RS Aggarwal solution class 8 chapter 2 Exponents Test Paper 2

Test Paper 2

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Q1 Test Paper 2 Class 8 RS AGGARWAL chapter 2 Exponents Solution

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Question 1:

Evaluate:
(i) 3⁻⁴
(ii) (−4)³
(iii) ${\left(\frac{3}{4}\right)}^{-2}$
(iv) ${\left(\frac{-2}{3}\right)}^{-5}$
(v) ${\left(\frac{5}{7}\right)}^0$

Answer 1:

(i) $3^{-4}=\frac{1}{3^4}=\frac{1}{81}$

(ii) ${\left(-4\right)}^3={\left(-1\right)}^3\times {\left(4\right)}^3=-1\times 64=-64$

(iii) ${\left(\frac{3}{4}\right)}^{-2}={\left(\frac{4}{3}\right)}^2=\frac{4^2}{3^2}=\frac{16}{9}$

(iv) ${\left(\frac{-2}{3}\right)}^{-5}={\left(\frac{3}{-2}\right)}^5=\frac{3^5}{-2^5}=\frac{243}{-32}=\frac{243\times -1}{-32\times -1}=\frac{-243}{32}$

(v) Using the property ${\left(\frac{a}{b}\right)}^0=1$, we have:
${\left({\displaystyle \frac{5}{7}}\right)}^0 = 1$

Q2 Test Paper 2 Class 8 RS AGGARWAL chapter 2 Exponents Solution

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Question 2:

Evaluate: ${\left\{{\left(\frac{-2}{3}\right)}^3\right\}}^{-2}.$

Answer 2:

${\left\{{\left(\frac{-2}{3}\right)}^3\right\}}^{-2}={\left(\frac{-2}{3}\right)}^{-6}={\left(\frac{3}{-2}\right)}^6=\frac{3^6}{-2^6}=\frac{729}{64}$

Q3 Test Paper 2 Class 8 RS AGGARWAL chapter 2 Exponents Solution

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Question 3:

Simplify: $(3^{-1}+6^{-1})÷{\left(\frac{3}{4}\right)}^{-1}.$

Answer 3:

$$\begin{gathered} \left(3^{-1}+6^{-1}\right)÷{\left(\frac{3}{4}\right)}^{-1} \\ =\left(\frac{1}{3}+\frac{1}{6}\right)÷{\left(\frac{4}{3}\right)}^1 \\ =\left(\left[{\displaystyle \frac{1\times 2}{3\times 2}}\right]+\left[\frac{1\times 1}{6\times 1}\right]\right)÷\left(\frac{4}{3}\right) \\ =\left(\frac{2+1}{6}\right)÷\left(\frac{4}{3}\right) \\ =\left(\frac{3}{6}\right)÷\left(\frac{4}{3}\right) \\ =\left(\frac{1}{2}\right)÷\left(\frac{4}{3}\right) \\ =\left(\frac{1}{2}\right)\times \left(\frac{3}{4}\right) \\ =\frac{3}{8} \end{gathered}$$

$\therefore \left(3^{-1}+6^{-1}\right)÷{\left(\frac{3}{4}\right)}^{-1}=\frac{3}{8}$

Q4 Test Paper 2 Class 8 RS AGGARWAL chapter 2 Exponents Solution

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Question 4:

By what number should ${\left(\frac{-2}{3}\right)}^{-3}$ be divided so that the quotient is ${\left(\frac{4}{9}\right)}^{-2}$?

Answer 4:

Let the number be x.

∴$\left(\frac{-2}{3}\right)^{-3} \div x=\left(\frac{4}{9}\right)^{-2}$

$\Rightarrow\left(\frac{3}{-2}\right)^{3}+x=\left(\frac{9}{4}\right)^{2}$

$\Rightarrow \frac{\left(\frac{3}{-2}\right)^{3}}{x}=\left(\frac{9}{4}\right)^{2}$

$\Rightarrow \frac{3^{3}}{-2^{3}}=\frac{9^{2}}{4^{2}}$

$\Rightarrow x=\frac{\left(\frac{3^{3}}{-2^{3}}\right)}{\left(\frac{9^{2}}{4^{2}}\right)}=\frac{\left(\frac{3^{3}}{-2^{3}}\right)}{\left(\frac{\left(3^{2}\right)^{2}}{\left(2^{2}\right)^{2}}\right)}$

