RS Aggarwal solution class 8 chapter 1 Rational Numbers Test Paper -1

Test Paper -1

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Q1 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 1:

Find the additive inverse of:
(i) $\frac{7}{-10}$
(ii) $\frac{8}{5}.$

Answer 1:

(i) $\frac{7}{-10}=\frac{7\times -1}{-10\times -1}=\frac{-7}{10}$

Additive inverse of $\frac{-7}{10} \mathrm{is} \frac{7}{10}$.

(ii) Additive inverse of $\frac{8}{5} \mathrm{is} \frac{-8}{5}$.

Q2 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 2:

The sum of two rational numbers is −4. If one of them is $\frac{-11}{5}$, find the other.

Answer 2:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{be} x. \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ x+\frac{-11}{5}=-4 \\ \Rightarrow x-\frac{11}{5}=-4 \\ \Rightarrow x=-4+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of} \frac{-11}{5}\right) \\ \Rightarrow x=-4+\frac{11}{5} \\ \Rightarrow x=\frac{-4}{1}+\frac{11}{5} \\ \Rightarrow x=\frac{\left(-4\times 5\right)+\left(11\times 1\right)}{5} \\ \Rightarrow x=\frac{-20+11}{5} \\ \Rightarrow x=\frac{-9}{5} \end{gathered}$$

Q3 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 3:

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

Answer 3:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{required} \mathrm{number} \mathrm{be} x. \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ x+\frac{(-3)}{5}=\frac{2}{3} \\ \Rightarrow x-\frac{3}{5}=\frac{2}{3} \\ \Rightarrow x=\frac{2}{3}+\frac{3}{5}=\frac{2\times 5+3\times 3}{15} \\ \Rightarrow x=\frac{10+9}{15} \\ \Rightarrow x=\frac{19}{15} \end{gathered}$$

Q4 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 4:

What number should be subtracted from $\frac{-3}{4}$ to get $\frac{-1}{2}$?

Answer 4:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{required} \mathrm{number} \mathrm{be} x. \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ \frac{-3}{4}-x=\frac{-1}{2} \\ \Rightarrow \frac{-3}{4}+\frac{1}{2}=x \\ \Rightarrow x=\frac{2-3}{4} \\ \Rightarrow x=\frac{-1}{4} \end{gathered}$$

Q5 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 5:

Find the multiplicative inverse of:
(i) $\frac{-3}{4}$
(ii) $\frac{11}{4}.$

Answer 5:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Multiplicative} \mathrm{inverse} \mathrm{of} \frac{-3}{4} \mathrm{is} \frac{4}{-3}, \mathrm{i}.\mathrm{e}., \frac{-4}{3}. \\ \left(\mathrm{ii}\right) \mathrm{Multiplicative} \mathrm{inverse} \mathrm{of} \frac{11}{4} \mathrm{is} \frac{4}{11}. \end{gathered}$$

Q6 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 6:

The product of two numbers is −8. If one of them is −12, find the other.

Answer 6:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{be} x. \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ -12 \times x=-8 \\ \Rightarrow x=(-8)÷(-12) \\ \Rightarrow x=-8\times \frac{1}{-12} \\ \Rightarrow x=\frac{8}{12} \\ \Rightarrow x=\frac{2}{3} \end{gathered}$$

Q7 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 7:

Evaluate:
(i) $\frac{-3}{5}\times \frac{10}{7}$
(ii) ${\left(\frac{-5}{8}\right)}^{-1}$
(iii) ${\left(-6\right)}^{-1}$

Answer 7:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \frac{-3}{5}\times \frac{10}{7} \\ =\frac{-3\times 10}{5\times 7} \\ =\frac{-30}{35} \\ =\frac{-6\times 5}{7\times 5} \\ =\frac{-6}{7} \\ \left(\mathrm{ii}\right) \\ (\frac{-5}{8}{)}^{-1} \\ =\frac{1}{\left({\displaystyle \frac{-5}{8}}\right)} \\ =1\times \frac{8}{-5} \\ =\frac{8}{-5} \\ =\frac{8\times -1}{-5\times -1} \\ =\frac{-8}{5} \\ (\mathrm{iii}\left) \\ \right(-6{)}^{-1} \\ =\frac{1}{-6} \\ =\frac{1\times -1}{-6\times -1} \\ =\frac{-1}{6} \end{gathered}$$

Q8 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 8:

Name the property of multiplication shown by each of the following statements:
(i) $\frac{-12}{5}\times \frac{3}{4}=\frac{3}{4}\times \frac{-12}{5}$
(ii) $\frac{-8}{15}\times 1=\frac{-8}{15}$
(iii) $\left(\frac{-2}{3}\times \frac{7}{8}\right)\times \frac{-5}{7}=\frac{-2}{3}\times \left(\frac{7}{8}\times \frac{-5}{7}\right)$
(iv) $\frac{-2}{3}\times 0=0$
(v) $\frac{2}{5}\times \left(\frac{-4}{5}+\frac{-3}{10}\right)=\left(\frac{2}{5}\times \frac{-4}{5}\right)+\left(\frac{2}{5}\times \frac{-3}{10}\right)$

Answer 8:

(i) Commutative law of multiplication

(ii) Existence of  multiplicative identity

(iii) Associative law of multiplication

(iv) Multiplicative property of 0

(v) Distributive law of multiplication over addition

Q9 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 9:

Find two rational numbers lying between $\frac{-1}{3} \mathrm{and} \frac{1}{2}.$

Answer 9:

$$\begin{gathered} \mathrm{Required} \mathrm{number}=\frac{1}{2}\times \left(\frac{-1}{3}+\frac{1}{2}\right) \\ =\frac{1}{2}\times \left(\frac{-2+3}{6}\right) \\ =\frac{1}{2}\times \frac{1}{6} \\ =\frac{1}{12} \\ \frac{-1}{3}<\frac{1}{12}<\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{Rational} \mathrm{number} \mathrm{between} \frac{-1}{3} \mathrm{and} \frac{1}{12}: \\ \frac{1}{2}\times \left(\frac{-1}{3}+\frac{1}{12}\right) \\ =\frac{1}{2}\times \left(\frac{1-4}{12}\right) \\ =\frac{1}{2}\times \frac{-3}{12} \\ =\frac{-3}{24}=\frac{-3÷3}{24÷3} \\ =\frac{-1}{8} \end{gathered}$$

$\mathrm{Thus}, \frac{1}{12} \mathrm{and} \frac{-1}{8} \mathrm{are }$the two rational numbers between $\frac{-1}{3}$ and $\frac{1}{2}$

Q10 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 10:

Mark (✓) against the correct answer
What should be added to $\frac{-3}{5} \mathrm{to} \mathrm{get} \frac{-1}{3}?$
(a) $\frac{4}{5}$
(b) $\frac{8}{15}$
(c) $\frac{4}{15}$
(d) $\frac{2}{5}$

Answer 10:

 (c) $\frac{4}{15}$

Let the number be $x$. 
Now,

$$\begin{gathered} \frac{-3}{5}+x=\frac{-1}{3} \\ \Rightarrow x=\frac{-1}{3}+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of} \frac{-3}{5}\right) \\ \Rightarrow x=\frac{-1}{3}+\frac{3}{5} \\ \Rightarrow x=\left(\frac{-1\times 5}{3\times 5}\right)+\left(\frac{3\times 3}{5\times 3}\right) \\ \Rightarrow x=\frac{-5}{15}+\frac{9}{15} \\ \Rightarrow x=\frac{-5+9}{15} \\ \Rightarrow x=\frac{4}{15} \end{gathered}$$

Q11 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 11:

Mark (✓) against the correct answer
What should be subtracted from $\frac{-2}{3}$ to get $\frac{3}{4}?$
(a) $\frac{-11}{12}$
(b) $\frac{-13}{12}$
(c) $\frac{-5}{4}$
(d) $\frac{-17}{12}$

Answer 11:

 (d) $\frac{-17}{12}$

Let the number be $x$.
Now,

$\frac{-2}{3}-a=\frac{3}{4}$

$\Rightarrow-1 \times\left(\frac{2}{3}+x\right)=\frac{3}{4}$

$\Rightarrow \frac{2}{3}+x=\frac{-3}{4}$

$\Rightarrow x=\frac{-3}{4}+\left(\right.$ Additive inverse of $\left.\frac{2}{3}\right)$

$\Rightarrow x=\frac{-3}{4}+\left(\frac{-2}{3}\right)$

$\Rightarrow x=\frac{-3}{4}+\frac{-2}{3}$

$\Rightarrow x=\frac{-3 \times 3}{4 \times 3}+\frac{-2 \times 4}{3 \times 4}$

$\Rightarrow x= \frac{-9}{12}+\frac{-8}{12} \Rightarrow x=\frac{-17}{12}$

Q12 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 12:

Mark (✓) against the correct answer
${\left(\frac{-5}{4}\right)}^{-1}=?$
(a) $\frac{4}{5}$
(b) $\frac{-4}{5}$
(c) $\frac{5}{4}$
(d) $\frac{3}{5}$

Answer 12:

 (b) $\frac{-4}{5}$

We have:

$$\begin{gathered} {\left(\frac{-5}{4}\right)}^{-1} \\ =\frac{1}{\left({\displaystyle \frac{-5}{4}}\right)} \\ =1\times \frac{4}{-5} \\ =\frac{4}{-5} \\ =\frac{4\times -1}{-5\times -1} \\ =\frac{-4}{5} \end{gathered}$$

Q13 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 13:

Mark (✓) against the correct answer
The product of two numbers is $\frac{-1}{4}$. If one of them is $\frac{-3}{10},$ then the other is
(a) $\frac{5}{6}$
(b) $\frac{-5}{6}$
(c) $\frac{4}{3}$
(d) $\frac{-8}{5}$

Answer 13:

 (a) $\frac{5}{6}$

Let the required number be $x$.
Now,

$$\begin{gathered} \frac{-3}{10}\times x=\frac{-1}{4} \\ \Rightarrow x=\frac{-1}{4}÷\left(\frac{-3}{10}\right) \\ \Rightarrow x=\frac{-1}{4}\times \frac{10}{-3} \\ \Rightarrow x=\frac{10}{12}=\frac{5}{6} \end{gathered}$$

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Q14 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 14:

Mark (✓) against the correct answer
$\left(\frac{-5}{6}÷\frac{-2}{3}\right)=?$
(a) $\frac{-5}{4}$
(b) $\frac{5}{4}$
(c) $\frac{-4}{5}$
(d) $\frac{4}{5}$

Answer 14:

(b)​ $\frac{5}{4}$

We have:
$$\begin{gathered} \left(\frac{-5}{6}÷\frac{-2}{3}\right) \\ =\frac{-5}{6}\times \frac{3}{-2} \\ =\frac{15}{12} \\ =\frac{5}{4} \end{gathered}$$

Q15 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 15:

Mark (✓) against the correct answer
$\frac{4}{3}÷?=\frac{-5}{2}$
(a) $\frac{-8}{5}$
(b) $\frac{8}{5}$
(c) $\frac{-8}{15}$
(d) $\frac{8}{15}$

Answer 15:

(c) $\frac{-8}{15}$

$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \frac{4}{3}÷x=\frac{-5}{2} \\ \Rightarrow \frac{4}{3}\times \frac{1}{x}=\frac{-5}{2} \\ \Rightarrow \frac{1}{x}=\frac{\left({\displaystyle \frac{-5}{2}}\right)}{\left({\displaystyle \frac{4}{3}}\right)} \\ \Rightarrow \frac{1}{x}=\left(\frac{-5}{2}\right)\times \left(\frac{3}{4}\right) \\ \Rightarrow \frac{1}{x}=\frac{-15}{8} \\ \Rightarrow x=\frac{8}{-15}=\frac{8\times -1}{-15\times -1}=\frac{-8}{15} \end{gathered}$$

Q16 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 16:

Mark (✓) against the correct answer
Reciprocal of $\frac{-7}{9}$ is
(a) $\frac{9}{7}$
(b) $\frac{-9}{7}$
(c) $\frac{7}{9}$
(d) none of these

Answer 16:

(b) $\frac{-9}{7}$
$$\begin{gathered} \mathrm{Reciprocal} \mathrm{of} \frac{-7}{9}={\left(\frac{-7}{9}\right)}^{-1} \\ \mathrm{Now}, \mathrm{we} \mathrm{have}: \\ \frac{1}{\left({\displaystyle \frac{-7}{9}}\right)} \\ =\frac{9}{-7} \\ =\frac{9\times -1}{-7\times -1} \\ =\frac{-9}{7} \end{gathered}$$

Q17 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 17:

A rational number between $\frac{-2}{3}$ and $\frac{1}{2}$ is
(a) $\frac{-1}{6}$
(b) $\frac{-1}{12}$
(c) $\frac{-5}{6}$
(d) $\frac{5}{6}$

Answer 17:

