RD Sharma Solutions Class 6 Fractions Exercise 6.9

RD Sharma Solutions Class 6 Chapter 6 Exercise 6.9

RD Sharma Class 6 Solutions Chapter 6 Ex 6.9 PDF Free Download

Exercise 6.9

 

Q 1. Add :

i) \(\frac{ 3 }{ 4 }\) and \(\frac{ 5 }{ 6 }\)

ii) \(\frac{ 7 }{ 10 }\) and \(\frac{ 2 }{ 15 }\)

iii) \(\frac{ 8 }{ 13 }\) and \(\frac{ 2 }{ 3 }\)

iv) \(\frac{ 4 }{ 5 }\) and \(\frac{ 7 }{ 15 }\)

soln:

i) Given : \(\frac{ 3 }{ 4 }\) and \(\frac{ 5 }{ 6 }\)

\(\frac{ 3 }{ 4 }\) + \(\frac{ 5 }{ 6 }\)

LCM of 4 and 6 is 12 , so we will convert each fraction into an equivalent fraction with denominator 12.

= \(\frac{ 3 x 3 }{ 4 x 3 }\) + \(\frac{ 5 x 2 }{ 6 x 2 }\)

= \(\frac{ 9 }{ 12 }\) + \(\frac{ 10 }{ 12 }\)

= \(\frac{ 9 + 10 }{ 12 }\) = \(\frac{ 19 }{ 12 }\)

ii) Given : \(\frac{ 7 }{ 10 }\) and \(\frac{ 2 }{ 15 }\)

\(\frac{ 7 }{ 10 }\) + \(\frac{ 2 }{ 15 }\)

LCM of 10 and 15 is 30 , so we will convert each fraction into an equivalent fraction with denominator 30.

= \(\frac{ 7 x 3 }{ 10 x 3 }\) + \(\frac{ 2 x 2 }{ 15 x 2 }\)

= \(\frac{ 21 }{ 30 }\) +\(\frac{ 4 }{ 30 }\)

= \(\frac{ 21 + 4 }{ 30 }\) = \(\frac{ 25 }{ 30 }\)

iii) Given : \(\frac{ 8 }{ 13 }\) and \(\frac{ 2 }{ 3 }\)

\(\frac{ 8 }{ 13 }\) + \(\frac{ 2 }{ 3 }\)

LCM of 13 and 3 is 39 , so we will convert each fraction into an equivalent fraction with denominator 39.

= \(\frac{ 8 x 3 }{ 13 x 3 }\) + \(\frac{ 2 x 13 }{ 3 x 13 }\)

= \(\frac{ 24 }{ 39 }\) +\(\frac{ 26 }{ 39 }\)

= \(\frac{ 24 + 26 }{ 39 }\) = \(\frac{ 50 }{ 39 }\)

iv) Given : \(\frac{ 4 }{ 5 }\) and \(\frac{ 7 }{ 15 }\)

\(\frac{ 4 }{ 5 }\) + \(\frac{ 7 }{ 15 }\)

LCM of 5 and 15 is 15 , so we will convert each fraction into an equivalent fraction with denominator 15.

= \(\frac{ 4 x 3 }{ 5 x 3 }\) + \(\frac{ 7 x 1 }{ 15 x 1 }\)

= \(\frac{ 12 }{ 15 }\) + \(\frac{ 7 }{ 15 }\)

= \(\frac{ 12 + 7 }{ 15 }\) = \(\frac{ 19 }{ 15 }\)

Q 2. Subtract :

i) \(\frac{ 2 }{ 7 }\) from \(\frac{ 19 }{ 21 }\)

ii) \(\frac{ 21 }{ 25 }\) from \(\frac{ 18 }{ 20 }\)

iii) \(\frac{ 7 }{ 16 }\) from \(\frac{ 2 }{ 1 }\)

iv) \(\frac{ 4 }{ 15 }\) from \(2\frac{ 1 }{ 5 }\)

soln:

i) Given : \(\frac{ 2 }{ 7 }\) and \(\frac{ 19 }{ 21 }\)

\(\frac{ 2 }{ 7 }\) + \(\frac{ 19 }{ 21 }\)

LCM of 21 and 7 is 21, so we will convert each fraction into an equivalent fraction with denominator 21.