$=\left(\frac{3^{3}}{-2^{3}}\right) \times\left(\frac{\left(2^{2}\right)^{2}}{\left(3^{2}\right)^{2}}\right)$

$=\left(\frac{3^{3}}{-2^{3}}\right) \times\left(\frac{2^{4}}{3^{4}}\right)$

$=\left(\frac{3^{3}}{-2^{3}}\right) \times\left(\frac{2^{3}}{3^{3}}\right) \times\left(\frac{2^{1}}{3^{1}}\right)$

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

Q5 Test Paper 2 Class 8 RS AGGARWAL chapter 2 Exponents Solution

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Question 5:

By what number should (−3)⁻¹ be multiplied so that the product becomes 6⁻¹?

Answer 5:

Let the number be $x$.

$\therefore$ ${\left(-3\right)}^{-1}\times x={\left(6\right)}^{-1}$
$$\begin{gathered} \Rightarrow \frac{1}{-3} \times x=\frac{1}{6} \\ \Rightarrow \frac{1 \times -1}{-3 \times -1} \times x=\frac{1}{6} \\ \therefore \frac{-x}{3} = \frac{1}{6} \\ \mathrm{On} \mathrm{cross} \mathrm{multiplying}: \\ -x \times 6 = 1 \times 3 \\ \Rightarrow -6x = 3 \\ \Rightarrow 6x = -3 \\ \therefore x = \frac{-3}{6} = \frac{-1}{2} \end{gathered}$$

Q6 Test Paper 2 Class 8 RS AGGARWAL chapter 2 Exponents Solution

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Question 6:

Express each of the following in standard form:
(i) 345
(ii) 180000
(iii) 0.000003
(iv) 0.000027

Answer 6:

(i) 345=3.45×100$=3.45 \times 10^{2}$

(ii) 180000=18×10000$=18 \times 10^{4}=1.8 \times 10 \times 10^{4}$
$=1.8 \times 10^{(1+4)}=1.8 \times 10^{5}$

(iii) 0.000003$=\frac{3}{1000000}=3 \times 10^{-6}$

(iv) 0.000027$=\frac{27}{100000}=\frac{27}{10^{6}}$

$=\frac{2.7 \times 10}{10^{6}}$

$=2.7 \times 10^{(1-6)}$

$=2.7 \times 10^{-5}$

Q7 Test Paper 2 Class 8 RS AGGARWAL chapter 2 Exponents Solution

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Question 7:

Mark (✓) against the correct answer
The value of (−3)⁻³ is
(a) −27
(b) 9
(c) $\frac{-1}{27}$
(d) $\frac{1}{27}$

Answer 7:

(c) $\frac{-1}{27}$

$$\begin{gathered} {\left(-3\right)}^{-3}={\left({\displaystyle \frac{1}{-3}}\right)}^3 \\ =\frac{1^3}{-3^3} \\ =\frac{1}{-27} \\ =\frac{1\times -1}{-27\times -1} \\ =\frac{-1}{27} \end{gathered}$$

Q8 Test Paper 2 Class 8 RS AGGARWAL chapter 2 Exponents Solution

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Question 8:

Mark (✓) against the correct answer
The value of ${\left(\frac{3}{4}\right)}^{-3}$ is
(a) $\frac{-27}{64}$
(b) $\frac{64}{27}$
(c) $\frac{-9}{4}$
(d) $\frac{27}{64}$

Answer 8:

(b) $\frac{64}{27}$

${\left(\frac{3}{4}\right)}^{-3}={\left(\frac{4}{3}\right)}^3=\frac{4^3}{3^3}=\frac{64}{27}$

Q9 Test Paper 2 Class 8 RS AGGARWAL chapter 2 Exponents Solution

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Question 9:

Mark (✓) against the correct answer
(3⁻⁶ ÷ 3⁴) = ?
(a) 3⁻²
(b) 3²
(c) 3⁻¹⁰
(d) 3¹⁰

Answer 9:

(c) $3^{-10}$

$$\begin{gathered} \left(3^{-6}÷3^4\right) \\ =\left(\frac{1}{3^6}÷3^4\right) \\ =\frac{1}{3^6}\times \frac{1}{3^4} \\ =\frac{1}{3^{\left(6+4\right)}} \\ =\frac{1}{3^{10}} \\ =3^{-10} \end{gathered}$$

