Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

7. What is f of i of 7 ?

OPERATION.

} X 1 X 1 = 4*4 = 535 Ans. 8. What is of i of of ? Ans. 3785= 176 9. Change 14 of of of zy of 7 to a simple fraction.

Ans. 2326 Note ). If there be numbers in the numerators and denominators, that be alike, an equal number of the same value may be cancelled. 10. Reduce of 4 of 5 of 1 to a simple fraction.

STATEMENT.

CANCELLED.

[ocr errors]

3 X 4 X 5 X 7 3 X 4 X 5 X 7 3

Ans. 4 X 5 X 7 X Il 4 X 5 X 7 X11 11 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 4 of 17 of }} of 5%.

STATEMENT.

3 x 4 x 11 x 17 x 23
5 x 11 x 17 x 23 x 4

CANCELLED.

3 X 4 X11 X 17 X 23 3

Ans. 5 X X X X 17 X 23 X 4 5 12. Reduce f of g of ii of of to a simple fraction.

[blocks in formation]

1 X 8 X 9 X 5 X 3 1 X 8 X 9 X $ x 3 3

Ans. 5 X 9 X11 X 8 X 7 5 X 9 X11 X 8 X 7 77 13. Reduce of t of 7 of 1 of 44 to a simple fraction.

Ans. 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 té of g of Y to a simple fraction.

[blocks in formation]

15 X 8 X 7 15 X 8 X 7 35

Ans. 16 X 9 X 11 16 X 9 X 11 66

2 3 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 l over the 8, and the 2 under the 16. We then multiply the 5, and I, 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 }} of 33 of 9 to a whole number.

STATEMENT.

CANCELLED.

3

=

11 8 X 22X15X77 8 X 22 X 15 X77 3

3 Ans. 11 X 35 X 22 X 8 11 X 35 X 22X 8 1

5 16. Divide the continued product of 18, 24, 27, and 30, by the continued product of 20, 21, 9, and 10.

[blocks in formation]

2 6 9 3 18X24 X 27 X 30 18X24 x 27 x 30 324 20X21 X 9 X 10

. 20 X 21X 9 X 10

=995 Ans.

35 5 7 1 1 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.

CANCELLED. 2

2 2

2 20 X 19 X 18 X 17 X 16 X 15 X 14 X 13 X 12 X 11 10 X 9 X 8 X 7 X 6 X 5 X 4 X 8 X 2 X I 1 1 1 1 1 1

184756 Ans. 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.

[blocks in formation]

4X9X3X 8 X 225 4X9X3X8 X 225 225 6X6X4X6 X 11 6X6XA X6X 11 11

202 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 quo. tients 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 It is evident, that 24 is a com2)4 2 3 3

posite number, and that it is com

posed of the factors 2, 2, 3, and 3) 2 1 3 3

2 ; and, therefore, it may be di2 1 1 1 vided by any number, which is the 2 X 2 X3 X 2 = 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.

First Method.

OPERATION.

4) 468 4x2x3= 24 common denominator. 2) 162

4 6x3=18 numerator for = 131

6 4x5=20 numerator for &=if.

8 3x7=21 numerator for ž=1. Having first obtained a 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 fractfyns required.

H

Second Method.

OPERATION.

160

168
92•

3 x 6 x8= 144 numerator for 19.
5 X 4 X8= 160 numerator for
7 x 4 x 6= 168 numerator for s

4 x 6 x8 = 192 common denominator. Note. It will be perceived, that this method does not express the fractions in so low terms as the other.

From the above illustration we deduce the following

RULE.

99

Let compound fractions be reduced to simple fractions, mixed numbers to improper fractions, and whole numbers to improper fractions, by writing a unit under them ; then find the least common multiple of all the denominators by the last rule, and it will be the denominator required. Di. vide the common multiple by each of the denominators, and multiply the quotients by the respective numerators of the fractions, and their products will be the numeralors required.

Or, multiply each numerator into mil t'e denominators except its own for a new numerator ; enu . ''nomina. tors into each other for a common denom ol 2. Reduce and to a commen denominator,

Ans. 18. 3. Reduce y, is, and H.

Ans. 140, 1 4. Reduce 4, it, and T.

Ans. 41, 95, 9. 5. Reduce , , and 4.

Ans. ਤੇ ਉ, ਝੱਭ ਤੋਂ 6. Change ţ, s, š, and 15. Ans. , 168, 7. Change 1, 4, 5, and Z. Ans. 1920 1929, 138, 128. 8. Changed, f, g, and in Ans.

1485. 792 9. Reduce }, %, and 7.

Ans. 18, 19, 10. Reduce 4, 1, 1, and 54. Ans. 1$, 4, , 12 . 11. Reduce 1, 1, 5, 5, 5, and B. Ans. 12, 13, 34, 35, 3+, . 12. Change 3, 5, }, }, }, and I'. Ans. 1888, 18, 36, 36, 36. 13. Reduce , $, and 17.

Ans. 38, 36, 3. 14. Change 74,5141,7, and 8. Ans. 34, 44, 302, 15. Change , 4, 5, 7, and 9. Ans. 1, 2, 3, 4, 6.

84

880 360. 980, 1980, 1980, 1980

[graphic]
« ΠροηγούμενηΣυνέχεια »