RD Sharma solution class 7 chapter 5 Operations On Rational numbers Exercise 5.2

Exercise 5.2

Page-5.7

Question 1:

Subtract the first rational number from the second in each of the following:
(i) $\frac{3}{8}, \frac{5}{8}$
(ii) $\frac{-7}{9}, \frac{4}{9}$
(iii) $\frac{-2}{11}, \frac{-9}{11}$
(iv) $\frac{11}{13}, \frac{-4}{13}$

Answer 1:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{5}{8}-\frac{3}{8}=\frac{2}{8} \\ \left(\mathrm{ii}\right) \frac{4}{9}-\frac{-7}{9}=\frac{11}{9} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right) \frac{-9}{11}-\frac{-2}{11}=\frac{-9+2}{11}=\frac{-7}{11} \\ \left(\mathrm{iv}\right) \frac{-4}{13}-\frac{11}{13}=\frac{-4-11}{13}=\frac{-15}{13} \end{gathered}$$

Question 2:

Evaluate each of the following:
(i) $\frac{2}{3}-\frac{3}{5}$
(ii) $-\frac{4}{7}-\frac{2}{-3}$
(iii) $\frac{4}{7}-\frac{-5}{-7}$
(iv) $-2-\frac{5}{9}$

Answer 2:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{2}{3}-\frac{3}{5}=\frac{2\times 5-3\times 3}{15}=\frac{10-9}{15}=\frac{1}{15} \\ \left(\mathrm{ii}\right) -\frac{4}{7}-\frac{2}{-3}=\frac{-4\times 3+2\times 7}{21}=\frac{-12+14}{21}=\frac{2}{21} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) \frac{4}{7}-\frac{-5}{-7}=\frac{4-5}{7}=\frac{-1}{7} \\ \left(\mathrm{iv}\right) -\frac{2}{1}-\frac{5}{9}=\frac{-2\times 9-5}{9}=\frac{=-18-5}{9}=\frac{-23}{9} \end{gathered}$$

Question 3:

The sum of the two numbers is $\frac{5}{9}$. If one of the numbers is $\frac{1}{3}$, find the other.

Answer 3:

First number = $\frac{1}{3}$
Let the second number = x.
According to the question, we have
$$\begin{gathered} \frac{1}{3}+x=\frac{5}{9} \\ \Rightarrow x=\frac{5}{9}-\frac{1}{3}=\frac{5-1\times 3}{9}=\frac{5-3}{9}=\frac{2}{9} \end{gathered}$$

Question 4:

The sum of two numbers is $\frac{-1}{3}$. If one of the numbers is $\frac{-12}{3}$, find the other.

Answer 4:

First number = $\frac{-12}{3}$
Let the second number = x.
Then, according to the question, we have
$$\begin{gathered} \frac{-12}{3}+x=\frac{-1}{3} \\ \Rightarrow x=\frac{-1}{3}-\frac{-12}{3}=\frac{-1+12}{3}=\frac{11}{3} \end{gathered}$$

Question 5:

The sum of two numbers is $\frac{-4}{3}$. If one of the numbers is −5, find the other.

Answer 5:

First number = $-5$
Let the second number = x.
Then, according to the question, we have
$$\begin{gathered} -5+x=\frac{-4}{3} \\ \Rightarrow x=\frac{-4}{3}-\frac{-5}{1}=\frac{-4+5\times 3}{3}=\frac{-4+15}{3}=\frac{11}{3} \end{gathered}$$

Question 6:

The sum of two rational numbers is −8. If one of the numbers is $\frac{-15}{7}$, find the other.

