# 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

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 )