Gujarat Board GSEB Textbook Solutions Class 6 Maths Chapter 7 Fractions Ex 7.5 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 6 Maths Chapter 7 Fractions Ex 7.5

Question 1.
Write these fractions appropriately as additions or subtractions:

Solution:
(a) The given figure represents an addition of (frac { 1 }{ 5 } ) and (frac { 2 }{ 5 } ) pictorially, i.e.
(frac { 1 }{ 5 } ) + (frac { 2 }{ 5 } ) = (frac { 3 }{ 5 } )
The given figure will be as under:

(b) Here, the given figure represents a subtraction (frac { 3 }{ 5 } ) of from 1, i.e.
1 – (frac { 3 }{ 5 } ) = (frac { 2 }{ 5 } )
The given figure will be as under:

(e) The given figure represents addition of (frac { 2 }{ 6 } ) and (frac { 3 }{ 6 } ), i.e (frac { 2 }{ 6 } ) + (frac { 3 }{ 6 } ) = (frac { 5 }{ 6 } )
The given figure is represented as under:

Question 2.
Solve:

Solution:

(g) Since 1 and (frac { 3 }{ 3 } ) are equivalent fractions.

(i) Since 3 = (frac { 3 }{ 1 } ) and equivalent fraction of (frac { 3 }{ 1 } ) having denominator as 5 is given by
(frac { 3 }{ 1 } ) x (frac { 5 }{ 5 } ) = (frac { 15 }{ 5 } )
3 – (frac { 12 }{ 5 } ) = (frac { 15 }{ 5 } ) – (frac { 12 }{ 5 } ) = (frac { 15-12 }{ 5 } ) = (frac { 3 }{ 5 } )

Question 3.
Shuhham painted (frac { 2 }{ 3 } ) of the wall space in his room. His sister Macihavi helped and painted (frac { 1 }{ 3 } ) of the wail space. How much did they paint together ?
Solution:
Portion of the wall painted by Shubham = (frac { 2 }{ 3 } )
Portion of the wall painted by Madhavi = (frac { 1 }{ 3 } )
The portion of wall painted = (frac { 2 }{ 3 } ) + (frac { 1 }{ 3 } ) = (frac { 2+1 }{ 3 } ) = (frac { 3 }{ 3 } ) or 1
Thus, Shubham and Madhavi together painted the complete wall.

Question 4.
Fill in the missing fractions.

Solution:

The ‘missing fraction’ is less than (frac { 7 }{ 10 } ) by (frac { 3 }{ 10 } )
The ‘missing fraction’ = (frac { 7 }{ 10 } ) – (frac { 3 }{ 10 } )
= (frac { 7-3 }{ 10 } ) = (frac { 4 }{ 10 } ) or (frac { 2 }{ 5 } )
We have

The ‘missing fractions is more than (frac { 3 }{ 21 } ) by (frac { 5 }{ 21 } )
The sum of (frac { 3 }{ 21 } ) and (frac { 5 }{ 21 } ) must be equal to the missing fraction.
Missing fraction = (frac { 3 }{ 21 } ) + (frac { 5 }{ 21 } ) = (frac { 3+5 }{ 21 } ) = (frac { 8 }{ 21 } )
We have:

The ‘missing fraction’ is more than (frac { 3 }{ 6 } ) by (frac { 3 }{ 6 } )
Missing fraction (frac { 3 }{ 6 } ) + (frac { 3 }{ 6 } ) = (frac { 3+3 }{ 6 } ) = (frac { 6 }{ 6 } ) = 1
We have:

Since, the sum of the ‘missing fraction’ and (frac { 5 }{ 27 } ) is (frac { 12 }{ 27 } ).
The missing fraction = (frac { 12 }{ 27 } ) – (frac { 5 }{ 27 } ) = (frac { 12-5 }{ 27 } ) = (frac { 7 }{ 27 } )
Thus,

Question 5.
Javed was given (frac { 5 }{ 7 } ) of a basket of oranges. What fraction of oranges was left in the basket?
Solution:
Let the basket full of oranges be denoted by 1.
Portion of oranges given to Javed = (frac { 5 }{ 7 } )
Portion of oranges left in the basket 1 – (frac { 5 }{ 7 } ) = (frac { 7 }{ 7 } ) – (frac { 5 }{ 7 } ) = (frac { 7-5 }{ 7 } ) = (frac { 2 }{ 7 } )
[1and (frac { 7 }{ 7 } ) are equivalent fractions]
(frac { 2 }{ 7 } ) of oranges was left in the basket.