RD Sharma solution class 7 chapter 5 Operations On Rational numbers Exercise 5.4

Exercise 5.4

Page-5.13

Question 1:

Divide:
(i) $1 \mathrm{by} \frac{1}{2}$
(ii) $5 \mathrm{by} \frac{-5}{7}$
(iii) $\frac{-3}{4} \mathrm{by} \frac{9}{-16}$
(iv) $\frac{-7}{8} \mathrm{by} \frac{-21}{16}$
(v) $\frac{7}{-4} \mathrm{by} \frac{63}{64}$
(vi) $0 \mathrm{by} \frac{-7}{5}$
(vii) $\frac{-3}{4} \mathrm{by} -6$
(viii) $\frac{2}{3} \mathrm{by} \frac{-7}{12}$

Answer 1:

$$\begin{gathered} \left(\mathrm{i}\right) 1÷\frac{1}{2}=1\times \frac{2}{1}=2 \\ \left(\mathrm{ii}\right) 5÷\frac{-5}{7}=5\times \frac{7}{-5}=-7 \\ \left(\mathrm{iii}\right) \frac{-3}{4}÷\frac{9}{-16}=\frac{-3}{4}\times \frac{-16}{9}=\frac{-3}{4}\times \frac{-4\times 4}{3\times 3}=\frac{4}{3} \\ \left(\mathrm{iv}\right) \frac{-7}{8}÷\frac{-21}{16}=\frac{-7}{8}\times \frac{-16}{21}=\frac{-7}{8}\times \frac{-8\times 2}{7\times 3}=\frac{2}{3} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right) \frac{7}{-4}÷\frac{63}{64}=\frac{7}{-4}\times \frac{64}{63}=\frac{-7}{4}\times \frac{4\times 16}{7\times 9}=\frac{-16}{9} \\ \left(\mathrm{vi}\right) 0÷\frac{-7}{5}=0\times \frac{-7}{5}=0 \\ \left(\mathrm{vii}\right) \frac{-3}{4}÷-6=\frac{-3}{4}\times \frac{-1}{6}=\frac{-3}{4}\times \frac{-1}{2\times 3}=\frac{1}{8} \\ \left(\mathrm{viii}\right) \frac{2}{3}÷\frac{-7}{12}=\frac{2}{3}\times \frac{-12}{7}=\frac{2}{3}\times \frac{-4\times 3}{7}=\frac{-8}{7} \end{gathered}$$

Question 2:

Find the value and express as a rational number in standard form:
(i) $\frac{2}{5}÷\frac{26}{15}$
(ii) $\frac{10}{3}÷\frac{-35}{12}$
(iii) $-6÷\left(\frac{-8}{17}\right)$
(iv) $\frac{40}{98}÷(-20)$

Answer 2:

Question 3:

The product of two rational numbers is 15. If one of the numbers is −10, find the other.

Answer 3:

Let the first rational number = x.
Second number = −10
Their product = 15

Then, we have
$$\begin{gathered} x\times -10=15 \\ \Rightarrow x=15\times \frac{1}{-10}=5\times 3\times \frac{-1}{2\times 5}=\frac{-3}{2} \end{gathered}$$

Question 4:

The product of two rational numbers is $\frac{-8}{9}$. If one of the numbers is $\frac{-4}{15}$, find the other.

Answer 4:

Let the first rational number = x
Second number             = $\frac{-4}{15}$
Their product              = $\frac{-8}{9}$

Then, we have
$$\begin{gathered} x\times \frac{-4}{15}=\frac{-8}{9} \\ \Rightarrow x=\frac{-8}{9}\times \frac{-15}{4}=\frac{-2\times 4}{3\times 3}\times \frac{-3\times 5}{4}=\frac{10}{3} \end{gathered}$$

Question 5:

By what number should we multiply $\frac{-1}{6}$ so that the product may be $\frac{-23}{9}$?