= \(\frac{ 19 x 1 }{ 21 x 1 }\)\(\frac{ 2 x 3 }{ 7 x 3 }\)

= \(\frac{ 19 }{ 21 }\)\(\frac{ 6 }{ 21 }\)

= \(\frac{ 19 – 6 }{ 21 }\) = \(\frac{ 13 }{ 21 }\)

ii) Given : \(\frac{ 21 }{ 25 }\) and \(\frac{ 18 }{ 20 }\)

\(\frac{ 18 }{ 20 }\)\(\frac{ 21 }{ 25 }\)

LCM of 20 and 25 is 100 , so we will convert each fraction into an equivalent fraction with denominator 100.

= \(\frac{ 18 x 5 }{ 20 x 5 }\)\(\frac{ 21 x 4 }{ 25 x 4 }\)

= \(\frac{ 90 }{ 100 }\)\(\frac{ 84 }{ 100 }\)

= \(\frac{ 90 – 84 }{ 100 }\) = \(\frac{ 6 }{ 100 }\)

iii) Given : \(\frac{ 2 }{ 1 }\) and \(\frac{ 7 }{ 16 }\)

\(\frac{ 2 }{ 1 }\)\(\frac{ 7 }{ 16 }\)

LCM of 1 and 16 is 16 , so we will convert each fraction into an equivalent fraction with denominator 16.

= \(\frac{ 16 x 2 }{ 16 x 1 }\)\(\frac{ 7 x 1 }{ 16 x 1 }\)

= \(\frac{ 32 }{ 16 }\)\(\frac{ 7 }{ 16 }\)

= \(\frac{ 32 – 7 }{ 16 }\) = \(\frac{ 25 }{ 16 }\)

iv) Given : \(2\frac{ 1 }{ 5 }\) and \(\frac{ 4 }{ 15 }\)

\(\frac{ ( 2 x 5 ) + 1 }{ 5 }\)\(\frac{ 4 }{ 15 }\)

LCM of 5 and 15 is 15 , so we will convert each fraction into an equivalent fraction with denominator 15.

= \(\frac{ 11 x 3 }{ 5 x 3 }\)\(\frac{ 4 x 1 }{ 15 x 1 }\)

= \(\frac{ 33 }{ 15 }\)\(\frac{ 4 }{ 15 }\)

= \(\frac{ 33 – 4 }{ 15 }\) = \(\frac{ 29 }{ 15 }\)

Q 3. Find the difference of :

i) \(\frac{ 13 }{ 24 }\) and \(\frac{ 7 }{ 16 }\)

i) \(\frac{ 5 }{ 18 }\) and \(\frac{ 4 }{ 15 }\)

i) \(\frac{ 1 }{ 12 }\) and \(\frac{ 3 }{ 4 }\)

i) \(\frac{ 2 }{ 3 }\) and \(\frac{ 6 }{ 7 }\)

soln:

i) Given : \(\frac{ 13 }{ 24 }\) and \(\frac{ 7 }{ 16 }\)

\(\frac{ 13 }{ 24 }\)\(\frac{ 7 }{ 16 }\)

= \(\frac{ 13 x 2 }{ 24 x 2 }\)\(\frac{ 7 x 3 }{ 16 x 3 }\)

= \(\frac{ 26 }{ 48 }\)\(\frac{ 21 }{ 48 }\) ( because LCM of 24 and 16 is 48 )

= \(\frac{ 26 – 21 }{ 48 }\) = \(\frac{ 5 }{ 48 }\)

ii) Given : \(\frac{ 5 }{ 18 }\) and \(\frac{ 4 }{ 15 }\)

\(\frac{ 5 }{ 18 }\)\(\frac{ 4 }{ 15 }\)

= \(\frac{ 5 x 5 }{ 18 x 5 }\)\(\frac{ 4 x 6 }{ 15 x 6 }\)

= \(\frac{ 25 }{ 90 }\)\(\frac{ 24 }{ 90 }\) ( because LCM of 18 and 15 is 90 )

= \(\frac{ 25 – 24 }{ 90 }\) = \(\frac{ 1 }{ 90 }\)

iii) Given : \(\frac{ 3 }{ 4 }\) and \(\frac{ 1 }{ 12 }\)