Answer 6:

First number = $\frac{-15}{7}$
Let the second number = x.
Then, according to the question, we have
$$\begin{gathered} \frac{-15}{7}+x=-8 \\ \Rightarrow x=\frac{-8}{1}-\frac{-15}{7}=\frac{-8\times 7+15}{7}=\frac{-56+15}{7}=\frac{-41}{7} \end{gathered}$$

Question 7:

What should be added to $\frac{-7}{8}$ so as to get $\frac{5}{9}?$

Answer 7:

Let x  be added to $\frac{-7}{8} \mathrm{to} \mathrm{get} \frac{5}{9}.$
Then, according to the question, we have
$$\begin{gathered} x+\frac{-7}{8}=\frac{5}{9} \\ =>x = \frac{5}{9}-\frac{-7}{8}=\frac{5\times 8+7\times 9}{72}=\frac{40+63}{72}=\frac{103}{72} \end{gathered}$$

Question 8:

What number should be added to $\frac{-5}{11}$ so as to get $\frac{26}{33}?$

Answer 8:

Let x  be added to $\frac{-5}{11} \mathrm{to} \mathrm{get} \frac{26}{33}.$
Then, according to the question, we have
$$\begin{gathered} x+\frac{-5}{11}=\frac{26}{33} \\ => x = \frac{26}{33}-\frac{-5}{11} \\ =\frac{26+5\times 3}{33} \\ =\frac{26+15}{33} \\ =\frac{41}{33} \end{gathered}$$

Question 9:

What number should be added to $\frac{-5}{7}$ to get $\frac{-2}{3}?$

Answer 9:

Let x  be added to $\frac{-5}{7} \mathrm{to} \mathrm{get} \frac{-2}{3}.$
Then, according to the question, we have
$$\begin{gathered} x+\frac{-5}{7}=\frac{-2}{3} \\ \Rightarrow x = \frac{-2}{3}-\frac{-5}{7}=\frac{-2\times 7+5\times 3}{21}=\frac{-14+15}{21}=\frac{1}{21} \end{gathered}$$

Page-5.8

Question 10:

What number should be suvtracted from $\frac{-5}{3}$ to get $\frac{5}{6}?$

Answer 10:

Let x be the number that should be subtracted from $\frac{-5}{3} \mathrm{to} \mathrm{get} \frac{5}{6}.$
Then, according to the question, we have
 $$\begin{gathered} \frac{-5}{3}-x=\frac{5}{6} \\ \Rightarrow x = \frac{-5}{3}-\frac{5}{6}=\frac{-5\times 2}{3\times 2}-\frac{5}{6}=\frac{-10-5}{6}=\frac{-15}{6}=\frac{-5}{2} \end{gathered}$$

Question 11:

What number should be subtracted from $\frac{3}{7}$ to get $\frac{5}{4}?$

Answer 11:

Let x be the number that should be subtracted from $\frac{3}{7} \mathrm{to} \mathrm{get} \frac{5}{4}.$
Then, according to the question, we have
$$\begin{gathered} \frac{3}{7}-x=\frac{5}{4} \\ \Rightarrow x=\frac{3}{7}-\frac{5}{4}=\frac{3\times 4-5\times 7}{28}=\frac{12-35}{28}=\frac{-23}{28} \end{gathered}$$

Question 12:

What should be added to $\left(\frac{2}{3}+\frac{3}{5}\right)$ to get $\frac{-2}{15}?$

Answer 12:

$\frac{2}{3}+\frac{3}{5}=\frac{2\times 5}{3\times 5}+\frac{3\times 3}{5\times 3}=\frac{10}{15}+\frac{9}{15}=\frac{19}{15}$
Let x be the number that should be added to $\frac{19}{15}$ to get $\frac{-2}{15}$
Then, we have
$$\begin{gathered} \frac{19}{15}+x=\frac{-2}{15} \\ =>x=\frac{-2}{15}-\frac{19}{15} \\ =\frac{-21}{15} \\ =\frac{-7\times 3}{5\times 3} \\ =\frac{-7}{5} \end{gathered}$$

Question 13:

What should be added to $\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}\right)$ to get 3?

