RD Sharma solution class 8 chapter 1 Rational Numbers Exercise 1.7

Exercise 1.7

Page-1.35

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) $\frac{2}{3} \mathrm{by} \frac{-7}{12}$

(ix) $-4 \mathrm{by} \frac{-3}{5}$

(x) $\frac{-3}{13} \mathrm{by} \frac{-4}{65}$

Answer 1:

$$\begin{gathered} \left(\mathrm{i}\right) 1÷\frac{1}{2}=1\times \frac{2}{1}=2 \\ \left(\mathrm{ii}\right) 5÷\frac{-5}{7}=5\times \frac{7}{-5}=-7 \\ \left(\mathrm{iii}\right) \frac{-3}{4}÷\frac{9}{-16}=\frac{-3}{4}\times \frac{-16}{9}=\frac{4}{3} \\ \left(\mathrm{iv}\right) \frac{-7}{8}÷\frac{-21}{16}=\frac{-7}{8}\times \frac{16}{-21}=\frac{2}{3} \\ \left(\mathrm{v}\right) \frac{7}{-4}÷\frac{63}{64}=\frac{7}{-4}\times \frac{64}{63}=\frac{-16}{9} \\ \left(\mathrm{vi}\right) 0÷\frac{-7}{5}=0\times \frac{5}{-7}=0 \\ \left(\mathrm{vii}\right) \frac{-3}{4}÷-6=\frac{-3}{4}\times \frac{1}{-6}=\frac{1}{8} \\ \left(\mathrm{viii}\right) \frac{2}{3}÷\frac{-7}{12}=\frac{2}{3}\times \frac{12}{-7}=\frac{-8}{7} \\ \left(\mathrm{ix}\right) -4÷\frac{-3}{5}=-4\times \frac{5}{-3}=\frac{20}{3} \\ \left(\mathrm{x}\right) \frac{-3}{13}÷\frac{-4}{65}=\frac{-3}{13}\times \frac{65}{-4}=\frac{15}{4} \end{gathered}$$

Page-1.36

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) $\frac{-40}{99}÷(-20)$

(v) $\frac{-22}{27}÷\frac{-110}{18}$

(vi) $\frac{-36}{125}÷\frac{-3}{75}$

Answer 2:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{2}{5}÷\frac{26}{15}=\frac{2}{5}\times \frac{15}{26}=\frac{3}{13} \\ \left(\mathrm{ii}\right) \frac{10}{3}÷\frac{-35}{12}=\frac{10}{3}\times \frac{12}{-35}=\frac{-8}{7} \\ \left(\mathrm{iii}\right) -6÷\frac{-8}{17}=-6\times \frac{17}{-8}=\frac{51}{4} \\ \left(\mathrm{iv}\right) \frac{-40}{99}÷(-20)=\frac{-40}{99}\times \frac{1}{-20}=\frac{2}{99} \\ \left(\mathrm{v}\right) \frac{-22}{27}÷\frac{-110}{18}=\frac{-22}{27}\times \frac{18}{-110}=\frac{2}{15} \\ \left(\mathrm{vi}\right) \frac{-36}{125}÷\frac{-3}{75}=\frac{-36}{125}\times \frac{75}{-3}=\frac{36}{5} \end{gathered}$$

Question 3:

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

Answer 3:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{other} \mathrm{number} be x. \\ \therefore x\times (-10)=15 \\ \mathrm{or} x=\frac{15}{-10}=\frac{3}{-2} \\ \mathrm{So}, \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{is} \frac{-3}{2}. \end{gathered}$$

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:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{other} \mathrm{number} be \mathrm{x}. \\ \therefore \mathrm{x}\times \frac{-4}{15}=\frac{-8}{9} \\ \mathrm{or} \mathrm{x}=\frac{-8}{9}÷\frac{-4}{15} \\ \mathrm{or} \mathrm{x}=\frac{-8}{9}\times \frac{15}{-4} \\ \mathrm{or} \mathrm{x}=\frac{10}{3} \\ Thus, \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{is} \frac{10}{3}. \end{gathered}$$

Question 5:

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

Answer 5:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} be \mathrm{x}. \\ \therefore \mathrm{x}\times \frac{-1}{6}=\frac{-23}{9} \\ \mathrm{x}=\frac{-23}{9}÷\frac{-1}{6} \\ \mathrm{x}=\frac{-23}{9}\times \frac{6}{-1}=\frac{46}{3} \\ \mathrm{Therefore}, \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{is} \frac{46}{3}. \end{gathered}$$

Question 6:

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

Answer 6:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{other} \mathrm{number} be \mathrm{x}. \\ \therefore \mathrm{x}\times \frac{-15}{28}=\frac{-5}{7} \\ \mathrm{or} \mathrm{x}=\frac{-5}{7}÷\frac{-15}{28} \\ \mathrm{or} \mathrm{x}=\frac{-5}{7}\times \frac{28}{-15} \\ \mathrm{or} \mathrm{x}=\frac{4}{3} \\ Thus, \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{is} \frac{4}{3}. \end{gathered}$$

Question 7:

