class 7 fraction and decimals exercise 2.3

ncert solutions for class 7 maths chapter 2 exercise 2.3

ncert solutions for class 7 maths chapter 2 exercise 2.3 in simple solutions are given here. ncert solutions for class 7 maths chapter 2 exercise 2.3 of NCERT Solutions for Maths Class 7 Chapter 2 contains all the topics containing the basic information of Fractions. ncert solutions for class 7 maths chapter 2 exercise 2.3 provide students the study of fractions including proper, improper and mixed fractions as well as their addition and subtraction. NCERT Solutions for ncert solutions for class 7 maths chapter 2 exercise 2.3 provides necessary material for solving different range of questions that test the students capability of understanding the concepts. It also helps the students of CBSE Class 7 Maths students

 

Question 1. 

Find:

(i) \(12 \div \dfrac{3}{4}\)

Solution:-

= 12 \div  \(\dfrac{3}{4}\)

= 12 × \(\dfrac{4}{3}\)

= 4 × 4

= 16

(ii) \(14 \div \dfrac{5}{6}\)

Solution:-

= 14 × \(\dfrac{6}{5}\)

= \(\dfrac{84}{5}\)

(iii) 8 ÷ \(\dfrac{7}{3}\)

Solution:-

= 8 × \(\dfrac{3}{7}\)

= \(\dfrac{24}{7}\)

(iv) 4 ÷ \(\dfrac{8}{3}\)

Solution:-

= 4 × \(\dfrac{3}{8}\)

=  \(\dfrac{3}{2}\)

= \(\dfrac{3}{2}\)

(v) 3 ÷ \(2\dfrac{1}{3}\)

Solution:-

= 3 ÷ \(\dfrac{7}{3}\)

= 3 × \(\dfrac{3}{7}\)

= \(\dfrac{9}{7}\)

(vi) 5 ÷ \(3\dfrac{4}{7}\)

Solution:-

= 5 ÷ \(\dfrac{25}{7}\)

= 5 × \(\dfrac{7}{25}\)

= 1 × \(\dfrac{7}{5}\)

= \(\dfrac{7}{5}\)

Question 2. 

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

(i) \(\dfrac{3}{7}\)

Solution:-

Reciprocal of \(\dfrac{3}{7}\) is \(\dfrac{7}{3}\) 

So, it is an improper fraction.

Improper fraction is that fraction in which numerator is greater than its denominator.

(ii) \(\dfrac{5}{8}\)

Solution:-

Reciprocal of \(\dfrac{5}{8}\) is \(\dfrac{8}{5}\)

So, it is an improper fraction.

Improper fraction is that fraction in which numerator is greater than its denominator.

(iii) \(\dfrac{9}{7}\)

Solution:-

Reciprocal of \(\dfrac{9}{7}\) is \(\dfrac{7}{9}\)

So, it is a proper fraction.

A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.

(iv) \(\dfrac{6}{5}\)

Solution:-

Reciprocal of \(\dfrac{6}{5}\) is \(\dfrac{5}{6}\)

So, it is a proper fraction.

A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.

(v) \(\dfrac{12}{7}\)

Solution:-

Reciprocal of \(\dfrac{12}{7}\) is \(\dfrac{7}{12}\)

So, it is a proper fraction.

A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.

(vi) \(\dfrac{1}{8}\)

Solution:-

Reciprocal of \(\dfrac{1}{8}\) is \(\dfrac{8}{1}\) or 8.

So, it is a whole number.

Whole numbers are collection of all positive integers including 0.

(vii) \(\dfrac{1}{11}\)

Solution:-

Reciprocal of \(\dfrac{1}{11}\) is \(\dfrac{11}{1}\) or 11 .

So, it is a whole number.

Whole numbers are collection of all positive integers including 0.

Question 3. 

Find:

(i) \(\dfrac{7}{3}\) ÷ 2

Solution:-

We have,

=\(\dfrac{7}{3}\)  × reciprocal of 2

= \(\dfrac{7}{3}  × \dfrac{1}{2}\)

= \(\dfrac{7 × 1 }{ 3 × 2}\) 

= \(\dfrac{7}{6}\) 

=\(2\dfrac{1}{3}\) 

(ii) \(\dfrac{4}{9}\) ÷ 5

Solution:-

= \(\dfrac{4}{9}\) × reciprocal of 5

= \(\dfrac{4}{9}\) × \(\dfrac{1}{5}\)

= \(\dfrac{4 × 1}{9 × 5}\)

= \(\dfrac{4}{45}\)

(iii) \(\dfrac{6}{13}\) ÷ 7

Solution:-

= \(\dfrac{6}{13}\) × reciprocal of 7

= \(\dfrac{6}{13}\) × \(\dfrac{1}{7}\)

