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.