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By resolving both fractions as above into their prime factors, and striking out for marking] all those that occur in both terms of the complete fraction, and are not found in the imperfect one, the required denominator appears to be 2•2•2.5=80.

By following a similar process for No. 8, the required numerator would also be found to be 80.

But processes like these may be rendered much more simple by cancelling while transferring the compound fraction to the slate, reserving, of course, such factors as are found in the imperfect fraction. Thus, 16 4•1•6•4

1.10.8.6 No. 7. 1.10-8-6

No. 8.

16-4.1.6.4 giving 80 as before for the required deficient term, when the two superfluous 6s are struck out.

9. Supply the denominator 10. Supply the numerator in 18 14:27

16.6 in

and
prove.

and
16.6

18 14:27'

prove.
2.3.3
Prime factors

2.7 X3:3•1
Prime factors

2.2.2.2X 2:3 2.2.2.2X 2

2:3:3 2:7 X3:3:3 Here, cancelling the marked 3s, and dividing by the superfluous 7, gives 44 as the required deficient term.

11. Supply the denominator 12. Supply the numerator in 24 16• 5 •14

18.35.16 in

and 18.35:16

prove.

2416.5.14' Prime factors for No. 11, 2•2•2•3_2•2•2•2x 5 x 2.7

2.3.3 X5.7 X 2.2.2.2 Here, as there is a 3 in the given term of the imperfect fraction, and none in the corresponding term of the perfect one, it must be supplied in both terms of the latter ; that is, the fraction must be multiplied by 3, giving, after proper cancellation, as marked above, 216 for the required deficient terms. Or the operation may be more simply performed by cancelling whilst transferring to the slate, as under :

24 8•1•7

9•7•8 By striking out the two 7s, and inserting 3, as before, we have again 216 as the deficient term.

and prove.

No. 12 does not differ from the above, except in the reversal of the terms.

13. Supply the denominator 14. Supply the numerator in 15 16•14:17

25.34•28 in

and
prove.

and

prove. 25•34.28'

1516-14.17' Here, as the corresponding terms of the equivalent fractions have no common factor, no advantage would result from resolving them into their prime factors. But, as 17 and 34, and also 14 and 28, admit of cancellation, and as the two 2s thence arising may be cancelled with 16, the complete fraction thus becomes 4, giving 93} for the deficient term.

15. Supply the denominator 16. Supply the numerator in 7 4•2•13

5•3•17 in

and
prove.

and

prove. 5•3•17'

4.2.13 Here, as no factor is common to the equivalent fractions, and the perfect fraction admits of no cancellation, the problem can only be solved as in Example 1 of this case, namely, by placing 7 in both terms of the perfect fraction ; that is, by multiplying by 7; and then dividing by the factors 4•2•13, or their product, 104.

Suggestive Questions. How can both terms of the perfect fraction be multiplied by 7? How can the other factors of the numerator on the left, and of the denominator on the right (4•2•13), be removed, thus leaving 7 as sole numerator (or denominator]? Will this solve the question ?

17. Supply denominators in 18. Supply the numerators each of the following pairs of in each of the following pairs

56 16.7 equivalent fractions:

of equivalent fractions : 48.3

56 (solely by division of perfect 48.3 18 27

(solely by division of perfractions);

(division

16:7 15

15 fect fraction); (division

20 12 and multiplication);

'1827 15

15 (division and multiplication);

and multiplication);

2012 (change mixed num-(division and multiplication); ber to improper fraction, and 9017

(change i to eighths, to eighths.) Prove as be- and mixed number to improper fore.

fraction.) Prove as before.

[ocr errors]

90 15

and prove.

19. Supply the denominator 20. Supply the numerator, 5 1.3.4.3.2

11.4•7•5.3

and

prove. 11.4.7.5.3

5

1•3•4•3•2' Multiply by 5, and divide by 3•4•6, or their product. Why? 21. Supply the denomina 22. Supply the numerators: 4 69.16 15 19 644• 3

16 tors :

;
;

;
644. 3

16'4 69.16' 15519' 504 504 5 12 5.18.2 7 9

6:30.5 4:18 ; ;

; 5

; 16.30-5 9

125.18•2 75 3.6 3. 6 350 9.15 16 5

12 6

6
;
;

;
4.18
12• 6 6'350- 9.15

165 5 7 5 4

686 320
;
Prove each

;

Prove each as 686 320

7 5 4 4 as before.

before. 23. Change 2 of 10 of 24. Change 24 of 1 of 1 of to an equivalent fraction, of to an equivalent fraction, with 16 as denominator, by in- with 16 as numerator, by inspection, and prove.

