# RD Sharma Solutions Class 6 Fractions Exercise 6.8

## RD Sharma Solutions Class 6 Chapter 6 Exercise 6.8

### RD Sharma Class 6 Solutions Chapter 6 Ex 6.8 PDF Free Download

#### Exercise 6.8

Q 1. Write these fractions appropriately as additions or subtraction :

Q 2. Solve :

i) $\frac{ 5 }{ 12 } + \frac{ 1 }{ 12 }$

ii) $\frac{ 3 }{ 15 } + \frac{ 7 }{ 15 }$

iii) $\frac{ 3 }{ 22 } + \frac{ 7 }{ 22 }$

iv)$\frac{ 1 }{ 4 } + \frac{ 0 }{ 4 }$

v) $\frac{ 4 }{ 13 } + \frac{ 2 }{ 13 }$ + $\frac{ 1 }{ 13 }$

vi) $\frac{ 0 }{ 15 } + \frac{ 2 }{ 15 }$ + $\frac{ 1 }{ 15 }$

vii) $\frac{ 7 }{ 31 } – \frac{ 4 }{ 31 }$ + $\frac{ 9 }{ 31 }$

viii) $3\frac{ 2 }{ 7 } + \frac{ 1 }{ 7 } – 2\frac{ 3 }{ 7 }$

ix) $2\frac{ 1 }{ 3 } – 1\frac{ 2 }{ 3 } – 4\frac{ 1 }{ 3 }$

x) $\frac{ 1 }{ 1 } – \frac{ 2 }{ 3 }$ + $\frac{ 7 }{ 3 }$

xi) $\frac{ 16 }{ 7 } – \frac{ 5 }{ 7 }$ + $\frac{ 9 }{ 7 }$

Sol :

i) The given fractions are:

$\frac{ 5 }{ 12 } + \frac{ 1 }{ 12 }$ = $\frac{ 1 + 2 }{ 5 }$ = $\frac{ 3 }{ 5 }$

Hence the answer is $\frac{ 3 }{ 5 }$

ii) The given fractions are:

$\frac{ 3 }{ 6 } + \frac{ 2 }{ 6 }$ = $\frac{ 3 + 2 }{ 6 }$ = $\frac{ 8 }{ 6 } = \frac{ 4 }{ 3 }$

Hence the answer is $\frac{ 4 }{ 3 }$

iii) the given fractions are :

$\frac{ 3 }{ 22 } + \frac{ 7 }{ 22 }$ = $\frac{ 3 + 7 }{ 22 }$ = $\frac{ 10 }{ 22 }$ = $\frac{ 5 }{ 11 }$

Hence the answer is $\frac{ 5 }{ 11 }$

iv) the given fractions are :

$\frac{ 1 }{ 4 } + \frac{ 0 }{ 4 }$

= $\frac{ 1 + 0 }{ 4 }$ = $\frac{ 1 }{ 4 }$

Hence the answer is $\frac{ 1 }{ 4 }$

v) The given fractions are :

$\frac{ 4 }{ 13 } + \frac{ 2 }{ 13 }$ + $\frac{ 1 }{ 13 }$

= $\frac{ 4 + 2 + 1 }{ 13 }$ = $\frac{ 7 }{ 13 }$

Hence the answer is $\frac{ 7 }{ 13 }$

vi) the given fractions are :

$\frac{ 0 }{ 15 } + \frac{ 2 }{ 15 }$ + $\frac{ 1 }{ 15 }$

= $\frac{ 0 + 2 + 1 }{ 15 }$ = $\frac{ 3 }{ 15 }$ = $\frac{ 1 }{ 5 }$

Hence the answer is $\frac{ 1 }{ 5 }$

vii) the given fractions are :

$\frac{ 7 }{ 31 } – \frac{ 4 }{ 31 }$ + $\frac{ 9 }{ 31 }$

= $\frac{ 7 – 4 + 9 }{ 31 }$ = $\frac{ 12 }{ 31 }$

Hence the answer is $\frac{ 12 }{ 31 }$

viii) the given fractions are :

$3\frac{ 2 }{ 7 } + \frac{ 1 }{ 7 } – 2\frac{ 3 }{ 7 }$

= $\frac{ 23 + 1 – 17 }{ 7 }$ = $\frac{ 7 }{ 7 }$ = $\frac{ 1 }{ 1 }$ = 1

Hence the answer is $\frac{ 1 }{ 1 }$ = 1

ix) the given fractions are :

$3\frac{ 2 }{ 7 } + \frac{ 1 }{ 7 } – 2\frac{ 3 }{ 7 }$

= $\frac{ 23 + 1 – 17 }{ 7 }$ = $\frac{ 35 }{ 7 }$ = $\frac{ 5 }{ 1 }$ = 5

Hence the answer is $\frac{ 5 }{ 1 }$ = 5

x) the given fractions are :

$\frac{ 1 }{ 1 } – \frac{ 2 }{ 3 }$ + $\frac{ 7 }{ 3 }$

= $\frac{ 3 – 2 + 7 }{ 3 }$ = $\frac{ 8 }{ 3 }$

Hence the answer is $\frac{ 8 }{ 3 }$

xi ) the given fractions are :

$\frac{ 16 }{ 7 } – \frac{ 5 }{ 7 }$ + $\frac{ 9 }{ 7 }$

= $\frac{ 16 – 5 + 9 }{ 7 }$ = $\frac{ 20 }{ 7 }$

Hence the answer is $\frac{ 20 }{ 7 }$

Q 3. Shikha painted $\frac{ 1 }{ 5 }$ of the wall space in her room. Her brother ravish helped and painted $\frac{ 3 }{ 5 }$ of the wall space. How much did they paint together? How much the room is left unpainted?