Answer 13:

Let x be added to $(\frac{1}{2}+\frac{1}{3}+\frac{1}{5})=(\frac{1\times 15}{2\times 15}+\frac{1\times 10}{3\times 10}+\frac{1\times 6}{5\times 6})=(\frac{15}{30}+\frac{10}{30}+\frac{6}{30})=\frac{31}{30} \mathrm{to} \mathrm{get} 3.$
Then, we have
$$\begin{gathered} \frac{31}{30}+x=3 \\ \Rightarrow x=3-\frac{31}{30}=\frac{3\times 30}{1\times 30}-\frac{31}{30}=\frac{90}{30}-\frac{31}{30}=\frac{59}{30} \end{gathered}$$

Question 14:

What should be subtracted from $\left(\frac{3}{4}-\frac{2}{3}\right)$ to get $\frac{-1}{6}?$

Answer 14:

Let  x be the number that should be subtracted from $\frac{3}{4}-\frac{2}{3}=\frac{3\times 3}{4\times 3}-\frac{2\times 4}{3\times 4}=\frac{9}{12}-\frac{8}{12}=\frac{1}{12} \mathrm{to} \mathrm{get} \frac{-1}{6}.$

Then, we have
$$\begin{gathered} \frac{1}{12}-x=\frac{-1}{6} \\ \Rightarrow x=\frac{1}{12}-\frac{-1}{6}=\frac{1}{12}-\frac{-1\times 2}{6\times 2}=\frac{1}{12}-\frac{-2}{12}=\frac{3}{12}=\frac{1}{4} \end{gathered}$$

Question 15:

Simplify:
(i) $\frac{-3}{2}+\frac{5}{4}-\frac{7}{4}$
(ii) $\frac{5}{3}-\frac{7}{6}+\frac{-2}{3}$
(iii) $\frac{5}{4}-\frac{7}{6}-\frac{-2}{3}$
(iv) $\frac{-2}{5}-\frac{-3}{10}-\frac{-4}{7}$

Answer 15:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{-3}{2}+\frac{5}{4}-\frac{7}{4}=\frac{-3\times 2}{2\times 2}+\frac{5}{4}-\frac{7}{4}=\frac{-6+5-7}{4}=\frac{-8}{4}=-2 \\ \left(\mathrm{ii}\right) \frac{5}{3}-\frac{7}{6}+\frac{-2}{3}=\frac{5\times 2}{3\times 2}-\frac{7}{6}+\frac{-2\times 2}{3\times 2}=\frac{10-7-4}{6}=\frac{-1}{6} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) \frac{5}{4}-\frac{7}{6}-\frac{-2}{3}=\frac{5\times 3}{4\times 3}-\frac{7\times 2}{6\times 2}-\frac{-2\times 4}{3\times 4}=\frac{15-14+8}{12}=\frac{9}{12}=\frac{3}{4} \\ \left(\mathrm{iv}\right) \frac{-2}{5}-\frac{-3}{10}-\frac{-4}{7}=\frac{-2\times 14}{5\times 14}-\frac{-3\times 7}{10\times 7}-\frac{-4\times 10}{7\times 10}=\frac{-28+21+40}{70}=\frac{33}{70} \end{gathered}$$

Question 16:

Fill in the blanks:
(i) $\frac{-4}{13}-\frac{-3}{26}=....$
(ii) $\frac{-9}{14}+....=-1$
(iii) $\frac{-7}{9}+....=3$
(iv) $....+\frac{15}{23}=4$

Answer 16:

(i) $\frac{-4}{13}-\frac{-3}{26}=\frac{-4\times 2}{13\times 2}-\frac{-3}{26}=\frac{-8+3}{26}=\frac{-5}{26}$
$$\begin{gathered} \left(\mathrm{ii}\right) \frac{-9}{14}+\mathrm{x}=-1 \\ \Rightarrow \mathrm{x}=-1-\frac{-9}{14}=\frac{-14+9}{14}=\frac{-5}{14} \\ \left(\mathrm{iii}\right) \frac{-7}{9}+\mathrm{x}=3 \\ \Rightarrow \mathrm{x}=\frac{3\times 9}{1\times 9}-\frac{-7}{9}=\frac{27+7}{9}=\frac{34}{9} \end{gathered}$$

(iv)
$$\begin{gathered} x+\frac{15}{23}=4 \\ \Rightarrow x=4-\frac{15}{23}=\frac{4\times 23}{1\times 23}-\frac{15}{23}=\frac{92-15}{23}=\frac{77}{23} \end{gathered}$$