S.chand publication New Learning Composite mathematics solution of class 7 Chapter 2 Fractions and Decimals Exercise 2D

Exercise 2D

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Q1 | Ex-2D | Class 7 | Schand New Learning Composite | Fractions and Decimals | Chapter 2| myhelper

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Question 1

Find the reciprocal of :

(a) $\frac{1}{3}$

(b) $\frac{7}{8}$

(c) $2 \frac{2}{3}$

(d) 5

Sol :

(a) Reciprocal of $\frac{1}{3}=3$

(b) Reciprocal of $\frac{7}{8}=\frac{8}{7}=1\frac{1}{7}$

(c) Reciprocal of 2$\frac{2}{3}=\frac{3}{8}$

(d) Reciprocal of $5=\frac{1}{5}$

(e) Reciprocal of $6\frac{1}{6}=\frac{6}{37}$

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Q2 | Ex-2D | Class 7 | Schand New Learning Composite | Fractions and Decimals | Chapter 2| myhelper

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Question 2

By taking following examples, show that the product of a number and its reciprocal

(a) $\frac{25}{37}$

(b) $5 \frac{1}{9}$

(c) $7 \frac{1}{7}$

(d) $1 \frac{11}{35}$

Sol :

(a) Reciprocal of $\frac{25}{37}=\frac{37}{25}$

∴$\frac{25}{37}\times \frac{37}{25}$

=1

(b) Reciprocal of $5\frac{1}{9}=\frac{46}{9}$

∴$\frac{46}{9}\times \frac{9}{46}$

=1

(c) $7\frac{1}{7}=\frac{50}{7}$

∴$\frac{50}{7}\times \frac{7}{50}$

=1

(d) $1\frac{11}{35}=\frac{46}{35}$

Reciprocal of $\frac{46}{35}=\frac{35}{46}$

=1

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Q3 | Ex-2D | Class 7 | Schand New Learning Composite | Fractions and Decimals | Chapter 2| myhelper

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Question 3

Think and answer- in defining reciprocal of we have imposed the condition $a\not =0,~ b\not=0$ why ? Give reason.

Sol :

Because division by zero is not defined