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Cancelling 3 from the 15 of the numerator and from the 9 of the denominator gives 5 in place of the former and 3 in place of the latter. As no further cancellation can be made, we multiply the remaiņing factors toge

1 X 5 ther, which gives

for the result. 3 X 2 6

5

The following is a convenient form of writing the work:

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(h.) Reduce the following to their lowest terms:

23.

21.

7 X 9 8 x 12

10 X 12 22.

8 X 15 4 X 7 3 x 14

5 X 9 24.

6 X 15

7 X 8 X 9 25.

14 x 12 x 18

8 X 9 X 10 26.

25 x 24 x 27

15 x 17 x 24 28.

51 x 48 x 30

21 x 16 x 32 x 49 29.

24 X 28 X 35 X 36

17 x 13 x 36 x 28 30.

38 x 40 x 51 x 39

10 X 11 X 12 X 9 X 8 31.

24 x 27 x 60 x 33 x 55
15 x 63 x 12 x 77
33 x 40 x 56 x 144

144 x 18 x 75 x 132
33.

150 X 1728 X 96 X 38*

32.

9 X 10 X 11 27.

15 x 55 x 3

299 x 77 x 374 x 262 34.

121 X 713 X 324 x 561

86. To find a Fractional Part of a Number.

1. What is & of 715 ?

1st Solution. — of 715 equals 5 times $ of 715. f of 715, found by dividing 715 by 6, is 119}, and 5 times 1194 is 5954. Hence, of 715 5956.

2D SOLUTION. — of 715 t of 5 times 715. 5 times 715 = 3575, and $ of 3575 is 5958. Hence, f of 715 5958, as before.

(a.) As of 715 equals 5 times & of 715, or ļ of 5 times 715, 80 of any number equals 5 times $ of that number, or $ of 5 times that number; .07 of any number equals 7 times .01 of that number, or .01 of 7 times that number, etc. Hence

(b.) To get a fractional part of a number, we either divide that number by the denominator of the fraction and multiply the quotient by the numerator, or we multiply the number by the numerator of the fraction and divide the product by the denominator.

(c.) What is -
2. $ of 1728? 5. i of 6181 ?

8. }} of 4284 ? 3. $ of 2946 ? 6. of 1359 ?

9. } of 1235 ? 4. of 4243 ? 7. f of 6453 ? 10. z of 1235 ? 11. What is 4 times 389 ? SOLUTION. -47 times 389 4 times 389 + f of 389

1556 + 259} = 1815 (d.) What is 12. 57 times 957 ?

16. 26} times 1856 ? 13. 25 times 879 ?

17. 3736 times 7431 ? 14. 9 times 745 ?

18. 1974 times 3886 ? 15. 8f times 1347 ?

19. 48 31 times 7932? 20. What is .009 of 32.45 ?

1st SOLUTION.—.009 of 32.45 equals 9 times .001 of 32.45. .001 of 32.45, found by removing the decimal point 3 places to the left (see 14), is .03245, and 9 times .03245 .29205.

20 SOLUTION.-.009 of 32.45 = .001 of 9 times 32.45. 9 times 32.45 292.05, and .001 of this, found by removing the decimal point 3 places further to the left, is .29205, as before.

(e.) What is

21. .05 of 375 ?
22. .13 of 87.3?
23. .016 of 245.03 ?
24. .25 of .25 ?

25. .623 of 4.86 ?
26. .001 of .001?
27. 1.3* of 34.2 ?
28. 2.5 of 25 ?

29. 2.03 of 832?
30. 1.54 of .761 ?
31. 3.7 of .896 ?
32. 2.003 of 2.003 ?

* The 1.3 should be read as an improper fraction, as should all other numbers similarly situated.

87. Reduction of Compound Fractions to Simple Ones.

(a.) To reduce a compound fraction to a simple one is merely to find a fractional part of a fraction.

1. Reduce the compound fraction 3 of Iz to a simple fraction, i. e. find 1% of pe

SOLUTION. — io of it may be found by writing pe, and then making 9 a factor of the numerator and 10 a factor of the denominator (see 83, ex. 15), cancelling and reducing as in the written work below:

3 9

5 x 0 10 12 12 X 10

4 2 2. Reduce #1 of 4of log to a simple fraction.

SOLUTION. — Writing 10g, or its equal, 37, as the number of which the fractional part is to be obtained, we find 18 of by making 49 a factor of the numerator and 54 a factor of the denominator; and {1 of this result by making 11 a factor of the numerator and 21 a factor of the denominator.

