S.chand publication New Learning Composite mathematics solution of class 7 Chapter 3 Rational numbers Exercise 3F

 Exercise 3F

Question 1

Find the reciprocal of each of the following.

(a) $\frac{8}{27}$

Sol :

$=\frac{27}{8}$

(b) -11

Sol :

$=\frac{-1}{11}$

(c) $\frac{-16}{45}$

Sol :

$=-\frac{45}{16}$

(d) $\frac{-3}{-19}$

Sol :

$=\frac{19}{3}$

(e) $\left|-2\frac{3}{27}\right|$

Sol :

$=-\frac{7}{17}$

(f) $-\left|1\frac{5}{6}\right|$

Sol :

$=\frac{-6}{11}$

(g) $|\frac{-4}{3} + \frac{8}{4}|$

Sol :

$=\left|-\frac{4}{3}-\frac{3}{4}\right|$

$=\frac{-16-9}{12}$

$=\frac{-25}{12}$

∴Reciprocal of $\left|-\frac{4}{3}+-\frac{3}{4}\right|$ is $-\frac{12}{25}$

Question 2

(a) $\frac{3}{4} \div \frac{1}{2}$

Sol :

$=\frac{3}{4} \times \frac{2}{1}$

$=\frac{3}{2}=1\frac{1}{2}$

(b) $-\frac{5}{8} \div \frac{3}{2}$

Sol :

$=\frac{-5}{8} \times \frac{2}{3}$

$=-\frac{5}{12}$

(c) $\frac{4}{5} \div -\frac{4}{25}$

Sol :

$= -\left(\frac{4}{5}\times \frac{25}{4} \right)$

=-5

(d) $-3\frac{1}{8} \div (-10)$

Sol :

$=\frac{25}{8} \times \frac{1}{10}$

$=\frac{5}{16}$

(e) $20 \div \frac{1}{-5}$

Sol :

=(-20×5)

=-100

(f) $-16 \div \left(-3\frac{1}{5}\right)$

Sol :

$=-16 \times \frac{-5}{16}$

=5

Question 3

(a) $1\frac{8}{49} \div \left(-13 \frac{4}{7}\right)$

Sol :

$=-\left(\frac{57}{49} \times \frac{7}{95}\right)$

$=-\frac{57}{665}=-\frac{3}{35}$

(b) $-5\frac \frac{5}{36} \div 4\frac{1}{9}$

 Sol :

$=-\left(\frac{185}{36} \times \frac{9}{37}\right)$

$=-\frac{5}{4} =-1\frac{1}{4}$

(c) $-3\frac{7}{9} \div \left(-4\frac{8}{15}\right)$

$=\frac{34}{9} \times \frac{15}{68}$

$=\frac{5}{6}$

Question 4

$2\frac{17}{54} \div \left(-4\frac{31}{36}\right)$

Sol :

$=-\left(\frac{125}{54} \times \frac{36}{175}\right)$

$=-\frac{10}{21}=-\frac{21}{10}=-2\frac{1}{10}$

Question 5

(a) $\left(\frac{2}{5} \div \frac{4}{-7}\right) \div \frac{1}{2}$

Sol :

$=-\left(\frac{2}{5} \times \frac{7}{4}\right) \div \frac{1}{-2}$

$=-\frac{7}{10}\div \frac{1}{-2}$

$=\frac{7}{10} \times \frac{2}{1}$

$=\frac{7}{5}=1\frac{2}{5}$

(b) $\frac{2}{5} \div \left(\frac{4}{-7}+\frac{1}{-2}\right)$

Sol :

$=\frac{2}{5} \div \left(\frac{4}{7} \div \frac{1}{-2}\right)$

$=\frac{2}{5} \div \left(\frac{4}{7} \times \frac{2}{1}\right)$

$=\frac{2}{5} \div \frac{8}{7}=\frac{7}{20}$

Ans : They are not equal