Sol :

Shikha painted $\frac{ 1 }{ 5 }$ of the wall space in her room

Ravish painted $\frac{ 3 }{ 5 }$ of the wall space

Wall space painted by both of them together = $\frac{ 1 }{ 5 }$ + $\frac{ 3 }{ 5 }$ = $\frac{ 1 + 3 }{ 5 }$ = $\frac{ 4 }{ 5 }$

Unpainted part of the room = 1 – $\frac{ 4 }{ 5 }$ =

$\frac{ 5 – 4 }{ 5 }$ = $\frac{ 1 }{ 5 }$

Q 4. Ramesh bought $2\frac{ 1 }{ 2 }$ kg sugar whereas rohit bought $3\frac{ 1 }{ 2 }$ kg of sugar. Find the total amount of sugar bought by both of them.

Sol :

Quantity of sugar bought by ramesh = $2\frac{ 1 }{ 2 }$ kg

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

Quantity of sugar bought by rohit = $3\frac{ 1 }{ 2 }$ kg

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

Total amount of sugar bought by them :

Quantity of sugar bought by rohit + Quantity of sugar bought by ramesh

= $\frac{ 5 }{ 2 }$ kg + $\frac{ 7 }{ 2 }$ kg

= 6 kg ( Dividing numerator and denominator by their HCF ( 6 ) )

Q 5. The teacher taught $\frac{ 3 }{ 5 }$ of the book, Vivek revised $\frac{ 1 }{ 5 }$ more on his own. How much does he still have to revise ?

Sol :

Fraction of the book taught by the teacher = $\frac{ 3 }{ 5 }$

Fraction of the book revised by vivek = $\frac{ 1 }{ 5 }$

Fraction of the book still left for revision by vivek :

$\frac{ 3 }{ 5 }$$\frac{ 1 }{ 5 }$ = $\frac{ 3 – 1 }{ 5 }$ = $\frac{ 2 }{ 5 }$

Therefore, Fraction of the book still left for revision by vivek is $\frac{ 2 }{ 5 }$

Q 6. Amit was given $\frac{ 5 }{ 7 }$ of a bucket of oranges. What fraction of oranges was left in the basket ?

Sol :

Fraction of oranges given to amit = $\frac{ 5 }{ 7 }$

Fraction of oranges left in the basket :

1 – $\frac{ 5 }{ 7 }$ = = $\frac{ 7 – 5 }{ 7 }$ = $\frac{ 2 }{ 7 }$

Therefore, Fraction of oranges left in the basket is $\frac{ 2 }{ 7 }$

Q 7. Fill in the missing fractions :

i) $\frac{ 7 }{ 10 }$$\frac{ * }{ * }$ = $\frac{ 3 }{ 10 }$

ii) $\frac{ * }{ * }$$\frac{ 3 }{ 21 }$ = $\frac{ 5 }{ 21 }$

iii) $\frac{ * }{ * }$$\frac{ 3 }{ 6 }$ = $\frac{ 3 }{ 6 }$

iv) $\frac{ * }{ * }$$\frac{ 5 }{ 27 }$ = $\frac{ 12 }{ 27 }$

sol :

i) Given :

$\frac{ 7 }{ 10 }$$\frac{ * }{ * }$ = $\frac{ 3 }{ 10 }$

$\frac{ 7 }{ 10 }$$\frac{ 3 }{ 10 }$ = $\frac{ * }{ * }$

$\frac{ 7 – 3 }{ 10 }$ = $\frac{ 2 }{ 5 }$

Therefore,

$\frac{ * }{ * }$ = $\frac{ 2 }{ 5 }$

ii) Given :

$\frac{ * }{ * }$$\frac{ 3 }{ 21 }$ = $\frac{ 5 }{ 21 }$

$\frac{ * }{ * }$ = – $\frac{ 3 }{ 21 }$ + $\frac{ 5 }{ 21 }$

$\frac{ 5 – 3 }{ 21 }$ = $\frac{ 2 }{ 21 }$

Therefore,

$\frac{ * }{ * }$ = $\frac{ 2 }{ 21 }$

iii) Given :

$\frac{ * }{ * }$$\frac{ 3 }{ 6 }$ = $\frac{ 3 }{ 6 }$

$\frac{ * }{ * }$ = $\frac{ 3 }{ 6 }$ + $\frac{ 3 }{ 6 }$

$\frac{ 3 + 3 }{ 6 }$ = $\frac{ 6 }{ 6 }$

Therefore,

$\frac{ * }{ * }$ = $\frac{ 6 }{ 6 }$ = $\frac{ 1 }{ 1 }$ = 1

iv) Given :

$\frac{ * }{ * }$$\frac{ 5 }{ 27 }$ = $\frac{ 12 }{ 27 }$

$\frac{ * }{ * }$ = $\frac{ 5 }{ 27 }$ + $\frac{ 12 }{ 27 }$

$\frac{ 5 + 12 }{ 27 }$ = $\frac{ 17 }{ 27 }$

Therefore,

$\frac{ * }{ * }$ = $\frac{ 17 }{ 27 }$