RS Aggarwal solution class 8 chapter 1 Rational Numbers Exercise 1E

Exercise 1E

Page-19

Q1 | Ex-1E | Rational Numbers | Class 8 | RS AGGARWAL | Chapter 1 | myhelper

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Question 1:

Simplify:
(i) $\frac{4}{9}÷\frac{-5}{12}$
(ii) $-8÷\frac{-7}{16}$
(iii) $\frac{-12}{7}÷\left(-18\right)$
(iv) $\frac{-1}{10}÷\frac{-8}{5}$
(v) $\frac{-16}{35}÷\frac{-15}{14}$
(vi) $\frac{-65}{14}÷\frac{13}{7}$

Answer 1:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \frac{4}{9}÷\frac{-5}{12} \\ =\frac{4}{9}\times \frac{12}{-5} \\ =\frac{4\times 12}{9\times -5} \\ =\frac{48}{-45} \\ =\frac{-48}{45} \\ =\frac{-16}{15} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \\ -8÷\frac{-7}{16} \\ =-8\times \frac{16}{-7} \\ =\frac{8\times 16}{7} \\ =\frac{128}{7} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) \\ \frac{-12}{7}÷(-18) \\ =\frac{-12}{7}\times \frac{1}{-18} \\ =\frac{12}{126} \\ =\frac{12÷3}{126÷3} \\ =\frac{4}{42} \\ =\frac{4÷2}{42÷2} \\ =\frac{2}{21} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right) \\ \frac{-1}{10}÷\frac{-8}{5} \\ =\frac{-1}{10}\times \frac{5}{-8} \\ =\frac{5}{80} \\ =\frac{5÷5}{80÷5} \\ =\frac{1}{16} \\ \left(\mathrm{v}\right) \\ \frac{-16}{35}÷\frac{-15}{14} \\ =\frac{-16}{35}\times \frac{14}{-15} \\ =\frac{32}{75} \\ \left(\mathrm{vi}\right) \\ \frac{-65}{14}÷\frac{13}{7} \\ =\frac{-65}{14}\times \frac{7}{13} \\ =\frac{-5}{2} \end{gathered}$$

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Question 2:

Verify whether the given statement is true or false:
(i) $\frac{13}{5}÷\frac{26}{10}=\frac{26}{10}÷\frac{13}{5}$
(ii) $-9÷\frac{3}{4}=\frac{3}{4}÷\left(-9\right)$
(iii) $\frac{-8}{9}÷\frac{-4}{3}=\frac{-4}{3}÷\frac{-8}{9}$
(iv) $\frac{-7}{24}÷\frac{3}{-16}=\frac{3}{-16}÷\frac{-7}{24}$

Answer 2:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \frac{13}{5}÷\frac{26}{10}=\frac{26}{10}÷\frac{13}{5} \end{gathered}$$

$$\begin{gathered} \text{LHS} \\ \frac{13}{5}÷\frac{26}{10} \\ =\frac{13}{5}\times \frac{10}{26} \\ =\frac{130}{130} \\ =1 \end{gathered}$$

$$\begin{gathered} \text{RHS} \\ \frac{26}{10}÷\frac{13}{5} \\ =\frac{26}{10}\times \frac{5}{13} \\ =\frac{130}{130} \\ =1 \end{gathered}$$

$\text{TRUE}$

(ii) $-9÷\frac{3}{4}=\frac{3}{4}÷\left(-9\right)$

$\text{LHS}-9÷\frac{3}{4}=-9\times \frac{4}{3}=\frac{-36}{3}=-12$

$\text{RHS}\frac{3}{4}\text{÷(-9)}\text{=}\frac{3}{4}\text{×}\frac{1}{-9}\text{=}\frac{3}{-36}\text{=}\frac{-1}{12}$

$\mathrm{FALSE}$

(iii) $\frac{-8}{9}÷\frac{-4}{3}=\frac{-4}{3}÷\frac{-8}{9}$

$\mathrm{LHS}\frac{-8}{9}÷\frac{-4}{3}=\frac{-8}{9}\times \frac{3}{-4}=\frac{24}{36}=\frac{2}{3}$

$\mathrm{RHS}\frac{-4}{3}÷\frac{-8}{9}=\frac{-4}{3}\times \frac{9}{-8}=\frac{36}{24}=\frac{3}{2}$