(b) $\frac{-1}{12}$

$$\begin{gathered} \mathrm{Number} \mathrm{between} \frac{-2}{3} \mathrm{and} \frac{1}{2}=\frac{1}{2}\times \left(\frac{-2}{3}+\frac{1}{2}\right) \\ =\frac{1}{2}\times \left(\frac{-2\times 2}{3\times 2}+\frac{1\times 3}{2\times 3}\right) \\ =\frac{1}{2}\times \left(\frac{-4}{6}+\frac{3}{6}\right) \\ =\frac{1}{2}\times \left(\frac{-4+3}{6}\right) \\ =\frac{-1}{12} \end{gathered}$$

Q18 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 18:

Fill in the blanks.
(i) $\frac{25}{8}÷\left(.......\right)=-10.$
(ii) $\frac{-8}{9}\times \left(.......\right)=\frac{-2}{3}.$
(iii) $\left(-1\right)+\left(.......\right)=\frac{-2}{9}.$
(iv) $\frac{2}{3}-\left(.......\right)=\frac{1}{15}.$

Answer 18:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} x. \\ \mathrm{Now}, \mathrm{we} \mathrm{have}: \\ \frac{25}{8}÷x=-10 \\ \Rightarrow \frac{25}{8}\times \frac{1}{x}=-10 \\ \Rightarrow \frac{1}{x}=-10÷\frac{25}{8} \\ \Rightarrow \frac{1}{x}=-10\times \frac{8}{25} \\ \Rightarrow \frac{1}{x}=\frac{-80}{25} \\ \Rightarrow x=\frac{25}{-80} \\ \Rightarrow x=\frac{25\times -1}{-80\times -1} \\ \Rightarrow x=\frac{-25}{80} \\ \Rightarrow x=\frac{-25÷5}{80÷5} \\ \Rightarrow x=\frac{-5}{16} \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} x. \\ \mathrm{Now}, \mathrm{we} \mathrm{have}: \\ \frac{-8}{9}\times x=\frac{-2}{3} \\ \Rightarrow x=\frac{-2}{3}÷\frac{-8}{9} \\ \Rightarrow x=\frac{-2}{3}\times \frac{9}{-8} \\ \Rightarrow x=\frac{18}{24}=\frac{18÷6}{24÷6} \\ \Rightarrow x=\frac{3}{4} \end{gathered}$$

(iii)
$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{blank} \mathrm{space} \mathrm{be} x. \\ \mathrm{Now}, \mathrm{we} \mathrm{have}: \\ (-1)+x=\frac{-2}{9} \\ \Rightarrow x=\frac{-2}{9}+1 \\ \Rightarrow x=\frac{-2+9}{9} \\ \Rightarrow x=\frac{7}{9} \end{gathered}$$

(iv)
$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{blank} \mathrm{space} \mathrm{be} x. \\ \mathrm{Now}, \mathrm{we} \mathrm{have}: \\ \frac{2}{3}-x=\frac{1}{15} \\ \Rightarrow -x=\frac{1}{15}-\frac{2}{3} \\ \Rightarrow -x=\frac{1-10}{15} \\ \Rightarrow -x=\frac{-9}{15} \\ \Rightarrow x=\frac{9}{15}=\frac{3}{5} \end{gathered}$$

Q19 Test paper 1 Class 8 RS AGGARWAL chapter 1 Rational Numbers Solution

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Question 19:

Write 'T' for true and 'F' for false for each of the following:
(i) Rational numbers are always closed under subtraction.
(ii) Rational numbers are always closed under division.
(iii) 1 ÷ 0 = 0.
(iv) Subtraction is commutative on rational numbers.
(v) $-\left(\frac{-7}{8}\right)=\frac{7}{8}.$

Answer 19:

(i) T

​If $\frac{a}{b}$ and $\frac{c}{d}$ are rational numbers, then $\frac{a}{b}-\frac{c}{d}=\frac{a d-b c}{b d}$ is also a rational number because ad, bc and bd are all rational numbers.

(ii) F

​Rational numbers are not always closed under division. They are closed under division only if the denominator is non-zero.

(iii) F

$1÷0$cannot be defined.
​
(iv) F

​Let $\frac{a}{b} \mathrm{and} \frac{c}{d}$ represent rational numbers. 

Now, we have:

$\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}$
$\frac{c}{d}-\frac{a}{b}=\frac{bc-ad}{bd}$

∴ $\frac{a}{b}-\frac{c}{d}\ne \frac{c}{d}-\frac{a}{b}$

(v) T
`
$-\left(\frac{-7}{8}\right)=-1\times \left(\frac{-7}{8}\right)=\frac{-1\times -7}{8}=\frac{7}{8}$