may be divided by 3, and that the quotients will be 5 and 3. We write the 5 above the 15, and the 3 below the 9. We also find an 8 among the numerators, and a 16 among the denominators, which may be divided by 8, and that the quotients will be 1 and 2. We write the 1 over the 8, and the 2 under the 16. We then multiply the 5, and 1, and 7 together for a new numerator, and the 2, and 3, and 11 together for a new denominator. That the result will be the same by this process as by the other, is evident from the fact, that the multiples of any number have the same ratio to each other, as the numbers themselves. This cancelling principle, when well understood, will often facilitate the operations of many questions, when the divisors and dividends have a common denominator. 15. Reduce of 3 of 15 of 9§ to a whole number. 16. Divide the continued product of 18, 24, 27, and 30, by the continued product of 20, 21, 9, and 10. 17. Divide the continued product of 20, 19, 18, 17, 16, 15, 14, 13, 12, and 11, by the continued product of 10, 9, 8, 7, 6, 5, 4, 3, 2, and 1. 2 2 CANCELLED. 2 3 2 2 20 X 19 X 18 X 17 X 16 X 15 X 14 X 13 X 12 X 11 10 × 9 × 8 X 7 X 6 × 5 × 4 × 3 × 2 × I 184756 Ans. 1 1 1 1 1 1 NOTE. In this question the product of the quotients of 2, 3, 2, and 2 is cancelled by the product of 4, 3, and 2 in the lower line. Any numbers may be cancelled, when their product is equal to the product of certain other numbers, as in the following question. 18. Divide the continued product of 4, 9, 3, 8, and 225 by the continued product 6, 6, 4, 6, and 11. STATEMENT. 4X9X3X8X225 6×6×4×6× 11 6×6×4×6× 11 11 20 Ans. As the product of 4 times 9 in the upper line is equal to the product of 6 times 6 in the under line, they cancel each other; and as the product of 3 times 8 in the upper line is equal to 4 times 6 in the under line, they cancel each other. VI. To find the least common multiple of two or more numbers, that is, to find the least number, that may be divided by them without a remainder. RULE. Divide by such a number, as will divide most of the given numbers without a remainder, and set the several quotients with the several undivided numbers in a line beneath, and so continue to divide, until no number, greater than unity, will divide two or more of them. Then multiply all the divisors, quotients, and undivided numbers together, and the product is the least common multiple. 1. What is the least common multiple of 8, 4, 3, 6 ? 2)8 4 3 6 2)4 2 3 3 3)2 1 3 3 It is evident, that 24 is a composite number, and that it is composed of the factors 2, 2, 3, and 2; and, therefore, it may be divided by any number, which is the 2111 2 X2 X3X2=24 Ans. product of any two of them; and, as the given numbers are either some one of these, or such a number as may be produced by the product of two or more of them, it is evident, therefore, that 24 may be divided by either of them without a remainder. Q. e. d. 2. What is the least common multiple of 7, 14, 21, and 15? Ans. 210. 3. What is the least common multiple of 3, 4, 5, 6, 7, and 8 ? Ans. 840. 4. What is the least number, that 10, 12, 16, 20, and 24 will divide without a remainder ? Ans. 240. 5. Five men start from the same place to go round a certain island. The first can go round it in 10 days; the second in 12 days; the third in 16 days; the fourth in 18 days; the fifth in 20 days. In what time will they all meet at the place from which they started? Ans. 720 days. VII. To reduce fractions to a common denominator; that is, to change fractions to other fractions, all having their denominators alike, yet retaining the same value. 1. Reduce, §, and to a common denominator. Having first obtained common multiple of all the denominators of the given fractions by the last rule, we assume this, as the common denominator required. This number (24) we divide by the denominators of the given fractions, 4, 6, and 8, and find their quotients to be 6, 4, and 3, which we place under the 24; these numbers we multiply by the numerators, 3, 5, and 7, and find their products to be 18, 20, and 21, and these numbers are the numerators of the fractions required. H |