$\mathrm{FALSE}$

(iv) $\frac{-7}{24}÷\frac{3}{-16}=\frac{3}{-16}÷\frac{-7}{24}$

$\mathrm{LHS}\frac{-7}{24}\times \frac{-16}{3}=\frac{112}{72}$

$\mathrm{RHS}\frac{3}{-16}÷\frac{-7}{24}=\frac{3}{-16}\times \frac{24}{-7}=\frac{72}{112}$

$\mathrm{FALSE}$

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Question 3:

Verify whether the given statement is true or false:
(i) $\left(\frac{5}{9}÷\frac{1}{3}\right)÷\frac{5}{2}=\frac{5}{9}÷\left(\frac{1}{3}÷\frac{5}{2}\right)$
(ii) $\left\{\left(-16\right)÷\frac{6}{5}\right\}÷\frac{-9}{10}=\left(-16\right)÷\left\{\frac{6}{5}÷\frac{-9}{10}\right\}$
(iii) $\left(\frac{-3}{5}÷\frac{-12}{35}\right)÷\frac{1}{14}=\frac{-3}{5}÷\left(\frac{-12}{35}÷\frac{1}{4}\right)$

Answer 3:

$$\begin{gathered} \left(\mathrm{i}\right) \\ (\frac{5}{9}÷\frac{1}{3})÷\frac{5}{2}=\frac{5}{9}÷(\frac{1}{3}÷\frac{5}{2}) \end{gathered}$$

$$\begin{gathered} \mathrm{LHS} \\ (\frac{5}{9}÷\frac{1}{3})÷\frac{5}{2} \\ =(\frac{5}{9}\times \frac{3}{1})\times \frac{2}{5} \\ =\frac{5\times 3\times 2}{9\times 1\times 5} \\ =\frac{30}{45} \\ =\frac{2}{3} \end{gathered}$$

$$\begin{gathered} \mathrm{RHS} \\ \frac{5}{9}÷\left(\frac{1}{3}÷\frac{5}{2}\right) \\ =\frac{5}{9}÷\left(\frac{1}{3}\times \frac{2}{5}\right) \\ =\frac{5}{9}÷\left(\frac{2}{15}\right) \\ =\frac{5}{9}\times \left(\frac{15}{2}\right)=\frac{75}{18} \\ =\frac{25}{6} \end{gathered}$$

$$\begin{gathered} \mathrm{LHS}\ne \mathrm{RHS} \\ \mathrm{FALSE} \end{gathered}$$

​(ii)
$\left[\right(-16)÷\frac{6}{5}]÷\frac{-9}{10}=(-16)÷[\frac{6}{5}÷\frac{-9}{10}]$

$$\begin{gathered} \mathrm{LHS} \\ =\left[\right(-16)÷\frac{6}{5}]÷\frac{-9}{10} \\ =\left[\right(-16)\times \frac{5}{6}]\times \frac{10}{-9} \\ =\frac{(-16)\times 5\times 10}{6\times (-9)} \\ =\frac{800}{54} \\ =\frac{400}{27} \end{gathered}$$

$$\begin{gathered} \mathrm{RHS} \\ (-16)÷(\frac{6}{5}÷\frac{-9}{10}) \\ =(-16)÷(\frac{6}{5}\times \frac{10}{-9}) \\ =-16÷\left\{\frac{-60}{45}\right\} \\ =-16\times \left(\frac{-45}{60}\right) \\ =-16\times \left(\frac{-3}{4}\right) \\ =\frac{48}{4} \\ =12 \end{gathered}$$

$$\begin{gathered} \mathrm{LHS} \ne \mathrm{RHS} \\ \mathrm{FALSE} \end{gathered}$$

(iii)
$(\frac{-3}{5}÷\frac{-12}{35})÷\frac{1}{14}=\frac{-3}{5}÷(\frac{-12}{35}÷\frac{1}{4})$

$$\begin{gathered} \mathrm{LHS} \\ =(\frac{-3}{5}\times \frac{35}{-12})\times 14 \\ =\frac{(-3)\times 35\times 14}{5\times (-12)} \\ =\frac{1470}{60} \\ =\frac{49}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{RHS} \\ =\frac{-3}{5}÷(\frac{-12}{35}÷\frac{1}{4}) \\ =\frac{-3}{5}÷(\frac{-12}{35}\times \frac{4}{1}) \\ =\frac{-3}{5}÷\left(\frac{-12\times 4}{35}\right) \\ =\frac{-3}{5}÷\left(\frac{-48}{35}\right) \\ =\frac{-3}{5}\times \frac{35}{-48} \\ =\frac{3\times 35}{5\times 48} \\ =\frac{105}{240} \\ =\frac{7}{16} \end{gathered}$$

$\mathrm{LHS}\ne \mathrm{RHS}$

$\mathrm{FALSE}$

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Question 4:

The product of two rational numbers is −9. If one of the numbers is −12, find the other.