\(\frac{ 3 }{ 4 }\)\(\frac{ 1 }{ 12 }\)

= \(\frac{ 3 x 3 }{ 4 x 3 }\)\(\frac{ 1 x 1 }{ 12 x 1 }\)

= \(\frac{ 9 }{ 12 }\)\(\frac{ 1 }{ 12 }\) ( because LCM of 4 and 12 is 12 )

= \(\frac{ 9 – 1 }{ 12 } = \frac{ 8 }{ 12 }\)

iv) Given : \(\frac{ 6 }{ 7 }\) and \(\frac{ 2 }{ 3 }\)

\(\frac{ 6 }{ 7 }\)\(\frac{ 2 }{ 3 }\)

= \(\frac{ 6 x 3 }{ 7 x 3 } – \frac{ 2 x 7 }{ 3 x 7 }\)

= \(\frac{ 18 }{ 21 } – \frac{ 14 }{ 21 }\) ( because LCM of 7 and 3 is 21 )

= \(\frac{ 18 – 14 }{ 21 } = \frac{ 4 }{ 21 }\)

Q 4. Subtract as indicated :

i) \(\frac{ 8 }{ 3 } – \frac{ 5 }{ 9 }\)

ii) \(4\frac{ 2 }{ 5 } – 2\frac{ 1 }{ 5 }\)

iii) \(5\frac{ 6 }{ 7 } – 2\frac{ 2 }{ 3 }\)

iv) \(4\frac{ 3 }{ 4 } – 2\frac{ 1 }{ 6 }\)

soln:

i) Given : \(\frac{ 8 }{ 3 }\) and \(\frac{ 5 }{ 9 }\)

\(\frac{ 8 }{ 3 }\)\(\frac{ 5 }{ 9 }\)

= \(\frac{ 8 x 3 }{ 3 x 3 }\)\(\frac{ 5 x 1 }{ 9 x 1 }\)

= \(\frac{ 24 }{ 9 }\)\(\frac{ 5 }{ 9 }\) ( because LCM of 3 and 9 is 9 )

= \(\frac{ 24 – 5 }{ 9 } = \frac{ 19 }{ 9 }\)

ii) Given : \(4\frac{ 2 }{ 5 } \; and \;2\frac{ 1 }{ 5 }\)

\(4\frac{ 2 }{ 5 } – 2\frac{ 1 }{ 5 }\)

\(\frac{ ( 5 x 4 ) + 2 }{ 5 }\)\(\frac{ ( 5 x 2 ) + 1 }{ 5 }\)

= \(\frac{ 22 }{ 5 }\)\(\frac{ 11 }{ 5 }\)

= \(\frac{ 22 – 11 }{ 5 }\) = \(\frac{ 11 }{ 5 }\)

iii) Given : \(5\frac{ 6 }{ 7 }\) and \(2\frac{ 2 }{ 3 }\)

\(5\frac{ 6 }{ 7 }\)\(2\frac{ 2 }{ 3 }\)

= \(\frac{ ( 5 x 7 ) + 6 }{ 7 }\)\(\frac{ ( 3 x 2 ) + 2 }{ 3 }\)

= \(\frac{ 41 }{ 7 } – \frac{ 8 }{ 3 }\)

= \(\frac{ ( 41 x 3 ) }{ 7 x 3 } – \frac{ ( 8 x 7 ) }{ 3 x 7 }\)

( because LCM of 7 and 3 is 21 )

= \(\frac{ 123 }{ 21 }\)\(\frac{ 56 }{ 21 }\)

= \(\frac{ 123 – 56 }{ 21 }\) = \(\frac{ 67 }{ 21 }\)

iv) Given : \(4\frac{ 3 }{ 4 }\) and \(2\frac{ 1 }{ 6 }\)

\(4\frac{ 3 }{ 4 }\)\(2\frac{ 1 }{ 6 }\)

= \(\frac{ ( 4 x 4 ) + 3 }{ 4 }\)\(\frac{ ( 6 x 2 ) + 1 }{ 6 }\)

= \(\frac{ 19 }{ 4 }\)\(\frac{ 13 }{ 6 }\)