spection, and prove. 25. Change 21 of 1 to an 26. Change of 4 to an equivalent fraction, with 56 as equivalent fraction, with 56 as denominator, by inspection, and numerator, by inspection, and prove.

prove. 27. Change 6 of 1934 to 28. Change of 1998 to an equivalent fraction, with 4 an equivalent fraction, with 4 as denominator, ascertaining as numerator, ascertaining the the sole divisor by inspection, sole divisor by inspection, and

prove. 29. Change 40 of to 30. Change 1 of 1 to an an equivalent fraction, with 4 equivalent fraction, with 4 as as denominator, ascertaining the numerator, ascertaining the sole sole divisor by inspection, and divisor by inspection, and prove. prove.

32. Change into an equiva31. Change into an equiva- lent fraction, with as nulent fraction, with jí as de- merator, and prove by again nominator, and prove by again resolving it into a simple fracresolving it into a simple fraction of lowest denomination. tion of lowest denomination.

5

and into 3 33. Change

and
into

2
$ 5

equivalent simple fractions, and equivalent simple fractions, and prove by again resolving the

and prove.

34. Change

36. Change

prove by again resolving the first into a fraction whose nufirst into a fraction whose de- merator shall be 4, and the nominator shall be , and the second into one whose numesecond into one whose denomi- rator shall be 5. nator shall be 5.

9 2

and

into 35. Change

and into
9

equivalent simple fractions, and equivalent simple fractions, and prove by again resolving the prove by again resolving the first into a fraction whose nufirst into the denomination of merator shall be ž, and the g, and the second into the de- second into one whose numenomination of 9ths.

Trator shall be 9. A remarkable property of numbers is developed by the above examples, namely, that any number whatever may be expressed by a common fraction, whose numerator (or whose denominator) shall consist of any specified number, whether whole or fractional.

II.—Addition and Subtraction of Common Fractions. Suggestive Questions.-Can numbers of different denominations be added together or subtracted ? See Oral Arithmetic, Chap. III., Sect. IV., p. 91. What previous operation is necessary ?

Exercises for the Slate or Black-board. 1. Find the sum and the difference of } and 20.

Ans. Sum ; Diff. . 2. Find the sum of the five following fractions, and prove the operation by subtracting from it the sum of the last four : 4, š, , 5, 6. What fraction will be left ?

3. Add 4, 15, 14, 20, 25, and prove by subtracting the sum of the first four.

4. Add 73, 33, 93, 54, and prove by subtracting the sum of the last three. [Change the fractional parts of these numbers to the same denomination, add them, carrying what integers they may contain to the given integers.] 5. What is the sum and difference of and

2

63 74

Ans. Sum, ; Diff., di58. 6. Add 4, 3, 15, and y, and prove the operation by giving

these 4 fractions a decimal form, and changing their sum into a common fraction, which, of course, will be equivalent to the sum of the four common fractions.

7. Add 3, 2, and 78, and prove by the same process as in the last example. III.--Multiplication and Division of Common Fractions. [See Oral Arithmetic, Chap. III., Sect. V., p. 93.]

Exercises for the Slate or Black-board. 1. Multiply by 4, 3 by 7, zda by 3, and by 7, all by division; or, which is the same thing, by cancelling what would otherwise be equal factors in both terms of the product. Prove the operations by performing them by multiplication, and bringing each fraction to its lowest denomination.

2. Divide & by B, or, which is the same thing, change into a simple fraction, or integer.

Ans. 2. 3. Divide by ti by 3; by.

De Remark, that in the above example, and in all other cases of division where the numerators (or where the denominators) are alike in the divisor and dividend, the quotient is formed by removing the like term in the dividend, and puttingthe unlike term of the divisor in its place. Thus, £=*=;; &=3= Why is this so ? Perform the above example in the usual method, and see.

4. Divide $ by t; } by 23 ; 1 by ; 43 by 4$; each at a glance.

5. Divide { by 15, by cancelling what would otherwise be equal factors in the quotient Solution. == x1= 532; or, omitting the superfluous steps, 8.15=$:

6. Divide 44 by 3t by inspection, first cancelling equal factors in the numerators, and equal factors in the denominators. Why?

7. Divide si by }; by lhi o by mg; by 24; by cancelling as in the preceding example

. Prove by reproducing the dividend as in division of integers; that is, by considering the divisor and quotient as factors of the dividend.

8. Change of of of of of ļ, by cancellation, into a simple fraction, by inspection, and prove by dividing the

Ans. .

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