Answer 4:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} \mathrm{x}. \\ \mathrm{Now}, \\ \mathrm{x}\times (-12)=-9 \\ \Rightarrow \mathrm{x}=-9÷(-12) \\ \Rightarrow \mathrm{x}=(-9)\times \frac{1}{-12} \\ \Rightarrow \mathrm{x}=\frac{-9}{-12} \\ \Rightarrow \mathrm{x}=\frac{3}{4} \end{gathered}$$

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Question 5:

The product of two rational numbers is $\frac{-16}{9}$. If one of the numbers is $\frac{-4}{3}$, find the other.

Answer 5:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} \mathrm{x}. \\ \mathrm{Now}, \\ \mathrm{x}\times \frac{-4}{3}=\frac{-16}{9} \\ \Rightarrow \mathrm{x}=\frac{-16}{9}÷\frac{-4}{3} \\ \Rightarrow \mathrm{x}=\frac{-16}{9}\times \frac{3}{-4} \\ \Rightarrow \mathrm{x}=\frac{-16\times 3}{9\times (-4)} \\ \Rightarrow \mathrm{x}=\frac{48}{36} \\ \Rightarrow \mathrm{x}=\frac{4}{3} \end{gathered}$$.

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Question 6:

By what rational number should we multiply $\frac{-15}{56}$ to get $\frac{-5}{7}$?

Answer 6:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} \mathrm{x}. \\ \mathrm{Now}, \\ \mathrm{x}\times \frac{-15}{56}=\frac{-5}{7} \\ \Rightarrow \mathrm{x}=\frac{-5}{7}÷\frac{-15}{56} \\ \Rightarrow \mathrm{x}=\frac{-5}{7}\times \frac{56}{-15} \\ \Rightarrow \mathrm{x}=\frac{280}{105} \\ \Rightarrow \mathrm{x}=\frac{280÷5}{105÷5} \\ \Rightarrow \mathrm{x}=\frac{56}{21} \\ \Rightarrow \mathrm{x}=\frac{56÷7}{21÷7} \\ \Rightarrow \mathrm{x}=\frac{8}{3} \end{gathered}$$

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Question 7:

By what rational number should $\frac{-8}{39}$ be multiplied to obtain $\frac{1}{26}$?

Answer 7:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} \mathrm{x}. \\ \mathrm{Now}, \\ \mathrm{x}\times \frac{-8}{39}=\frac{1}{26} \\ \Rightarrow \mathrm{x}=\frac{1}{26}÷\frac{-8}{39} \\ \Rightarrow \mathrm{x}=\frac{1}{26}\times \frac{39}{-8} \\ \Rightarrow \mathrm{x}=\frac{39}{-208} \\ \Rightarrow \mathrm{x}=\frac{39\times -1}{-208\times -1} \\ \Rightarrow \mathrm{x}=\frac{-39}{208} \\ \Rightarrow \mathrm{x}=\frac{-39÷13}{208÷13} \\ \Rightarrow \mathrm{x}=\frac{-3}{16} \end{gathered}$$

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Question 8:

By what number should $\frac{-33}{8}$ be divided to get $\frac{-11}{2}$?

Answer 8:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} \mathrm{x}. \\ \mathrm{Now}, \\ \frac{-33}{8}÷\mathrm{x}=\frac{-11}{2} \\ \Rightarrow \frac{-33}{8}\times \frac{1}{\mathrm{x}}=\frac{-11}{2} \\ \Rightarrow \frac{1}{\mathrm{x}}=\frac{-11}{2}÷\frac{-33}{8} \\ \Rightarrow \frac{1}{\mathrm{x}}=\frac{-11}{2}\times \frac{8}{-33} \\ \Rightarrow \frac{1}{\mathrm{x}}=\frac{88}{66} \\ \Rightarrow \frac{1}{\mathrm{x}}=\frac{4}{3} \\ \Rightarrow \mathrm{x}=\frac{3}{4} (\mathrm{Reciprocal} \mathrm{of} \frac{4}{3}) \end{gathered}$$

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Question 9:

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

Answer 9:

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

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Question 10:

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

Answer 10:

 $$\begin{gathered} \left(\frac{65}{12}+\frac{8}{3}\right)÷\left(\frac{65}{12}-\frac{8}{3}\right) \\ =\left(\frac{65}{12}+\frac{32}{12}\right)÷\left(\frac{65}{12}-\frac{32}{12}\right) \\ =\left(\frac{97}{12}\right)÷\left(\frac{33}{12}\right) \\ =\frac{97}{12}\times \frac{12}{33} \\ =\frac{97}{33} \end{gathered}$$

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Question 11:

Fill in the blanks:
(i) $\frac{9}{8}÷\left(...\right)=\frac{-3}{2}$
(ii) $\left(...\right)÷\left(\frac{-7}{5}\right)=\frac{10}{19}$
(iii) $\left(...\right)÷\left(-3\right)=\frac{-4}{15}$
(iv) $\left(-12\right)÷\left(...\right)=\frac{-6}{5}$

Answer 11:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Let} \frac{9}{8}÷\mathrm{x}=\frac{-3}{2} \\ \Rightarrow \frac{9}{8}\times \frac{1}{\mathrm{x}}=\frac{-3}{2} \\ \Rightarrow \frac{1}{\mathrm{x}}=\frac{-3}{2}÷\frac{9}{8} \\ \Rightarrow \frac{1}{\mathrm{x}}=\frac{-3}{2}\times \frac{8}{9} \\ \Rightarrow \frac{1}{\mathrm{x}}=\frac{-24}{18} \\ \Rightarrow \frac{1}{\mathrm{x}}=\frac{-4}{3} \\ \Rightarrow \mathrm{x}=\frac{-3}{4} [\mathrm{Reciprocal} \mathrm{of} \frac{-4}{3}] \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \\ \mathrm{Let} \mathrm{x}÷\left(\frac{-7}{5}\right)=\frac{10}{19} \\ \Rightarrow \mathrm{x}\times \left(\frac{5}{-7}\right)=\frac{10}{19} \\ \Rightarrow \mathrm{x}=\left(\frac{10}{19}\right)÷\left(\frac{5}{-7}\right) \\ \Rightarrow \mathrm{x}=\frac{10}{19}\times \frac{-7}{5} \\ \Rightarrow \mathrm{x}=\frac{-70}{95} \\ \Rightarrow \mathrm{x}=\frac{-14}{19} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) \\ \mathrm{Let} \mathrm{x}÷(-3)=\frac{-4}{15} \\ \Rightarrow \mathrm{x} \times \left({\displaystyle \frac{1}{-3}}\right)=\frac{-4}{15} \\ \Rightarrow \mathrm{x}=\frac{-4}{15}\times (-3) \\ \Rightarrow \mathrm{x}=\frac{12}{15} \\ \Rightarrow \mathrm{x}=\frac{4}{5} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right) \\ \mathrm{Let} (-12)÷\mathrm{x}=\frac{-6}{5} \\ \Rightarrow (-12)\times \frac{1}{\mathrm{x}}=\frac{-6}{5} \\ \Rightarrow \frac{1}{\mathrm{x}}=\frac{-6}{5}÷(-12) \\ \Rightarrow \frac{1}{\mathrm{x}}=\frac{-6}{5}\times \frac{1}{-12} \\ \Rightarrow \frac{1}{\mathrm{x}}=\frac{1}{10} \\ \Rightarrow \mathrm{x}=10 \end{gathered}$$

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Question 12:

(i) Are rational numbers always closed under division?
(ii) Are rational numbers always commutative under division?
(iii) Are rational numbers always associative under division?
(iv) Can we divide 1 by 0?

Answer 12:

​(i)  No, rational numbers are not closed under division in general.
 
$\frac{a}{0}=\infty$; it is not a rational number.

(ii) No

 $$\begin{gathered} \frac{a}{b}÷\frac{c}{d}=\frac{a}{b}\times \frac{d}{c}=\frac{ad}{bc} \\ \mathrm{Also}, \\ \frac{c}{d}÷\frac{a}{b}=\frac{c}{d}\times \frac{b}{a}=\frac{cb}{da}\mathrm{Thus}, \frac{a}{b}÷\frac{c}{d}\ne \frac{c}{d}÷\frac{a}{b} \end{gathered}$$

Therefore, division is not commutative.

(iii) No, rational numbers are not associative under division. 

$\frac{a}{b}÷\left(\frac{c}{d}÷\frac{e}{f}\right)\ne \left(\frac{a}{b}÷\frac{c}{d}\right)÷\frac{e}{f}$

(iv) No, we cannot divide 1 by 0. The answer will be$\infty$, which is not defined.