Q10 Test Paper 2 Class 8 RS AGGARWAL chapter 2 Exponents Solution

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Question 10:

Mark (✓) against the correct answer
If ${\left(\frac{5}{12}\right)}^{-4}\times {\left(\frac{5}{12}\right)}^{3x}={\left(\frac{5}{12}\right)}^5, \mathrm{then} x=?$
(a) −1
(b) 1
(c) 2
(d) 3

Answer 10:

(d) 3

$$\begin{gathered} {\left(\frac{5}{12}\right)}^{-4}\times {\left(\frac{5}{12}\right)}^{3x}={\left(\frac{5}{12}\right)}^5 \\ \Rightarrow {\left(\frac{5}{12}\right)}^{-4+3x}={\left(\frac{5}{12}\right)}^5 \\ \Rightarrow -4+3x=5 \\ \Rightarrow 3x=5+4=9 \\ \Rightarrow x=\frac{9}{3}=3 \end{gathered}$$

Q11 Test Paper 2 Class 8 RS AGGARWAL chapter 2 Exponents Solution

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Question 11:

Mark (✓) against the correct answer
${\left(\frac{3}{5}\right)}^0=?$
(a) $\frac{5}{3}$
(b) $\frac{3}{5}$
(c) 1
(d) 0

Answer 11:

(c) 1

Using the law of exponents, which says ${\left(\frac{a}{b}\right)}^0=1,$ we get:
${\left(\frac{3}{5}\right)}^0=1$

Q12 Test Paper 2 Class 8 RS AGGARWAL chapter 2 Exponents Solution

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Question 12:

Mark (✓) against the correct answer
${\left(\frac{-6}{5}\right)}^{-1}=?$
(a) $\frac{6}{5}$
(b) $\frac{-6}{5}$
(c) $\frac{5}{6}$
(d) $\frac{-5}{6}$

Answer 12:

(d) $\left(\frac{-5}{6}\right)$

${\left(\frac{-6}{5}\right)}^{-1}={\left(\frac{5}{-6}\right)}^1=\frac{5}{-6}=\frac{5\times -1}{-6\times -1}=\frac{-5}{6}$

Q13 Test Paper 2 Class 8 RS AGGARWAL chapter 2 Exponents Solution

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Question 13:

Mark (✓) against the correct answer
${\left(\frac{-1}{3}\right)}^3=?$
(a) $\frac{-1}{9}$
(b) $\frac{1}{9}$
(c) $\frac{-1}{27}$
(d) $\frac{1}{27}$

Answer 13:

(c) $\frac{-1}{27}$

${\left(\frac{-1}{3}\right)}^3=\frac{-1^3}{3^3}=\frac{-1}{27}$

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Q14 Test Paper 2 Class 8 RS AGGARWAL chapter 2 Exponents Solution

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Question 14:

Fill in the blanks.
(i) 360000 written in standard form is .........
(ii) 0.0000123 written in standard form is .........
(iii) ${\left(\frac{-2}{3}\right)}^{-2}=.........$
(iv) 3 × 10⁻³ in usual form is .........
(v) 5.32 × 10⁻⁴ in usual form is .........

Answer 14:

(i) The standard form of 36000 is $3.6 \times 10^{5}$

$360000=36 \times 10^{4}=3.6 \times 10 \times 10^{4}$

$=3.6 \times 10^{(1+4)}=3.6 \times 10^{5}$

(ii) The standard form of 0.0000123 is $1.23 \times 10^{-5}$

$0.0000123=\frac{123}{10000000}=\frac{123}{10^{7}}$

$=\frac{1.23 \times 100}{10^{7}}=\frac{1.23 \times 10^{2}}{10^{7}}$

$=1.23 \times 10^{(2-7)}=1.23 \times 10^{-5}$

(iii) $\left(\frac{-2}{3}\right)^{-2}=\frac{9}{4}$
$\left(\frac{-2}{3}\right)^{-2}=\left(\frac{3}{-2}\right)^{2}=\frac{3^{2}}{-2^{2}}=\frac{9}{4}$

(iv) The usual form of $3 \times 10^{-3}$ is 0.003.
$3 \times 10^{-3}=\frac{3}{10^{3}}=\frac{3}{1000}$=0.003​

(v) The usual form of $5.32 \times 10^{-4}$ is 0.000532.
$5.32 \times 10^{-4}=\frac{5.32}{10^{4}}=\frac{5.32}{10000}$=0.000532