RS Aggarwal solution class 8 chapter 2 Exponents Exercise 2C

Exercise 2C

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Q1 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 1:

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The value of ${\left(\frac{2}{5}\right)}^{-3}$ is
(a) $-\frac{8}{125}$
(b) $\frac{25}{4}$
(c) $\frac{125}{8}$
(d) $-\frac{2}{5}$

Answer 1:

(c) $\frac{125}{8}$

${\left(\frac{2}{5}\right)}^{-3}={\left(\frac{5}{2}\right)}^3=\frac{5^3}{2^3}=\frac{125}{8}$

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Q2 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 2:

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The value of (−3)⁻⁴ is
(a) 12
(b) 81
(c) $-\frac{1}{12}$
(d) $\frac{1}{81}$

Answer 2:

(d) $\frac{1}{81}$

${\left(-3\right)}^{-4}=\frac{1}{{\left(-3\right)}^4}=\frac{1}{{\left(-1\right)}^4\times {\left(3\right)}^4}=\frac{1}{{\left(3\right)}^4}=\frac{1}{81}$

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Q3 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 3:

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The value of (−2)⁻⁵ is
(a) −32
(b) $\frac{-1}{32}$
(c) 32
(d) $\frac{1}{32}$

Answer 3:

(b) $\frac{-1}{32}$

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

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Question 4:

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(2⁻⁵ ÷ 2⁻²) = ?
(a) $\frac{1}{128}$
(b) $\frac{-1}{128}$
(c) $-\frac{1}{8}$
(d) $\frac{1}{8}$

Answer 4:

(d) $\frac{1}{8}$

$\left(2^{-5} \div 2^{-2}\right)=\left(\frac{1}{2^{5}} \div \frac{1}{2^{2}}\right)$

$=\left(\frac{1}{32} \div \frac{1}{4}\right)=\left(\frac{1}{32} \times 4\right)$

$=\frac{4}{32}=\frac{1}{8}$

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Q5 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 5:

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The value of (3⁻¹ + 4⁻¹)⁻¹ ÷ 5⁻¹ is
(a) $\frac{7}{10}$
(b) $\frac{60}{7}$
(c) $\frac{7}{5}$
(d) $\frac{7}{15}$

Answer 5:

(b) $\frac{60}{7}$

$\left(3^{-1}+4^{-1}\right)^{-1} \div 5^{-1}$

$=\left(\frac{1}{3}+\frac{1}{4}\right)^{-1} \div \frac{1}{5}$

$=\left(\frac{4+3}{12}\right)^{-1} \div \frac{1}{5}=\left(\frac{7}{12}\right)^{-1} \div \frac{1}{5}$

$=\left(\frac{12}{7}\right) \div \frac{1}{5}=\frac{12}{7} \times 5=\frac{60}{7}$

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Question 6:

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${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}=?$
(a) $\frac{61}{144}$
(b) $\frac{144}{61}$
(c) 29
(d) $\frac{1}{29}$

Answer 6:

(c) 29

$$\begin{gathered} {\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}= {\left(\frac{2}{1}\right)}^2+{\left(\frac{3}{1}\right)}^2+{\left(\frac{4}{1}\right)}^2 \\ =2^2+3^2+4^2 \\ =4+9+16 \\ =29 \end{gathered}$$

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Question 7:

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$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}=?$
(a) $\frac{19}{64}$
(b) $\frac{27}{16}$
(c) $\frac{64}{19}$
(d) $\frac{16}{25}$

Answer 7:

(a) $\frac{19}{64}$

$$\begin{gathered} \left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3} \\ =\left\{3^3-2^3\right\}÷4^3 \\ =\left\{27-8\right\}÷64 \\ =19÷64 \\ =\frac{19}{64} \end{gathered}$$

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Q8 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 8:

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${\left[{\left\{{\left(-\frac{1}{2}\right)}^2\right\}}^{-2}\right]}^{-1}=?$
(a) $\frac{1}{16}$
(b) 16
(c) $-\frac{1}{16}$
(d) −16

Answer 8:

(a) $\frac{1}{16}$

$$\begin{gathered} {\left[{\left\{{\left(-\frac{1}{2}\right)}^2\right\}}^{-2}\right]}^{-1} \\ ={\left[{\left\{-\frac{1}{2}\right\}}^{-4}\right]}^{-1} \\ ={\left(-\frac{1}{2}\right)}^{(-4\times -1)} \\ ={\left(-\frac{1}{2}\right)}^4 \\ =\frac{1}{16} \end{gathered}$$

