Page 75 - classs 6 a_Neat

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LCM of 2, 3, 6, 9 = (2 × 3 × 3) = 18. EXAMPLE 13: Compare the fractions and ,
So.we convert each of the given fractions into an equivalent fraction with 18 as the denominator. 8 12
SOLUTION: By cross multiplying, we get:
Thus, we have: 3 × 12 = 36 and 8 × 5 = 40.

1 = 1 9 = 9 2 2 6 = 12 Clearly, 36 < 40. 5
×
×
; =
3
×
×
2 2 9 18 3 3 6 18 Hence, 8 < 12
×
5 = 5 3 = 15 and 4 = 4 2 = 18
×
×
×
6 6 3 18 9 9 2 18 EXAMPLE 14: Compare the fraction 5 and 6 .
9 12 15 8 9 11
Hence, the required like fractions are , , and SOLUTION: By cross multiplying, we get:
18 18 18 18
5 × 11 = 55 and 9 × 6 = 54
Clearly, 55 > 54.
COMPARISON OF FRACTIONS 5 6
Hence, >
RULE 1 Among two fractions with the same denominator, the one with the greater numerator is the greater of 9 11
the two. II. METHOD OF CONVERTING THE GIVEN FRACTIONS INTO LIKE FRACTIONS
EXAMPLE 11: (i) 8 > 5 (ii) 7 > 6 (iii) 9 > 7
9 9 11 11 10 10 RULE Change each of the given fractions into an equivalent fraction with the denominator equal to the
LCM of the denominators of the given fractions. Now the new fraction are like fractions, which may be com-

COMPARISON OF FRACTIONS WITH THE SAME NUMERATOR pared by Rule.1
5 8
RULE Among two fractions with the same numerator, the one with the smaller denominator is the greater of the EXAMPLE 15: Compare the fractions 6 and 9
two.
SOLUTION: LCM of 6 and 9 = (3 × 2 × 3) = 18.
EXAMPLE 12: (i) 5 > 5 (ii) 3 > 3 (iii) 9 > 9 Now, we convert each one of 5 and 8 into an equivalent fraction having 18 as
6 8 5 7 10 11 denominator. 6 9

GENERAL METHODS OF COMPARING TWO FRACTIONS 5 = 5 × 3 = 15 = 8 = 8 × 2 = 16
6 6 × 3 18 9 9 × 2 18
I. METHOD OF CROSS MULTIPLICATION 15 16
Clearly, <
Let a and c be the two given fractions. 18 18
b d a c 5 8
Cross multiply, as shown Hence, 6 < 9
b d
Find cross products ad and bc. EXAMPLE 16: Compare the fractions 7 and 9

a c 12 16
(i) if ad > bc then > SOLUTION: LCM of 12 and 16 = (4 × 3 × 4) = 48.
b d 7 9
a c Now, we convert each one of and into an equivalent fraction having 48 as
(ii) if ad < bc then < denominator. 12 16
b d
a c 17 = 7 × 4 = 28 = 28 and 9 = 9 × 2 = 27
(iii) if ad = bc then = 12 12 × 4 48 48 16 16 ×3 48
b d 28 27 7 9
Clearly, 48 > 48 Hence, 12 > 16