RD Sharma solution class 8 chapter 2 Powers Exercise (MCQs)

Exercise (MCQs)

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Question 1:

Square of $\left(\frac{-2}{3}\right)$ is
(a) $-\frac{2}{3}$
(b) $\frac{2}{3}$
(c) $-\frac{4}{9}$
(d) $\frac{4}{9}$

Answer 1:

(d) 4/9
To square a number is to raise it to the power of 2. Hence, the square of (−2/3) is
$\frac{(-2{)}^2}{3^2} = \frac{4}{9}$      ---> ( (a/b)ⁿ =  (aⁿ)/(bⁿ) )               

Question 2:

Cube of $\frac{-1}{2}$ is
(a) $\frac{1}{8}$
(b) $\frac{1}{16}$
(c) $-\frac{1}{8}$
(d) $\frac{-1}{16}$

Answer 2:

(c) -1/8
The cube of a number is the number raised to the power of 3. Hence the cube of −1/2 is

$\frac{(-1{)}^3}{2^3}$        ---> ( (a/b)ⁿ = (aⁿ)/(bⁿ)
  $=\frac{-1}{8}$

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Question 3:

Which of the following is not equal to ${\left(\frac{-3}{5}\right)}^4?$
(a) $\frac{(-3{)}^4}{5^4}$
(b) $\frac{3^4}{(-5{)}^4}$
(c) $-\frac{3^4}{5^4}$
(d) $\frac{-3}{5}\times \frac{-3}{5}\times \frac{-3}{5}\times \frac{-3}{5}$

Answer 3:

(c)  −(3⁴/5⁴)

${\left(\frac{-3}{5}\right)}^4 = \frac{(-3{)}^4}{5^4} = \frac{3^4}{(-5{)}^4} = \frac{-3}{5}\times \frac{-3}{5}\times \frac{-3}{5}\times \frac{-3}{5}$
$\mathrm{It} \mathrm{is} \mathrm{not} \mathrm{equal} \mathrm{to} -\frac{3^4}{5^4}$.

Question 4:

Which  of the following is not reciprocal of ${\left(\frac{2}{3}\right)}^4?$
(a) ${\left(\frac{3}{2}\right)}^4$
(b) ${\left(\frac{2}{3}\right)}^{-4}$
(c) ${\left(\frac{3}{2}\right)}^{-4}$
(d) $\frac{3^4}{2^4}$

Answer 4:

(c) (3/2)⁻⁴
The reciprocal of is .
Therefore, option (a) is the correct answer.
Option (b) is just re-expressing the number with a negative exponent.
Option (d) is obtained by working out the exponent.
Hence,option (c) is not the reciprocal of  .

Question 5:

Which of the following numbers is not equal to $\frac{-8}{27}?$
(a) ${\left(\frac{2}{3}\right)}^{-3}$
(b) $-{\left(\frac{2}{3}\right)}^3$
(c) ${\left(-\frac{2}{3}\right)}^3$
(d) $\left(\frac{-2}{3}\right)\times \left(\frac{-2}{3}\right)\times \left(\frac{-2}{3}\right)$

Answer 5:

(a) (2/3)^-3

We can write as . It can be written in the forms given below.


            ---> work out the minuses
                              
                                                  
Hence, option (b) is equal to .

We can also write:

                              
Hence, option (c) is also equal to .

We can also write:

Hence, option (d) is also equal to $-\frac{8}{27}$.

This leaves out option (a) as the one not equal to $-\frac{8}{27}$.

Question 6:

${\left(\frac{2}{3}\right)}^{-5}$ is equal to
(a) ${\left(\frac{-2}{3}\right)}^5$
(b) ${\left(\frac{3}{2}\right)}^5$
(c) $\frac{2x-5}{3}$
(d) $\frac{2}{3\times 5}$

Answer 6:

(b)${\left(\frac{3}{2}\right)}^5$

Rearrange (2/3)⁻⁵ to get a positive exponent.
$$\begin{gathered} {\left(\frac{2}{3}\right)}^{-5}=\frac{1}{{\left({\displaystyle \frac{2}{3}}\right)}^5} \left(a^{-n}=\frac{1}{a^n}\right) \\ =\frac{1}{{\displaystyle \frac{2^5}{3^5}}} \left\{ {\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n}\right\} \\ =\frac{3^5}{2^5} \\ ={\left(\frac{3}{2}\right)}^5 \end{gathered}$$

Question 7:

