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,

\(\frac{ * }{ * }\) = \(\frac{ 17 }{ 27 }\)

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