Answer 5:

Let x be the number by which we should multiply $\frac{-1}{6} \mathrm{to} \mathrm{get} \frac{-23}{9}.$
Then, according to the question, we have
$$\begin{gathered} \frac{-1}{6}\times x=\frac{-23}{9} \\ \Rightarrow x=\frac{-23}{9}\times (-6)=\frac{46}{3} \end{gathered}$$

Question 6:

By what number should we multiply $\frac{-15}{28}$ so that the product may be $\frac{-5}{7}$?

Answer 6:

Let x  be the number by which we multiply $\frac{-15}{28} \mathrm{to} \mathrm{get} \mathrm{the} \mathrm{product} \frac{-5}{7}.$
Then, we have

$$\begin{gathered} x\times \frac{-15}{28}=\frac{-5}{7} \\ \Rightarrow x=\frac{-5}{7}\times \frac{28}{-15}=\frac{-5}{7}\times \frac{7\times 4}{-5\times 3}=\frac{4}{3} \end{gathered}$$

Page-5.14

Question 7:

By what number should we multiply $\frac{-8}{13}$ so that the product may be 24?

Answer 7:

Let x be the number required. Then, we have
$$\begin{gathered} x\times \frac{-8}{13}=24 \\ \Rightarrow x=24\times \frac{-13}{8}=-13\times 3=-39 \end{gathered}$$

Question 8:

By what number should $\frac{-3}{4}$ be multiplied in order to produce $\frac{2}{3}?$

Answer 8:

Let x  be the number by which we should multiply $\frac{-3}{4} \mathrm{to} \mathrm{get} \frac{2}{3}.$
Then, we have
$\frac{-3}{4}\times x=\frac{2}{3}\Rightarrow x=\frac{2}{3}\times \frac{4}{-3}=\frac{-8}{9}$

Question 9:

Find (x + y) ÷ (x + y), if
(i) $x=\frac{2}{3}, y=\frac{3}{2}$
(ii) $x=\frac{2}{5}, y=\frac{1}{2}$
(iii) $x=\frac{5}{4}, y=\frac{-1}{3}$

Answer 9:

(i) x = $\frac{2}{3}, y = \frac{3}{2}$
Then, (x+y)  = $\frac{2}{3}+\frac{3}{2}=\frac{2\times 2}{3\times 2}+\frac{3\times 3}{2\times 3}=\frac{4}{6}+\frac{9}{6}=\frac{13}{6}$
$(x-y) = \frac{2}{3}-\frac{3}{2} = \frac{4}{6}- \frac{9}{6}= \frac{-5}{6}$
Then, $(x+y)÷(x-y) = \frac{13}{6}÷\frac{-5}{6}=\frac{13}{6}\times \frac{6}{-5}=\frac{-13}{5}$.

(ii) x = $\frac{2}{5}, y = \frac{1}{2}$
Then, (x+y) = $\frac{2}{5}+\frac{1}{2}=\frac{2\times 2}{5\times 2}+\frac{1\times 5}{2\times 5}=\frac{4}{10}+\frac{5}{10}=\frac{9}{10}$
$(x-y) = \frac{2}{5}-\frac{1}{2} = \frac{4}{10}- \frac{5}{10}= \frac{-1}{10}$

Then, $(x+y)÷(x-y) = \frac{9}{10}÷\frac{-1}{10}=-9$

(iii) x = $\frac{5}{4}, y = \frac{-1}{3}$
Then, (x+y) = $\frac{5}{4}+\frac{-1}{3}=\frac{5\times 3}{4\times 3}+\frac{-1\times 4}{3\times 4}=\frac{15}{12}+\frac{-4}{12}=\frac{11}{12}$
$(x-y) = \frac{5}{4}-\frac{-1}{3} = \frac{5}{4}+ \frac{1}{3}= \frac{19}{12}$
Then, $(x+y)÷(x-y) = \frac{11}{12}÷\frac{19}{12}=\frac{11}{12}\times \frac{12}{19}=\frac{11}{19}.$

Question 10:

The cost of $7\frac{2}{3}$ metres of rope is Rs $12\frac{3}{4}$. Find its cost per metre.

