NCERT Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4

NCERT Solutions Class 7 Mathematics 
Chapter – 2 (Fractions and Decimals)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 2 Fractions and Decimals Exercise 2.4 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 2 Fractions and Decimals

Exercise – 2.4

1. Find:
(i) 12 ÷ \mathbf{\frac{3}{4}}                   (ii) 14 ÷ \mathbf{\frac{5}{6}}              (iii) 8 ÷ \mathbf{\frac{7}{3}}   

(iv) 4 ÷ \mathbf{\frac{8}{3}}                   (v) 3 ÷ \mathbf{2\frac{1}{3}}              (vi) 5 ÷ \mathbf{3\frac{4}{7}}

Solution –
(i) 12 ÷ \mathbf{\frac{3}{4}}
= 12 × reciprocal of  \frac{3}{4}
= 12 × \frac{4}{3}
= 4 × 4
= 16

(ii) 14 ÷ \mathbf{\frac{5}{6}}
= 14 × reciprocal of  \frac{5}{6}
= 14 × \frac{6}{5}
= \frac{84}{5} = 16\frac{4}{5}

(iii) 8 ÷ \mathbf{\frac{7}{3}}
= 8 × reciprocal of \frac{7}{3}
= 8 × \frac{3}{7}

= \frac{24}{7} = 3\frac{3}{7}

(iv) 4 ÷ \mathbf{\frac{8}{3}}
= 4 × reciprocal of  \frac{8}{3}
= 4 × \frac{3}{8}

= 1 × \frac{3}{2}  = \frac{3}{2}

(v) 3 ÷ \mathbf{2\frac{1}{3}}

= 3 \div \frac{7}{3}
= 3 × reciprocal of \frac{7}{3}
= 3 × \frac{3}{7}

= \frac{9}{7} = 1\frac{2}{7}

(vi) 5 ÷ \mathbf{3\frac{4}{7}}

= 5 ÷ \frac{25}{7}
= 5 × reciprocal of \frac{25}{7}

= 5 × \frac{7}{25}

= 1 × \frac{7}{5} = \frac{7}{5}

2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.

(i) \mathbf{\frac{3}{7}}           (ii) \mathbf{\frac{5}{8}}            (iii) \mathbf{\frac{9}{7}}           (iv) \mathbf{\frac{6}{5}}     

(v) \mathbf{\frac{12}{7}}        (vi) \mathbf{\frac{1}{8}}          (vii) \mathbf{\frac{1}{11}}

Solution –
(i) \mathbf{\frac{3}{7}}
Reciprocal of \frac{3}{7} is \frac{7}{3}
[∵ \frac{3}{7}  × \frac{7}{3}  = 1]
So, it is an improper fraction.
Improper fraction is that fraction in which numerator is greater than its denominator.

(ii) \mathbf{\frac{5}{8}}
Reciprocal of \frac{5}{8} is \frac{8}{5}
[∵ \frac{5}{8} × \frac{8}{5} = 1]
So, it is an improper fraction.
Improper fraction is that fraction in which numerator is greater than its denominator.

(iii) \mathbf{\frac{9}{7}}
Reciprocal of \frac{9}{7} is \frac{7}{9}
[∵ \frac{9}{7} × \frac{7}{9} = 1]
So, it is a proper fraction.
A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.

(iv) \mathbf{\frac{6}{5}}
Reciprocal of \frac{6}{5} is \frac{5}{6}
[∵ \frac{6}{5} × \frac{5}{6} = 1]
So, it is a proper fraction.
A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.

(v) \mathbf{\frac{12}{7}}
Reciprocal of \frac{12}{7} is \frac{7}{12}
[∵ \frac{12}{7} × \frac{7}{12} = 1]
So, it is a proper fraction.
A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.

(vi) \mathbf{\frac{1}{8}}
Reciprocal of \frac{1}{8} is \frac{8}{1} or 8
[∵ \frac{1}{8} × \frac{8}{1} = 1]
So, it is a whole number.
Whole numbers are collection of all positive integers including 0.

(vii) \mathbf{\frac{1}{11}}
Reciprocal of \frac{1}{11} is \frac{11}{1} or 11
[∵ \frac{1}{11} × \frac{11}{1} = 1]
So, it is a whole number.
Whole numbers are collection of all positive integers including 0.

