NCERT Solutions for Class 7 Maths Exercise 2.4 Chapter 2 Fractions and Decimals are the best study materials for those students who find difficulty in solving problems. This exercise of NCERT Solutions for Class 7 Maths Chapter 2 contains the topics division of fractions, division of whole numbers by a fraction, division of a fraction by a whole number and division of a fraction by another fraction. Students who wish to score good marks in Maths should practise NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals.
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals – Exercise 2.4
Access Answers to Maths NCERT Solutions for Class 7 Chapter 2 – Fractions and Decimals Exercise 2.4
1. Find.
(i) 12 ÷ ¾
Solution:-
We have
= 12 × reciprocal of ¾
= 12 × (4/3)
= 4 × 4
= 16
(ii) 14 ÷ (5/6)
Solution:-
We have
= 14 × reciprocal of (5/6)
= 14 × (6/5)
= 84/5
(iii) 8 ÷ (7/3)
Solution:-
We have
= 8 × reciprocal of (7/3)
= 8 × (3/7)
= (24/7)
(iv) 4 ÷ (8/3)
Solution:-
We have
= 4 × reciprocal of (8/3)
= 4 × (3/8)
= 1 × (3/2)
= 3/2
(v) 3 ÷ 
Solution:-
While dividing a whole number by a mixed fraction, first convert the mixed fraction into an improper fraction.
We have
=
= 7/3
Then,
= 3 ÷ (7/3)
= 3 × reciprocal of (7/3)
= 3 × (3/7)
= 9/7
(vi) 5 ÷ 
Solution:-
While dividing a whole number by a mixed fraction, first convert the mixed fraction into an improper fraction.
We have
=
= 25/7
Then,
= 5 ÷ (25/7)
= 5 × reciprocal of (25/7)
= 5 × (7/25)
= 1 × (7/5)
= 7/5
2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.
(i) 3/7
Solution:-
Reciprocal of (3/7) is (7/3) [∵ ((3/7) × (7/3)) = 1]
So, it is an improper fraction.
An improper fraction is a fraction in which the numerator is greater than its denominator.
(ii) 5/8
Solution:-
Reciprocal of (5/8) is (8/5) [∵ ((5/8) × (8/5)) = 1]
So, it is an improper fraction.
An improper fraction is a fraction in which the numerator is greater than its denominator.
(iii) 9/7
Solution:-
Reciprocal of (9/7) is (7/9) [∵ ((9/7) × (7/9)) = 1]
So, it is a proper fraction.
A proper fraction is a fraction in which the denominator is greater than the numerator of the fraction.
(iv) 6/5
Solution:-
Reciprocal of (6/5) is (5/6) [∵ ((6/5) × (5/6)) = 1]
So, it is a proper fraction.
A proper fraction is a fraction in which the denominator is greater than the numerator of the fraction.
(v) 12/7
Solution:-
Reciprocal of (12/7) is (7/12) [∵ ((12/7) × (7/12)) = 1]
So, it is a proper fraction.
A proper fraction is a fraction in which the denominator is greater than the numerator of the fraction.
(vi) 1/8
Solution:-
Reciprocal of (1/8) is (8/1) or 8 [∵ ((1/8) × (8/1)) = 1]
So, it is a whole number.
Whole numbers are collections of all positive integers, including 0.
(vii) 1/11
Solution:-
Reciprocal of (1/11) is (11/1) or 11 [∵ ((1/11) × (11/1)) = 1]
So, it is a whole number.
Whole numbers are collections of all positive integers, including 0.
3. Find:
(i) (7/3) ÷ 2
Solution:-
We have
= (7/3) × reciprocal of 2
= (7/3) × (1/2)
= (7 × 1) / (3 × 2)
= 7/6
=
(ii) (4/9) ÷ 5
Solution:-
We have
= (4/9) × reciprocal of 5
= (4/9) × (1/5)
= (4 × 1) / (9 × 5)
= 4/45
(iii) (6/13) ÷ 7
Solution:-
We have
= (6/13) × reciprocal of 7
= (6/13) × (1/7)
= (6 × 1) / (13 × 7)
= 6/91
(iv) 
÷ 3
Solution:-
First, convert the mixed fraction into an improper fraction.
We have,
=
= 13/3
Then,
= (13/3) × reciprocal of 3
= (13/3) × (1/3)
= (13 × 1) / (3 × 3)
= 13/9
(iv) 3 ½ ÷ 4
Solution:-
First, convert the mixed fraction into an improper fraction.
We have
= 3 ½ = 7/2
Then,
= (7/2) × reciprocal of 4
= (7/2) × (1/4)
= (7 × 1) / (2 × 4)
= 7/8
(iv) 
÷ 7
Solution:-
First, convert the mixed fraction into an improper fraction.
We have
=
= 31/7
Then,
= (31/7) × reciprocal of 7
= (31/7) × (1/7)
= (31 × 1) / (7 × 7)
= 31/49
4. Find.
(i) (2/5) ÷ (½)
Solution:-
We have
= (2/5) × reciprocal of ½
= (2/5) × (2/1)
= (2 × 2) / (5 × 1)
= 4/5
(ii) (4/9) ÷ (2/3)
Solution:-
We have
= (4/9) × reciprocal of (2/3)
= (4/9) × (3/2)
= (4 × 3) / (9 × 2)
= (2 × 1) / (3 × 1)
= 2/3
(iii) (3/7) ÷ (8/7)
Solution:-
We have
= (3/7) × reciprocal of (8/7)
= (3/7) × (7/8)
= (3 × 7) / (7 × 8)
= (3 × 1) / (1 × 8)
= 3/8
(iv) 
÷ (3/5)
Solution:-
First, convert the mixed fraction into an improper fraction.
We have,
=
= 7/3
Then,
= (7/3) × reciprocal of (3/5)
= (7/3) × (5/3)
= (7 × 5) / (3 × 3)
= 35/9
(v) 3 ½ ÷ (8/3)
Solution:-
First, convert the mixed fraction into an improper fraction.
We have
= 3 ½ = 7/2
Then,
= (7/2) × reciprocal of (8/3)
= (7/2) × (3/8)
= (7 × 3) / (2 × 8)
= 21/16
(vi) (2/5) ÷ 1 ½
Solution:-
First, convert the mixed fraction into an improper fraction.
We have
= 1 ½ = 3/2
Then,
= (2/5) × reciprocal of (3/2)
= (2/5) × (2/3)
= (2 × 2) / (5 × 3)
= 4/15
(vii) 
÷ 
Solution:-
First, convert the mixed fraction into an improper fraction.
We have
=
= 16/5
=
= 5/3
Then,
= (16/5) × reciprocal of (5/3)
= (16/5) × (3/5)
= (16 × 3) / (5 × 5)
= 48/25
(viii) 
÷ 
Solution:-
First, convert the mixed fraction into an improper fraction.
We have
=
= 11/5
=
= 6/5
Then,
= (11/5) × reciprocal of (6/5)
= (11/5) × (5/6)
= (11 × 5) / (5 × 6)
= (11 × 1) / (1 × 6)
= 11/6
Access other exercises of NCERT Solutions for Class 7 Chapter 2 – Fractions and Decimals
Also, explore –Â
Comments