Answer 10:

The cost of $7\frac{2}{3}=\frac{23}{3}\mathrm{met}\text{res}$  of rope = Rs. $12\frac{3}{4}=\frac{51}{4}$.
Then, the cost of 1 metre of rope = Rs. $\frac{51}{4}÷\frac{23}{3}=\frac{51}{4}\times \frac{3}{23}=\frac{153}{92}= \mathrm{Rs}. 1\frac{61}{92}.$

Question 11:

The cost of $2\frac{1}{3}$ metres of cloth is Rs $75\frac{1}{4}$. Find the cost of cloth per metre.

Answer 11:

The cost of $2\frac{1}{3}=\frac{7}{3} \mathrm{met}\text{res of} \mathrm{cloth}$= $\mathrm{Rs}. 75\frac{1}{4}=\frac{301}{4}$.
The cost of 1 metre of cloth = Rs. $\frac{301}{4}÷\frac{7}{3}=\frac{301}{4}\times \frac{3}{7}=\frac{43\times 7}{4}\times \frac{3}{7}=\frac{129}{4}=32\frac{1}{4}.$

Question 12:

By what number should $\frac{-33}{16}$ be divided to got $\frac{-11}{4}$?

Answer 12:

Let x  be the number required.

Then, we have

$$\begin{gathered} \frac{-33}{16}÷x=\frac{-11}{4} \\ \Rightarrow \frac{-33}{16}\times \frac{1}{x}=\frac{-11}{4} \\ \Rightarrow \frac{-33}{16}\times \frac{4}{-11}=x \\ x=\frac{-3\times 11}{4\times 4}\times \frac{4}{-11}=\frac{3}{4} \end{gathered}$$

Question 13:

Divide the sum of $\frac{-13}{5}$ and $\frac{12}{7}$ by the product of $\frac{-31}{7} \mathrm{and} \frac{-1}{2}.$

Answer 13:

$$\begin{gathered} \mathrm{The} \mathrm{sum} \mathrm{of} \frac{-13}{5} \mathrm{and} \frac{12}{7} \mathrm{is} \\ \frac{-13}{5}+\frac{12}{7}=\frac{-13\times 7}{5\times 7}+\frac{12\times 5}{7\times 5}=\frac{-91}{35}+\frac{60}{35} \\ =\frac{-91+60}{35}=\frac{-31}{35} \end{gathered}$$

$$\begin{gathered} \mathrm{The} \mathrm{product} \mathrm{of} \frac{-31}{7} \mathrm{and} \frac{-1}{2} \mathrm{is} \\ \frac{-31}{7}\times \frac{-1}{2}=\frac{31}{14} \end{gathered}$$

Then, according to the question, we have

$\frac{-31}{35}÷\frac{31}{14}=\frac{-31}{35}\times \frac{14}{31}=\frac{-2}{5}$

Question 14:

Divide the sum of $\frac{65}{12}$  and $\frac{8}{3}$ by their difference.

Answer 14:

$$\begin{gathered} \mathrm{The} \mathrm{sum} \mathrm{of} \frac{65}{12} \mathrm{and} \frac{8}{3} \mathrm{is} \\ \frac{65}{12}+\frac{8}{3}=\frac{65}{12}+\frac{8\times 4}{3\times 4}=\frac{65}{12}+\frac{32}{12}=\frac{65+32}{12}=\frac{97}{12} \\ \mathrm{The} \mathrm{difference} \mathrm{of} \frac{65}{12} \mathrm{and} \frac{8}{3} \mathrm{is} \\ \frac{65}{12}-\frac{8}{3}=\frac{65}{12}-\frac{8\times 4}{3\times 4}=\frac{65}{12}-\frac{32}{12}=\frac{65-32}{12}=\frac{33}{12} \end{gathered}$$

According to the question, we need to divide the first figure by the second:

$\frac{97}{12}÷\frac{33}{12}=\frac{97}{12}\times \frac{12}{33}=\frac{97}{33}$

Question 15:

If 24 trousers of equal size can be prepared in 54 metres of cloth, what length of cloth is required for each trouser?

Answer 15:

Total cloth given = 54 metres
Total number of pairs of trousers made = 24
Length of cloth required for each pair of trousers = $\frac{54}{24}=\frac{9\times 6}{4\times 6}=\frac{9}{4} \mathrm{metres}.$