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Question 9:

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The value of x for which ${\left(\frac{7}{12}\right)}^{-4}\times {\left(\frac{7}{12}\right)}^{3x}={\left(\frac{7}{12}\right)}^5,$ is
(a) −1
(b) 1
(c) 2
(d) 3

Answer 9:

(d) 3

$$\begin{gathered} {\left(\frac{7}{12}\right)}^{-4}\times {\left(\frac{7}{12}\right)}^{3x}={\left(\frac{7}{12}\right)}^5 \\ \Rightarrow {\left(\frac{7}{12}\right)}^{-4+3x}={\left(\frac{7}{12}\right)}^5 \\ \Rightarrow 3x-4=5 \\ 3x=9 \\ \mathrm{or} x=\frac{9}{3}=3 \end{gathered}$$

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Question 10:

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If (2^{3x − 1} + 10) ÷ 7 = 6, then x is equal to
(a) −2
(b) 0
(c) 1
(d) 2

Answer 10:

(d) 2

$$\begin{gathered} \left(2^{3x-1}+10\right)÷7=6 \\ \Rightarrow \frac{\left(2^{3x-1}+10\right)}{7}=\frac{6}{1} \\ \mathrm{On} \mathrm{cross} \mathrm{multiplying}: \\ \left(2^{3x-1}+10\right)\times 1=6\times 7=42 \end{gathered}$$

⇒ 2^3x-1  = 42 -10
⇒2^3x-1 = 32
⇒ 2^3x-1  = 2⁵
⇒ 3x-1 = 5
⇒ 3x = 6
Therefore, x =  2

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Question 11:

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${\left(\frac{2}{3}\right)}^0=?$
(a) $\frac{3}{2}$
(b) $\frac{2}{3}$
(c) 1
(d) 0

Answer 11:

(c) 1

Using the law of exponents ${\left(\frac{a}{b}\right)}^0=1$:
$\therefore$ ${\left(\frac{2}{3}\right)}^0=1$

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Question 12:

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${\left(\frac{-5}{3}\right)}^{-1}=?$
(a) $\frac{5}{3}$
(b) $\frac{3}{5}$
(c) $\frac{-3}{5}$
(d) none of these

Answer 12:

(c) $\frac{-3}{5}$

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

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Question 13:

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${\left(\frac{-1}{2}\right)}^3=?$
(a) $\frac{-1}{6}$
(b) $\frac{1}{6}$
(c) $\frac{1}{8}$
(d) $\frac{-1}{8}$

Answer 13:

(d) $\frac{-1}{8}$

${\left(\frac{-1}{2}\right)}^3=\frac{-1^3}{2^3}=\frac{-1}{8}$

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Question 14:

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${\left(\frac{-3}{4}\right)}^2=?$
(a) $\frac{-9}{16}$
(b) $\frac{9}{16}$
(c) $\frac{16}{9}$
(d) $\frac{-16}{9}$

Answer 14:

(b) $\frac{9}{16}$

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

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Q15 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 15:

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3670000 in standard form is
(a) 367 × 10⁴
(b) 36.7 × 10⁵
(c) 3.67 × 10⁶
(d) none of these

Answer 15:

(c) $3.67 \times 10^{6}$

$3670000=367 \times 10^{4}$

$=3.67 \times 100 \times 10^{4}$

$=3.67 \times 10^{2} \times 10^{4}$

$=3.67 \times 10^{(2+4)}$

$=3.67 \times 10^{6}$

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Question 16:

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0.0000463 in standard form is
(a) 463 × 10⁻⁷
(b) 4.63 × 10⁻⁵
(c) 4.63 × 10⁻⁹
(d) 46.3 × 10⁻⁶

Answer 16:

(b) $4.63 \times 10^{-5}$

$0.0000463=\frac{463}{10^{7}}=\frac{4.63 \times 10^{2}}{10^{7}}$ 

$=4.63 \times 10^{(2-7)}=4.63 \times 10^{-5}$

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Question 17:

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0.000367 × 10⁴ in usual form is
(a) 3.67
(b) 36.7
(c) 0.367
(d) 0.0367

Answer 17:

(a) 3.67

$0.000367 \times 10^{4}=\frac{367}{10^{6}} \times 10^{4}$

$=367 \times 10^{(4-6)}=367 \times 10^{-2}$

$=\frac{367}{10^{2}}=\frac{367}{100}=3.67$