= \(\frac{ ( 19 x 3 ) }{ 4 x 3 }\)\(\frac{ ( 13 x 2 ) }{ 6 x 2 }\)

( because LCM of 4 and 6 is 12 )

= \(\frac{ 57 }{ 21 }\)\(\frac{ 26 }{ 21 }\)

= \(\frac{ 57 – 26 }{ 21 }\) = \(\frac{ 31 }{ 12 }\)

 

Q 5. simplify :

 

i) \(\frac{ 2 }{ 3 } + \frac{ 3 }{ 4 }\) + \(\frac{ 1 }{ 2 }\)

ii) \(\frac{ 5 }{ 8 } + \frac{ 2 }{ 5 }\) + \(\frac{ 3 }{ 4 }\)

iii) \(\frac{ 3 }{ 10 } + \frac{ 7 }{ 15 }\) + \(\frac{ 3 }{ 5 }\)

iv) \(\frac{ 3 }{ 4 } + \frac{ 7 }{ 16 }\) + \(\frac{ 5 }{ 8 }\)

v) \(4\frac{ 2 }{ 3 } + 3\frac{ 1 }{ 4 }\) + \(7\frac{ 1 }{ 2 }\)

vi) \(7\frac{ 1 }{ 3 } + 3\frac{ 2 }{ 3 }\) + \(5\frac{ 1 }{ 6 }\)

vii) \(\frac{ 7 }{ 1 } + \frac{ 7 }{ 4 }\) + \(5\frac{ 1 }{ 6 }\)

viii) \(\frac{ 5 }{ 6 } + \frac{ 3 }{ 1 }\) + \(\frac{ 3 }{ 4 }\)

ix) \(\frac{ 7 }{ 18 } + \frac{ 5 }{ 6 }\) + \(1\frac{ 1 }{ 12 }\)

soln:

i) given : \(\frac{ 2 }{ 3 } + \frac{ 3 }{ 4 }\) + \(\frac{ 1 }{ 2 }\)

= \(\frac{ 2 x 4 }{ 3 x 4 } + \frac{ 3 x 3 }{ 4 x 3 }\) + \(\frac{ 1 x 6 }{ 2 x 6 }\) ( because LCM of 3, 4 and 2 is 12 )

= \(\frac{ 8 }{ 12 } + \frac{ 9 }{ 12 }\) + \(\frac{ 6 }{ 12 }\)

= \(\frac{ 8 + 9 + 6 }{ 12 }\) = \(\frac{ 23 }{ 12 }\)

ii) given : \(\frac{ 5 }{ 8 } + \frac{ 2 }{ 5 }\) + \(\frac{ 3 }{ 4 }\)

= \(\frac{ 5 x 5 }{ 8 x 5 } + \frac{ 2 x 8 }{ 5 x 8 }\) + \(\frac{ 3 x 10 }{ 4 x 10 }\) ( because LCM of 8, 5 and 4 is 40 )

= \(\frac{ 25 }{ 40 } + \frac{ 16 }{ 40 }\) + \(\frac{ 30 }{ 40 }\)

= \(\frac{ 25 + 16 + 30 }{ 40 }\) = \(\frac{ 71 }{ 40 }\)

iii) given : \(\frac{ 3 }{ 10 } + \frac{ 2 }{ 5 }\) + \(\frac{ 3 }{ 4 }\)

= \(\frac{ 5 x 5 }{ 8 x 5 } + \frac{ 2 x 8 }{ 5 x 8 }\) + \(\frac{ 3 x 10 }{ 4 x 10 }\) ( because LCM of 8, 5 and 4 is 40 )

= \(\frac{ 25 }{ 40 } + \frac{ 16 }{ 40 }\) + \(\frac{ 30 }{ 40 }\)

= \(\frac{ 25 + 16 + 30 }{ 40 }\) = \(\frac{ 71 }{ 40 }\)

iv) given : \(\frac{ 3 }{ 4 } + \frac{ 7 }{ 16 }\) + \(\frac{ 5 }{ 8 }\)

= \(\frac{ 3 x 4 }{ 4 x 4 } + \frac{ 7 x 1 }{ 16 x 1 }\) + \(\frac{ 5 x 2 }{ 8 x 2 }\) ( because LCM of 4, 16 and 8 is 16 )