${\left(\frac{-1}{2}\right)}^5\times {\left(\frac{-1}{2}\right)}^3$ is equal to
(a) ${\left(\frac{-1}{2}\right)}^8$
(b) $-{\left(\frac{1}{2}\right)}^8$
(c) ${\left(\frac{1}{4}\right)}^8$
(d) ${\left(-\frac{1}{2}\right)}^{15}$

Answer 7:

(a) (−1/2)⁸

We have:


                             

Question 8:

${\left(\frac{-1}{5}\right)}^3÷{\left(\frac{-1}{5}\right)}^8$ is equal to
(a) ${\left(-\frac{1}{5}\right)}^5$
(b) ${\left(-\frac{1}{5}\right)}^{11}$
(c) $(-5{)}^5$
(d) ${\left(\frac{1}{5}\right)}^5$

Answer 8:

(c)  (−5)⁵

We have:


                                        
                                        
                                        
                                        
                                        
                                        

Question 9:

${\left(\frac{-2}{5}\right)}^7÷{\left(\frac{-2}{5}\right)}^5$ is equal to
(a) $\frac{4}{25}$
(b) $\frac{-4}{25}$
(c) ${\left(\frac{-2}{5}\right)}^{12}$
(d) $\frac{25}{4}$

Answer 9:

(a) 4/25

We have:


                                        
                                        
                                        

Question 10:

${\left\{{\left(\frac{1}{3}\right)}^2\right\}}^4$ is equal to
(a) ${\left(\frac{1}{3}\right)}^6$
(b) ${\left(\frac{1}{3}\right)}^8$
(c) ${\left(\frac{1}{3}\right)}^{24}$
(d) ${\left(\frac{1}{3}\right)}^{16}$

Answer 10:

(b) (1/3)⁸

We have:

          ---> ( (a^m)ⁿ = a^mxn)
                      

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Question 11:

${\left(\frac{1}{5}\right)}^0$ is equal to
(a) 0
(b) $\frac{1}{5}$
(c) 1
(d) 5

Answer 11:

(c) 1

We have:

         ---> (a⁰ = 1, for every non-zero rational number a.)

Question 12:

${\left(\frac{-3}{2}\right)}^{-1}$ is equal to
(a) $\frac{2}{3}$
(b) $-\frac{2}{3}$
(c) $\frac{3}{2}$
(d) none of these

Answer 12:

We have:

          --> (a⁻¹ = 1/a)
                    
                   

Question 13:

${\left(\frac{2}{3}\right)}^{-5}\times {\left(\frac{5}{7}\right)}^{-5}$ is equal to
(a) ${\left(\frac{2}{3}\times \frac{5}{7}\right)}^{-10}$
(b) ${\left(\frac{2}{3}\times \frac{5}{7}\right)}^{-5}$
(c) ${\left(\frac{2}{3}\times \frac{5}{7}\right)}^{25}$
(d) ${\left(\frac{2}{3}\times \frac{5}{7}\right)}^{-25}$

Answer 13:

We have:

          ---> ((a x b)ⁿ = aⁿ x bⁿ)

Question 14:

${\left(\frac{3}{4}\right)}^5÷{\left(\frac{5}{3}\right)}^5$ is equal to

Answer 14:

(a)  ${\left(\frac{3}{4}÷\frac{5}{3}\right)}^5$

We have:

            ---> $a^n ÷ b^n={\left(a÷b\right)}^{n^{}}$

Question 15:

For any two non-zero rational numbers a and b, a⁴ ÷ b⁴ is equal to
(a) (a ÷ b)¹
(b) (a ÷ b)⁰
(c) (a ÷ b)⁴
(d) (a ÷ b)⁸

Answer 15:

This is one of the basic exponential formulae, i.e. .

Question 16:

For any two rational numbers a and b, a⁵ × b⁵ is equal to
(a) (a × b)⁰
(b) (a × b)¹⁰
(c) (a × b)⁵
(d) (a × b)²⁵

Answer 16:

(c) (a x b)⁵
aⁿ x bⁿ = (a x b)ⁿ
Hence,
a⁵ x b⁵ = (a x b)⁵

Question 17:

For a non-zero rational number a, a⁷ ÷ a¹² is equal to
(a) a⁵
(b) a⁻¹⁹
(c) a⁻⁵
(d) a¹⁹

Answer 17:

(c) a⁻⁵

Hence,

Question 18:

For a non zero rational number a, (a³)⁻² is equal to
(a) a⁹
(b) a⁻⁶
(c) a⁻⁹
(d) a¹

Answer 18:

(b) a⁻⁶

We have:

          ---> ((a^m)ⁿ = a^{m x n})