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.