= \(\frac{ 12 }{ 16 } + \frac{ 7 }{ 16 }\) + \(\frac{ 10 }{ 16 }\)

= \(\frac{ 12 + 7 + 10 }{ 16 }\) = \(\frac{ 29 }{ 16 }\)

v) given : \(4\frac{ 2 }{ 3 } + 3\frac{ 1 }{ 4 }\) + \(7\frac{ 1 }{ 2 }\)

= \(\frac{ ( 4 x 3 ) + 2 }{ 3 } + \frac{ ( 3 x 4 ) + 1 }{ 4 }\) + \(\frac{ ( 7 x 2 ) + 1 }{ 2 }\)

= \(\frac{ 14 }{ 3 } + \frac{ 13 }{ 4 }\) + \(\frac{ 15 }{ 2 }\)

= \(\frac{ 14 x 4 }{ 3 x 4 } + \frac{ 13 x 3 }{ 4 x 3 }\) + \(\frac{ 15 x 6 }{ 2 x 6 }\) ( because LCM of 3, 4 and 2 is 12 )

= \(\frac{ 56 }{ 12 } + \frac{ 39 }{ 12 }\) + \(\frac{ 90 }{ 12 }\)

= \(\frac{ 56 + 39 + 90 }{ 12 }\) = \(\frac{ 185 }{ 12 }\)

vi) given : \(7\frac{ 1 }{ 3 } + 3\frac{ 2 }{ 4 }\) + \(5\frac{ 1 }{ 6 }\)

= \(\frac{ ( 7 x 3 ) + 1 }{ 3 } + \frac{ ( 3 x 4 ) + 2 }{ 4 }\) + \(\frac{ ( 5 x 6 ) + 1 }{ 6 }\)

= \(\frac{ 22 }{ 3 } + \frac{ 14 }{ 4 }\) + \(\frac{ 31 }{ 6 }\)

= \(\frac{ 22 x 4 }{ 3 x 4 } + \frac{ 14 x 3 }{ 4 x 3 }\) + \(\frac{ 31 x 2 }{ 6 x 2 }\) ( because LCM of 3, 4 and 6 is 12 )

= \(\frac{ 88 }{ 12 } + \frac{ 42 }{ 12 }\) + \(\frac{ 62 }{ 12 }\)

= \(\frac{ 88 + 42 + 62 }{ 12 }\) = \(\frac{ 16 }{ 1 }\)

( HCF of numerator and denominator is 12 )

vii) given : \(\frac{ 7 }{ 1 } + \frac{ 7 }{ 4 }\) + \(5\frac{ 1 }{ 6 }\)

= \(\frac{ 7 x 12 }{ 1 x 12 } + \frac{ 7 x 3 }{ 4 x 3 }\) + \(\frac{ 31 x 2 }{ 6 x 2 }\) ( because LCM of 1, 4 and 6 is 12 )

= \(\frac{ 84 }{ 12 } + \frac{ 21 }{ 12 }\) + \(\frac{ 62 }{ 12 }\)

= \(\frac{ 84 + 21 + 62 }{ 12 }\) = \(\frac{ 167 }{ 12 }\)

viii) given : \(\frac{ 5 }{ 6 } + \frac{ 3 }{ 1 }\) + \(\frac{ 3 }{ 4 }\)

= \(\frac{ 5 x 2 }{ 6 x 2 } + \frac{ 3 x 12 }{ 1 x 12 }\) + \(\frac{ 3 x 3 }{ 4 x 3 }\) ( because LCM of 6, 1 and 4 is 12 )

= \(\frac{ 10 }{ 12 } + \frac{ 36 }{ 12 }\) + \(\frac{ 9 }{ 12 }\)

= \(\frac{ 10 + 36 + 9 }{ 12 }\) = \(\frac{ 55 }{ 12 }\)

ix) given : \(\frac{ 7 }{ 18 } + \frac{ 5 }{ 6 }\) + \(1\frac{ 1 }{ 12 }\)

= \(\frac{ 7 }{ 18 } + \frac{ 5 }{ 6 }\) + \(\frac{ 13 }{ 12 }\)

= \(\frac{ 7 x 2 }{ 18 x 2 } + \frac{ 5 x 6 }{ 6 x 6 }\) + \(\frac{ 13 x 3 }{ 12 x 3 }\)