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

Answer 7:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} be \mathrm{x}. \\ \therefore \mathrm{x}\times \frac{-8}{13}=24 \\ \mathrm{or} \mathrm{x}=24÷\frac{-8}{13} \\ \mathrm{or} \mathrm{x}=24\times \frac{13}{-8} \\ \mathrm{or} x=-39 \\ Thus, \mathrm{the} \mathrm{number} \mathrm{is} -39. \end{gathered}$$

Question 8:

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

Answer 8:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{that} \mathrm{should} \mathrm{be} \mathrm{multiplied} \mathrm{with} \frac{-3}{4} \mathrm{to} \mathrm{produce} \frac{2}{3} be \mathrm{x}. \\ \therefore \mathrm{x}\times \frac{-3}{4}=\frac{2}{3} \\ \mathrm{or} \mathrm{x}=\frac{2}{3}÷\frac{-3}{4} \\ \mathrm{or} \mathrm{x}=\frac{2}{3}\times \frac{4}{-3} \\ \mathrm{or} x=\frac{-8}{9} \\ Thus, \mathrm{the} \mathrm{number} \mathrm{is} \frac{-8}{9}. \end{gathered}$$

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) $x=\frac{5}{4}, y=\frac{-1}{3}$
(iv) $x=\frac{2}{7}, y=\frac{4}{3}$
(v) $x=\frac{1}{4}, y=\frac{3}{2}$

Answer 9:

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:

$$\begin{gathered} \mathrm{The} \mathrm{cost} \mathrm{of} 7\frac{2}{3} \mathrm{metres} \mathrm{of} \mathrm{rope} \mathrm{is} \mathrm{Rs} 12\frac{3}{4}. \\ \therefore C\mathrm{ost} \mathrm{per} \mathrm{metre} =12\frac{3}{4}÷7\frac{2}{3} \\ =\frac{51}{4}÷\frac{23}{3} \\ =\frac{51}{4}\times \frac{3}{23} \\ =\frac{153}{92} \\ =\mathrm{Rs} 1\frac{61}{92} \end{gathered}$$

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:

$$\begin{gathered} \mathrm{The} \mathrm{cost} \mathrm{of} 2\frac{1}{3} \mathrm{metres} \mathrm{of} \mathrm{cloth} \mathrm{is} \mathrm{Rs} 75\frac{1}{4}. \\ \therefore C\mathrm{ost} \mathrm{per} \mathrm{metre}=75\frac{1}{4}÷2\frac{1}{3} \\ =\frac{301}{4}÷\frac{7}{3} \\ =\frac{301}{4}\times \frac{3}{7} \\ =\frac{129}{4} \\ =\mathrm{Rs} 32\frac{1}{4} \\ Thus, \mathrm{Rs} 32\frac{1}{4} \mathrm{or} \mathrm{Rs} 32.25 \mathrm{is} \mathrm{the} \mathrm{cost} \mathrm{of} \mathrm{cloth} \mathrm{per} \mathrm{metre}. \end{gathered}$$

Question 12:

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

Answer 12:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} be \mathrm{x}. \\ \therefore \frac{-33}{16}÷\mathrm{x}=\frac{-11}{4} \\ \mathrm{or} \frac{-33}{16}\times \frac{1}{\mathrm{x}}=\frac{-11}{4} \\ \mathrm{or} \frac{1}{\mathrm{x}}=\frac{-11}{4}\times \frac{16}{-33} \\ \mathrm{or} \frac{1}{\mathrm{x}}=\frac{4}{3} \\ \mathrm{or} \mathrm{x}=\frac{3}{4} \\ Thus, \mathrm{the} \mathrm{number} \mathrm{is} \frac{3}{4}. \end{gathered}$$

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:

$$\begin{gathered} (\frac{-13}{5}+\frac{12}{7})÷(\frac{-31}{7}\times \frac{-1}{2}) \\ =\frac{-13\times 7+12\times 5}{35}÷\frac{31}{14} \\ =\frac{-91+60}{35}÷\frac{31}{14} \\ =\frac{-31}{35}\times \frac{14}{31} \\ =\frac{-2}{5} \end{gathered}$$

Question 14:

Divide the sum of $\frac{65}{12} \mathrm{and} \frac{12}{7}$ by their difference.

Answer 14:

$$\begin{gathered} (\frac{65}{12}+\frac{12}{7})÷(\frac{65}{12}-\frac{12}{7}) \\ =\frac{65\times 7+12\times 12}{84}÷\frac{65\times 7-12\times 12}{84} \\ =\frac{455+144}{84}÷\frac{455-144}{84} \\ =\frac{599}{84}÷\frac{311}{84} \\ =\frac{599}{84}\times \frac{84}{311} \\ =\frac{599}{311} \end{gathered}$$

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:

$$\begin{gathered} \mathrm{Cloth} \mathrm{needed} \mathrm{to} \mathrm{prepare} 24 \mathrm{trousers}=54 \mathrm{m} \\ \therefore L\mathrm{ength} \mathrm{of} the \mathrm{cloth} \mathrm{required} \mathrm{for} \mathrm{each} \mathrm{trousers}= 54÷24=\frac{54}{24}=\frac{9}{4} \mathrm{m} = 2\frac{1}{4} \mathrm{metres} \end{gathered}$$