Gujarat BoardĀ 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 āmissing fractionā is less than \(\frac { 7 }{ 10 } \) by \(\frac { 3 }{ 10 } \)

The āmissing fractionā = \(\frac { 7 }{ 10 } \) – \(\frac { 3 }{ 10 } \)

= \(\frac { 7-3 }{ 10 } \) = \(\frac { 4 }{ 10 } \) or \(\frac { 2 }{ 5 } \)

We have

The āmissing 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 āmissing fractionā 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 āmissing fractionā 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.