= \(\frac{ 14 }{ 36 } + \frac{ 30 }{ 36 }\) + \(\frac{ 39 }{ 36 }\)

= \(\frac{ 14 + 30 + 39 }{ 36 }\) = \(\frac{ 83 }{ 36 }\)

Q 6. Replace * with a correct number :

i) * – \(\frac{ 5 }{ 8 }\) = \(\frac{ 1 }{ 4 }\)

ii) * – \(\frac{ 1 }{ 5 }\) = \(\frac{ 1 }{ 2 }\)

iii) \(\frac{ 1 }{ 2 }\) – * = \(\frac{ 1 }{ 6 }\)

soln:

i) * – \(\frac{ 5 }{ 8 }\) = \(\frac{ 1 }{ 4 }\)

* = \(\frac{ 5 }{ 8 }\) + \(\frac{ 1 }{ 4 }\)

* = \(\frac{ 1 x 2 }{ 4 x 2 }\) + \(\frac{ 5 x 1 }{ 8 x 1 }\)

* = \(\frac{ 2 }{ 8 }\) + \(\frac{ 5 }{ 8 }\) = \(\frac{ 2 + 5 }{ 8 }\) = \(\frac{ 2 }{ 8 }\)

Therefore, \(\frac{ 7 }{ 8 }\)

ii) * – \(\frac{ 1 }{ 5 }\) = \(\frac{ 1 }{ 2 }\)

* = \(\frac{ 1 }{ 2 }\) + \(\frac{ 1 }{ 5 }\)

* = \(\frac{ 1 x 5 }{ 5 x 2 }\) + \(\frac{ 1 x 2 }{ 2 x 5 }\)

= \(\frac{ 5 }{ 10 }\) + \(\frac{ 2 }{ 10 }\)

= \(\frac{ 5 + 2 }{ 10 }\) = \(\frac{ 7 }{ 10 }\)

iii) \(\frac{ 1 }{ 2 }\) – * = \(\frac{ 1 }{ 6 }\)

* = \(\frac{ 1 }{ 2 }\)\(\frac{ 1 }{ 6 }\)

* = \(\frac{ 1 x 3 }{ 2 x 3 }\)\(\frac{ 1 x 1 }{ 6 x 1 }\) ( because LCM of 2 and 6 is 6 )

= \(\frac{ 3 }{ 6 }\)\(\frac{ 1 }{ 6 }\)

= \(\frac{ 1 }{ 3 }\)

Q 7. Savita bought \(\frac{ 2 }{ 5 }\) m of ribbon and kavita \(\frac{ 3 }{ 4 }\) m of ribbon. What was the total length of the ribbon they bought ?

Soln: length of the ribbon bought by savita = \(\frac{ 2 }{ 5 }\) metres

Length of the ribbon bought by kavita = \(\frac{ 3 }{ 4 }\) metres

Total length of the ribbon bought by them:

\(\frac{ 2 }{ 5 }\) metres + \(\frac{ 3 }{ 4 }\) metres

= \(\frac{ 2 x 4 }{ 5 x 4 }\) metres + \(\frac{ 3 x 5 }{ 4 x 5 }\) metres

( because LCM of 5 and 4 is 20 )

= \(\frac{ 8 }{ 20 }\) metres + \(\frac{ 15 }{ 20 }\) metres = \(\frac{ 8 + 15 }{ 20 }\) metres

= \(\frac{ 23 }{ 20 }\) metres

Q 8. Ravish takes \(2\frac{ 1 }{ 5 }\) minutes to walk across the school ground. Rahul takes \(\frac{ 7 }{ 4 }\) minutes to do the same. Who takes less time and by what fraction ?