= \(\dfrac{6 × 1}{13 × 7}\)

= \(\dfrac{6}{91}\)

(iv) \(4\dfrac{1}{3}\)÷ 3

Solution:-

= \(\dfrac{13}{3}\) × reciprocal of 3

= \(\dfrac{13}{3}\) × \(\dfrac{1}{3}\)

= \(\dfrac{13 × 1}{3 × 3}\)

= \(\dfrac{13}{9}\)

(v) \(3\dfrac{1}{2}\) ÷ 4

Solution:-

= \(\dfrac{7}{2}\) × reciprocal of 4

= \(\dfrac{7}{2}\) × \(\dfrac{1}{4}\)

= \(\dfrac{7 × 1}{2 × 4}\)

= \(\dfrac{7}{8}\)

(vi)  \(4\dfrac{3}{7}\)÷ 7

Solution:-

=\(\dfrac{31}{7}\) 

= \(\dfrac{31}{7}\) × reciprocal of 7

= \(\dfrac{31}{7}\)  × \(\dfrac{1}{7}\) 

= \(\dfrac{31 × 1}{7 × 7}\)

= \(\dfrac{31}{49}\) 

4. Find:

(i) \(\dfrac{2}{5} ÷ \dfrac{1}{2}\)

Solution:-

We have,

= \(\dfrac{2}{5}\) × reciprocal of \(\dfrac{1}{2}\)

= \(\dfrac{2}{5}\)  × \(\dfrac{2}{1}\) 

= \(\dfrac{2 × 2}{5 × 1}\)

= \(\dfrac{4}{5}\) 

(ii) \(\dfrac{4}{9}\)  ÷ \(\dfrac{2}{3}\) 

Solution:-

We have,

= \(\dfrac{4}{9}\)× reciprocal of \(\dfrac{2}{3}\)

= \(\dfrac{4}{9}\) × \(\dfrac{3}{2}\)

= \(\dfrac{4 × 3}{9 × 2}\)

= \(\dfrac{2 × 1}{3 × 1}\)

= \(\dfrac{2}{3}\)

(iii) \(\dfrac{3}{7}\) ÷ \(\dfrac{8}{7}\)

Solution:-

We have,

= \(\dfrac{3}{7}\) × reciprocal of \(\dfrac{8}{7}\)

= \(\dfrac{3}{7}\) × \(\dfrac{7}{8}\)

= \(\dfrac{3 × 7}{7 × 8}\)

= \(\dfrac{3 × 1}{1 × 8}\)

= \(\dfrac{3}{8}\)

(iv) \(2\dfrac{1}{3}\) ÷ \(\dfrac{3}{5}\)

Solution:-

= \(\dfrac{7}{3}\) × reciprocal of \(\dfrac{3}{5}\)

= \(\dfrac{7}{3}\) × \(\dfrac{5}{3}\)

= \(\dfrac{7 × 5}{3 × 3}\)

= \(\dfrac{35}{9}\)

(v) \(3\dfrac{1}{2}\) ÷ \(\dfrac{8}{3}\)

Solution:-

= \(\dfrac{7}{2}\) × reciprocal of \(\dfrac{8}{3}\)

= \(\dfrac{7}{2}\) × \(\dfrac{3}{8}\)

= \(\dfrac{7 × 3}{2 × 8}\)

= \(\dfrac{21}{16}\)

(vi) \(\dfrac{2}{5}\) ÷ \(1\dfrac{1}{2}\)

Solution:-

= \(\dfrac{2}{5}\) × reciprocal of \(\dfrac{3}{2}\)

= \(\dfrac{2}{5}\) × \(\dfrac{2}{3}\)

= \(\dfrac{2 × 2}{5 × 3}\)

= \(\dfrac{4}{15}\)

(vii) \(3\dfrac{1}{5}\) ÷ \(\dfrac{2}{5}\)

Solution:-

= \(\dfrac{16}{5}\) × reciprocal of \(\dfrac{5}{3}\)

= \(\dfrac{16}{5}\) × \(\dfrac{3}{5}\)

= \(\dfrac{16 × 3}{5 × 5}\)

= \(\dfrac{48}{25}\)

(viii) \(\dfrac{2}{5}\div 1\dfrac{1}{2}\)

Solution:-

= \(\dfrac{2}{5}\) × reciprocal of \(\dfrac{3}{2}\)

= \(\dfrac{2}{5} × \dfrac{2}{3}\)

= \(\dfrac{2 × 2) }{ (5 × 3}\)

= \(\dfrac{4}{15}\)