The written work would take the following form :

5

3

of

8

[blocks in formation]

(b.) These solutions show that, to reduce a compound fraction to a simple one, we first reduce the mixed numbers, if there are any, to improper fractions, then make all the numerators of the compound fraction factors of the new numerator, and all the denominators factors of the new denominator, and then cancel and reduce.

(c.) Reduce each of the following compound fractions to simple

ones.

3. of .
4. šof .
5. 18 of .
6. f of .
17. of
8. of 3

9. 11 of 17

10. i of of . 11. of f of t. 12. $ off of 23. 13. of 11 of. 14. & of is of . 15. 4 of of 25. 16. of Y of .

17. % of f of z of 4 of f.
18. i of off of 10 of 11.
19. of li of 4 of 7.
20. f of of of 7 of 45.
21. 1 of f of f of 13.
22. of sof i of $of & off of } of f of yo.

88. Reduction of Vulgar Fractions to Decimals.

(a.) To reduce a vulgar fraction to an equivalent decimal, we have only to get the part of 1 which the given fraction indicates, carrying out the work to the required number of decimal places.

(b.) Practically, the reduction is made by annexing zeroes to the numerator and dividing by the depominator.

NOTE.—In most cases it will be sufficiently exact to carry out the division to 3 or 4 decimal places, bat sometimes it is necessary to carry it out much further.

1. Reduce to a decimal fraction. 14)5.0000).35713

SOLUTION. — * or of 1 it of 5 80

- 14 of 50 tenths 14 of 500 hundredths 100

1 of 5000 thousandths, etc. which may be

found as in the written work opposite. Hence, 20

1 .3571 +. (c.) Reduce the following to decimal fractions, carrying out the division, when only approximate values can be obtained, to 6 places of decimals.

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1. What is the value of 77 of a mile in units of lower denomi nations, i. e. in fur. rd. yd. etc.?

1st SOLUTION.—of a mile 4 of 7 miles, which is found by dividing 7 m. O fur. 0 rd. O yd, 0 ft. O in. by 12, as in compound division.

2ND SOLUTION.-Since 1 m. 8 fur., je of a mile must equal íg of 8 fur.

4; fur. But of a fur. = of 40 rd. 26ş rd. But şof a rd. } of 5 yd.

šof Į yd. 3}yd. But f of a yd. = f of 3 ft. 2 ft. Hence, te of a mile 4 fur. 26 rd. 3 yd. 2 ft.

3D Solution. — There must be 8 times as many furlongs as miles, or in Te of a mile, 8 times the

fur. 44 fur. But there must be 40 times as many rods as furlongs, or in fur. 40 times rods = 26rods. But there must be 54 or jų times as many yards as rods, or in f of a rod į times fyd.

3ş yd. But there must be three times as many feet as yards, or in of a yard, 3 times f ft. 2 ft. Hence, te of a mile 4 fur. 26 rd. 3 yde. 2 ft.

12

(a.) Reduce each of the following to units of a lower denomination: 2. § of £1.

6. i of 1 acre. 3. šof 1 gal.

7. of 1 ton. 4. 9 of 1 wk.

8. 23 of 1 mile. 5. f of 1 cord.

9. of 1 degree. 10. Reduce .5135 of a pound Troy to units of a lower denomination. SOLUTION. - There must be 12 times as many ounces as

.5135 pounds, or, in this case, 12 times .5135 of an ounce = 6.1620 oz. = 6.162 oz. But there must be 20 times as many

6.1620 pennyweights as ounces, or, in .162 of an ounce, 20 times

20 .162 of a pennyweight 3.240 dwt. 3.24 dwt. But there must be 24 times as many grains as pennyweights, or, in

3.240 .24 of a pennyweight, 24 times .24 of a grain

24 Hence, .5135 of a lb. 6 oz. 3 dwt. 5.76 gr.

6.76 NOTE. -- It will be seen that, practically, the reductions in the last example, as well as in those immediately preceding it, were made by multiplying the fractional part of each denomination by the number which expresses how many units of the next lower denomination are equal to a unit of the denomination of the fraction. (b.) Reduce each of the following to lower denominations: 11. .4595 of a £.

16. .00568 of a cord. 12. .3572 of a ton.

17. .475 of a bushel. 13. .076 of a week.

18. .396 of a lb. 14. .00276 of a league.

19. .0005 of a mile. 15. ,1824 of a rood.

20. ,1876 of an acre.

5.76 gr.

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