Soln: Time taken by ravish = \(2\frac{ 1 }{ 5 }\) = \(\frac{ ( 5 x 2 )+ 1 }{ 5 }\) = \(\frac{ 11 }{ 5 }\) minutes

Time taken by rahul = \(\frac{ 7 }{ 4 }\) minutes

Comparing \(\frac{ 11 }{ 5 }\) minutes and \(\frac{ 7 }{ 4 }\) minutes, we get :

\(\frac{ 11 x 4 }{ 5 x 4 }\) minutes , \(\frac{ 7 x 5 }{ 4 x 5 }\) minutes

( LCM of 4 and 5 is 20, so will we convert each fraction into an equivalent fraction with denominator 20 )

\(\frac{ 44 }{ 20 }\) > \(\frac{ 35 }{ 20 }\)

Rahul takes less time ,

i.e, \(\frac{ 44 }{ 20 }\)\(\frac{ 35 }{ 20 }\) = \(\frac{ 44 – 35 }{ 20 }\) = \(\frac{ 9 }{ 20 }\) minutes.

Q 9. A piece of a wire \(\frac{ 7 }{ 8 }\) metres long broke into two pieces. One piece was \(\frac{ 1 }{ 4 }\) meter long. How long is the other piece ?

Ans: Length of the wire = \(\frac{ 7 }{ 8 }\) metres

Length of one piece of wire = \(\frac{ 1 }{ 4 }\) metres

Let the length of the second piece of wire be x m.

Therefore, Length of the wire = length of one piece + length of the second piece

\(\frac{ 7 }{ 8 }\) metres = \(\frac{ 1 }{ 4 }\) metres + x

X = \(\frac{ 7 }{ 8 }\) metres – \(\frac{ 1 }{ 4 }\) metres

X = \(\frac{ 7 x 1 }{ 8 x 1 }\) metres – \(\frac{ 1 x 2 }{ 4 x 2 }\) metres

= \(\frac{ 7 }{ 8 }\) metres – \(\frac{ 2 }{ 8 }\) metres

= \(\frac{ 7 – 2 }{ 8 }\) metres

X = \(\frac{ 5 }{ 8 }\) metres

Therefore, the length of the second piece is \(\frac{ 5 }{ 8 }\) metres

Q 10. Shikha and priya have bookshelves of the same size shikha’s shelf is \(\frac{ 5 }{ 6 }\) full of book and priya’s shelf is \(\frac{ 2 }{ 5 }\) full. Whose bookshelf is more full ? By what fraction ?

Soln:

Fraction of shikha’s filled bookshelf = \(\frac{ 5 }{ 6 }\)

Fraction of Priya’s filled bookshelf = \(\frac{ 2 }{ 5 }\)

Comparing \(\frac{ 5 }{ 6 }\) and \(\frac{ 2 }{ 5 }\) , we get :

LCM of 5 and 6 is 30, so will convert each fraction into an equivalent fraction with denominator 30.

= \(\frac{ 5 x 5 }{ 6 x 5 }\) metres , \(\frac{ 2 x 6 }{ 5 x 6 }\) metres

\(\frac{ 25 }{ 30 }\) > \(\frac{ 12 }{ 30 }\)

Shikha’s shelf is more full.

Therefore,

\(\frac{ 25 }{ 30 }\)\(\frac{ 12 }{ 30 }\) = \(\frac{ 25 – 12 }{ 30 }\) = \(\frac{ 13 }{ 30 }\)

Q 11. Ravish’s house is \(\frac{ 9 }{ 10 }\) Km from his school. He walked some distance and then took a bus for \(\frac{ 1 }{ 2 }\) Km. How far did he walk?

Soln:

Total distance between the house and the school = \(\frac{ 9 }{ 10 }\) Km

Distance covered in the bus = \(\frac{ 1 }{ 2 }\) Km

Distance covered by walking + distance covered in the bus = total distance between the house and the school

Distance covered by walking = total distance between the house and the school – Distance covered in the bus

Distance covered by walking:

\(\frac{ 9 }{ 10 }\) Km – \(\frac{ 1 }{ 2 }\) Km

LCM of 10 and 2 is 10, so we convert each fraction into an equivalent fraction with denominator 10

= \(\frac{ 9 x 1 }{ 10 x 1 }\)\(\frac{ 1 x 5 }{ 2 x 5 }\) = \(\frac{ 9 }{ 10 }\)\(\frac{ 5 }{ 10 }\)

= \(\frac{ 9 – 5 }{ 10 }\) = \(\frac{ 4 }{ 10 }\) km = \(\frac{ 2 }{ 5 }\) km ( HCF of numerator and denominator is 2 )

Leave a Comment

Your email address will not be published. Required fields are marked *