NCERT Solutions for Class 7 Maths Exercise 2.4
NCERT Solutions for Class 7 Maths Exercise 2.4 Chapter 2 Fractions and Decimals in simple solutions are given here. Class 7 Fractions and Decimals NCERT Exercise 2.4 of NCERT Solutions for Maths Class 7 Chapter 2 contains all the topics containing the basic information of Fractions. Class 7 Fractions and Decimals NCERT Exercise 2.4 provide students the study of fractions including proper, improper and mixed fractions as well as their addition and subtraction. NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals ex 2.4 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:-
We have,
= 12 × reciprocal of \(\dfrac{3}{4}\)
= 12 × \(\dfrac{4}{3}\)
= 4 × 4
= 16
(ii) \(14 \div \dfrac{5}{6}\)
Solution:-
= 14 × reciprocal of \(\dfrac{5}{6}\)
= 14 × \(\dfrac{6}{5}\)
= \(\dfrac{84}{5}\)
(iii) 8 ÷ \(\dfrac{7}{3}\)
Solution:-
= 8 × reciprocal of \(\dfrac{7}{3}\)
= 8 × \(\dfrac{3}{7}\)
= \(\dfrac{24}{7}\)
(iv) 4 ÷ \(\dfrac{8}{3}\)
Solution:-
We have,
= 4 × reciprocal of \(\dfrac{8}{3}\)
= 4 × \(\dfrac{3}{8}\)
= 1 × \(\dfrac{3}{2}\)
= \(\dfrac{3}{2}\)
(v) 3 ÷ \(2\dfrac{1}{3}\)
Solution:-
= 3 ÷ \(\dfrac{7}{3}\)
= 3 × reciprocal of \(\dfrac{3}{7}\)
= 3 × \(\dfrac{3}{7}\)
= \(\dfrac{9}{7}\)
(vi) 5 ÷ \(3\dfrac{4}{7}\)
Solution:-
= 5 ÷ \(\dfrac{25}{7}\)
= 5 × reciprocal of \(\dfrac{25}{7}\)
= 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:
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 \(\dfrac{2}{3}\)
= \(\dfrac{4}{9}\) × \(\dfrac{3}{2}\)
= \(\dfrac{2 × 1}{3 × 1}\)
= \(\dfrac{2}{3}\)
(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 \(\dfrac{3}{2}\)
= \(\dfrac{2}{5} × \dfrac{2}{3}\)
= \(\dfrac{2 × 2) }{ (5 × 3}\)
= \(\dfrac{4}{15}\)