3. Find:

(i) \mathbf{\frac{7}{3}} ÷ 2            (ii) \mathbf{\frac{4}{9}} ÷ 5                    (iii) \mathbf{\frac{6}{13}} ÷ 7

(iv) \mathbf{4\frac{1}{3}} ÷ 3      (v) \mathbf{3\frac{1}{2}} ÷ 4                  (vi) \mathbf{4\frac{3}{7}} ÷ 7         

Solution –

(i) \mathbf{\frac{7}{3}} ÷ 2

= \frac{7}{3} × reciprocal of 2

= \frac{7}{3} × \frac{1}{2}

= \frac{7}{6} = 1\frac{1}{6}

(ii) \mathbf{\frac{4}{9}} ÷ 5

= \frac{4}{9} × reciprocal of 5

= \frac{4}{9} × \frac{1}{5}

= \frac{4}{45}

(iii) \mathbf{\frac{6}{13}} ÷ 7

= \frac{6}{13} × reciprocal of 7

= \frac{6}{13} × \frac{1}{7}

= \frac{6}{91}

(iv) \mathbf{4\frac{1}{3}} ÷ 3

= \frac{13}{3} ÷ 3

= \frac{13}{3} × reciprocal of 3

= \frac{13}{3} × \frac{1}{3}

= \frac{13}{9} = 1\frac{4}{9}

(v) \mathbf{3\frac{1}{2}} ÷ 4

= \frac{7}{2} ÷ 4

= \frac{7}{2} × reciprocal of 4

= \frac{7}{2} × \frac{1}{4}

= \frac{7}{8}

(vi) \mathbf{4\frac{3}{7}} ÷ 7

= \frac{31}{7} ÷ 7

= \frac{31}{7} × reciprocal of 7

= \frac{31}{7} × \frac{1}{7}

= \frac{31}{49}

4. Find:

(i) \mathbf{\frac{2}{5}} ÷ \mathbf{\frac{1}{2}}          (ii) \mathbf{\frac{4}{9}} ÷ \mathbf{\frac{2}{3}}                 (iii) \mathbf{\frac{3}{7}} ÷ \mathbf{\frac{8}{7}}                  (iv) \mathbf{2\frac{1}{3}} ÷ \mathbf{\frac{3}{5}}       

(v) \mathbf{3\frac{1}{2}} ÷ \mathbf{\frac{8}{3}}      (vi) \mathbf{\frac{2}{5}} ÷ \mathbf{1\frac{1}{2}}             (vii) \mathbf{3\frac{1}{5}} ÷ \mathbf{1\frac{2}{3}}           (viii) \mathbf{2\frac{1}{5}} ÷ \mathbf{1\frac{1}{5}}

Solution –

(i) \mathbf{\frac{2}{5}} ÷ \mathbf{\frac{1}{2}}     

= \frac{2}{5} × reciprocal of \frac{1}{2}

= \frac{2}{5} × \frac{2}{1}

= \frac{4}{5}

(ii) \mathbf{\frac{4}{9}} ÷ \mathbf{\frac{2}{3}}

= \frac{4}{9} × reciprocal of \frac{2}{3}

= \frac{4}{9} × \frac{3}{2}

= \frac{2}{3}

(iii) \mathbf{\frac{3}{7}} ÷ \mathbf{\frac{8}{7}}

= \frac{3}{7} × reciprocal of \frac{8}{7}

= \frac{3}{7} × \frac{7}{8}

= \frac{3}{8}

(iv) \mathbf{2\frac{1}{3}} ÷ \mathbf{\frac{3}{5}}

= \frac{7}{3} ÷ \frac{3}{5}

= \frac{7}{3} × reciprocal of \frac{3}{5}

= \frac{7}{3} × \frac{5}{3}

= \frac{35}{9} = 3\frac{8}{9}

(v) \mathbf{3\frac{1}{2}} ÷ \mathbf{\frac{8}{3}}

= \frac{7}{2} ÷ \frac{8}{3}

= \frac{7}{2} × reciprocal of  \frac{8}{3}

= \frac{7}{2} × \frac{3}{8}

= \frac{21}{16}

(vi) \mathbf{\frac{2}{5}} ÷ \mathbf{1\frac{1}{2}}

= \frac{2}{5} ÷ \frac{3}{2}

= \frac{2}{5} × reciprocal of \frac{3}{2}

= \frac{2}{5} × \frac{2}{3}

= \frac{4}{15}

(vii) \mathbf{3\frac{1}{5}} ÷ \mathbf{1\frac{2}{3}}

= \frac{16}{5} ÷ \frac{5}{3}

= \frac{16}{5} × reciprocal of \frac{5}{3}

= \frac{16}{5} × \frac{3}{5}

= \frac{48}{25} = 1\frac{23}{25}

(viii) \mathbf{2\frac{1}{5}} ÷ \mathbf{1\frac{1}{5}}

= \frac{11}{5} ÷ \frac{6}{5}

= \frac{11}{5} × reciprocal of \frac{6}{5}

= \frac{11}{5} × \frac{5}{6}

= \frac{11}6{}

 

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