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I. Cancelling applied to Compound Fractions.

Rule 1. — If there be numbers in the numerators and denominators, that be alike, an equal number of the same value may be cancelled.

1. Reduce } of off of 3 of 5 to a simple fraction.

STATEMENT.

CANCELLED.

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OPERATION.

2X3X4X7X8 2X8X4X7X8 14

Ans. 3X4X5X8X9 8X4X5X8X9 45 In this question, we find a 3, 4, and 8 among the numerators, and also the same numbers among the denominators. These we cancel before we commence the operation. 2. What is the value of lof i of of it?

We find in this question, 7X 8 X11X17 7

8, 11, and 17 among the nu

Ans. 8X11X17 X 19 19

merators, also the same num.
bers
among

the denominators.

These we cancel. 3. What is the value of of is of is of 16 of 1 of $25. 7X8 X13X 7 X15 X 25 1225

$75 Ans. 8 x 13 x 15 X 10 X 17 X 1 170

4. Reduce ii of li of iî of if of 4 to a simple fraction. 5 XULX12X17 X19 5

14 Ans. 11X12X17X19X4 4

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5. Required the value of 1 of 1 of 18 of of 40. 7X9XXØX 18 X 40 280

= 11; Ans. 9X10 X 13 X 24 24

6. Reduce l} of 1 of 2 to its equivalent value. 11X15X16 11

14 Ans. 5X16X7 7

7. What is the value of i of 14 of 75 of 34 of $ 18? 5 XIIX 7 XX9 X 18 18

$ 18 Ans. 11x x 19 x $ x 1 1

8. What is the value of 11 of li of si of $ 71? 7 XII X 25 X 31 7

$ 1.75 Ans. 11 X 25 X 31 X 4 4

9. What is of of lof 3gallons ?

4X 9 X 17 X18 4
9X17 X18X 5

5 gal. Ans.

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.

OPERATION.

1. Reduce 4 of 1 of is of i to its lowest terms.

In performing this question, 2 7 1

we find that 14 among the 4X14X21 X 5 56 numerators, and 7 among the

Ans. denominators, may be divided 7 x 27 x 25 x 11 495

by 7, and that their quotients 9 5

will 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

56, and the 1, 9, 5, and 11 for a denominator = 495. The answer will therefore be 56 2. Reduce 1 of 1 of ii of Å to a simple fraction. 2 6 2

1
14X18XXØX 8 24

Ans.
15 X 25 X 11x2x 275
5 5

8

numerator

495

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3. What is the value of of 2 of 1 of 1 of $34?
1

3 2 2
4X 9 XA$X14X34 27

$6.75 Ans.
7 X 20 X 16 X 17 X 1 4

1 4 4 1 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?

1
8X4X9X2X12X16X5 1

Ans.
40 X 6X6X3X8X4 X20 5

5 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 and 5. We write the 1 above the 8 and the 5 below the 40. By the usual process, we now find our answer is ž.

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 X 13X14X15X16X 18 X 28 X 21 X 24 26208

=238211 2X3 X4 X5 X6X7 X8X9XXØX11 11

[Ans. 1 1 1 1 1 1 1 1 1

II. In finding the common multiple of two or more numbers, any one number that will measure another

may

be cancelled.

1. What is the least common multiple of 4, 6, 8, 12, 16, 10, and 20 ? 4) 4 6 8 12 16 10 20

4X3X4X5=240 Ans. 3 4 5

By examining this question, we find that 8 may be divided by 4, 12 by 6, 16 by 8, and 20 by 10; therefore we cancel 4, 6, 8, and 10.

2. What is the least common multiple of 5, 15, 30, 7, 14, and 28 ? 2) 5 15 30 7 14 28

2X15X14=420 Ans. 15

14

In this question, we find that 15 may be measured by 5, 30 by 15, 14 by 7, and 28 by 14; we therefore cancel 5, 15, 7, and 14.

3. What is the least common multiple of 1, 2, 3, 4, 5, 6, 7, 8, and 9? 2) 1 2 3 4 5 6 7 8 9 2X3X5X7X4X3=2520 3) 5 3 7 4 9

[Ans. 5 1 7 4 3

4. What is the least common múltiple of 9, 8, 12, 18, 24, 36, and 72 ? 9 8 12 18 24 86 72

72 Ans.

5. What is the least number that 18, 24, 36, 12, 6, 20, and 48 will measure ?

4) 18 24 36 12 6 20 48 4X3X3X5X4=720 Ans. 3)9

5 12 3

5 4

III. SINGLE PROPORTION,

PERFORMED BY CANCELLING.

RULE. When the first and second terms, or the first and third terms, can be divided by any number without a remainder, their quotients may be used in the operation of the questions instead of the terms themselves.

1. If 14cwt. of logwood cost $ 56, what cost 95cwt. ?

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2. If 23 men, in one month, can dig a ditch 19 rods long, 8 feet wide, and 3 feet deep, how many men would it require to dig a ditch 57 rods long, 4 feet wide, and 6 feet deep, in the same time?

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