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CANCELLED.

2 2

2 8 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 6 X 4 X 3 X 2 X X 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.

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4X9X3X8X225 4X9X3X8X225 225 6X6X4 X6 X 11 6 X 6X4 XBX 11 11

2041 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 give en 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 X 3 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 =.

8 3x7=21 numerator for }={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 fractions required.

H

Second Method.

OPERATION.

3 x 6 x8= 144 numerator for 194.
5 X 4 X8= 160 numerator for
7 x 4 x 6 = 168 numerator for s
4 X 6 X8= 192 common denominator.

192 168 192

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.

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. Die 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 numerators required.

Or, multiply each numerator into all the denominators except its own for a new numerator; and all the denominators into each other for a common denominator. 2. Reduce and to a common denominator.

Ans. 12, 18. 3. Reduce }, 75, and .

Ans. 148, 18, 19. 4. Reduce 4, it, and

Ans. 4, 4s, 18. 5. Reduce ,, and 1.

Ans. } , 3's, 6. Change d, is, f, and 15. Ans. Po, 1, 160,

Tor, 180 7. Change 1, 5, }, and .

. , ,

120, 1920, 120, 125 . 8. Change , , , and i. Ans. 1983, 1983, 1988, 1998 9. Reduce %, , and 7*.

Ans. , 46, 4 10. Reduce 4, i, it, and 54. Ans. 1, il, 15, a. 11. Reduce 1, , , , , and P. Ans. 14,23,24, 24, 31, 31. 12. Change $,$, , 1, 4, and I's. Ans. 18, 14, 18, 36, 366,36 13. Reduce , f, and 72.

Ans. 38, 36, ft. 14. Change 73,541, 7, and 8. Ans. 341, 24, 302, 15. Change I, 4, 5, 7, and 9. Ans. 1, 46, 4, 4, se

84

360

308 3 5 2

VIII. To reduce fractions of a lower denominaticn to a higher. 1. Reduce of a farthing to the fraction of a pound.

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OPERATION.

1 x gr. =*=jd. This question may be anI X jd. = 10gs.

alyzed thus ; since 4 far

things make a penny, there 25 X Togs. = zbo£.

will be as many pence as farthings ; therefore of of a farthing is = f of a penny. Again, as 12 pence make a shilling, there will be iz as many shillings as pence, therefore it off of a penny is TỚg of a shilling. As 20 shillings make a pound, there will be an as many pounds as shillings, therefore zb of idg of a shilling is too of a pound. Q. e. d.

The operation of this question may be abridged thus :

OPERATION.

4 1 1 1
X-X Х

12 20

1

Ans. 2160

Hence the following

RULE.

Let the given fraction be reduced to a compound one by comparing it with all the denominations between the given one and the one to which it is required to reduce it ; then reduce this compound fraction to a simple one. 2. Reduce of a grain Troy to the fraction of a pound. 4 X 1 X 1 X 1

1

Ans. 7 x 24 x 20 x 12 10080

3. What part of an ounce is p of a scruple ? 8 X1 X1 1

Ans. 10 X 8 X 8 80

4. What part of a ton is of an ounce ? 4 X 1 X 1 X1 x 1 1

Ans. 5 X 16 X 28 X 4 X 20 44800

5. What part of a mile is g of a rod ? 8 X1 X1 1

Ans. 9 X 40 X 8 360 6. What part of 3 acres is g of a square foot ? 4 X 1 X 1 X1 X1 1

Ans. 9 X 272 X 40 X 4 X 3 294030 7. What part of 3hhds. is # of a quart ? 4 X1 X 1 X1 1

Ans. 7 X 4 X 63 X 3 1323 8. What part of 3 yards square, are 3 square yards ?

Ans. s. 9. What part of t of a solid foot is t of a foot solid ?

Ans. . IX. To reduce fractions of a higher denomination to a lower. 1. Reduce Too of a pound to the fraction of a farthing.

Ans. . We explain this question in the following manner.

As shillings are twenleto X = Tido = tys. tieths of a pound, there to X *= = d.

will be 20 times as many 385 X*= 34qr. Ans.

parts of a shilling in

Tato of a pound, as there are parts of a pound ; therefore Tado of a pound is equal to tudo of ze = 1= of a shilling. And as pence are twelfths of shillings, there will be twelve times as many parts of a penny in / of a shilling, as there are parts of a shilling ; therefore o of a shilling is equal to 7 of 1 = + = of a penny. Again, as farthings are fourths of a penny, there will be 4 times as many parts of a farthing in 35 of a penny, as there are parts of a penny ; therefore 45 of a penny are equal to 5 of 1 =

of a farthing. Q. e. d. The operation of this question may be facilitated by the following manner.

OPERATION.

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