Page 16 - SM inner class 7.cdr

The LCM (least common multiple) of the = 11 An swer
denominators 5 and 6 is 30. 48
So, we convert the given fractions into Ex am ple 7. Sim plify: 4 2 - 3 1 + 2 1
equivalent fractions with denominator 30. 3 4 6
We have, So lu tion : We have,
14 ( 14 ´ 6) 84 2 1 1
= = = [since 30 ¸ 5 = 4 - 3 + 2
5 5 ( ´ 6) 30 3 4 6
6] First we convert mixed fractions into improper
23 ´ 5 115 84 115 fractions.
and, = = + [since 30 ¸ 6 = 5]
6 ´ 5 30 30 30
2
(4 ´ 3 + ) (3 ´ 4 + ) (2 ´ 6 + )
1
1
14 23 84 115 = - +
Thus, + = + 3 4 6
5 6 30 30
1
2
(12 + ) (12 + ) (12 + )
1
(84 + 115 ) = – +
= 3 4 6
30
14 13 13
199 = – +
= 3 4 6
30
19 The LCM (least common multiple) of the
= 6 An swer denominators 3, 4 and 6 is 12.
30
17 15 2 3, 4, 6
Ex am ple 6. Find the dif fer ence of and .
24 16 2 3, 2, 3
So lu tion : The LCM (least com mon mul ti ple) 3 3, 1, 3
of the de nom i na tors 24 and 16 is 48. 1, 1, 1
So, we convert the given fractions 2 24, 16 [Therefore, LCM = ´ ´ =2 2 3 12]
into equivalent fractions with 2 12, 8 So, we convert the given fractions into
denominator 48. 2 6, 4 equivalent fractions with denominator 12.
We have, 2 3, 2 We have,
3 3, 1
17 ( 17 ´ 2) 34
3
2
4
= = = 1, 1 = (14 ´ ) – (13 ´ ) + (13 ´ )
24 ( 24 ´ 2) 48 (3 ´ ) (4 ´ ) (6 ´ )
4
3
2
[Therefore, LCM = 2 ´ 2 ´ 2 ´ 2 ´ 3 = 48] 56 39 26
= – +
[since 48 ¸ 24 = 2] 12 12 12
–
15 ( 15 ´ 3) 45 (56 39 + 26 )
and, = = [since =
16 ( 16 ´ 3) 48 12
–
48 16 = 3] (82 39 )
¸
=
45 34 12
Clearly, >
48 48 43
=
15 17 12
Therefore, >
16 24 (Converting it to mixed fraction)
15 17
Hence, difference = - 7
16 24 = 3 An swer
12
45 34
= - Comparison of Fractions
48 48
(45 - 34 ) When two given fractions are unlike i e. . one
= positive and the other negative, it is to be
48
noted that :
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Mathematics In Focus - 7