# RD Sharma Solutions Class 8 Powers Exercise 2.1

## RD Sharma Solutions Class 8 Chapter 2 Exercise 2.1

#### Exercise 2.1

Question 1. Express each of the following as a rational number of the form $\frac{p}{q}$, where p and q are integers and $q \neq 0:$.

(i) $2 ^{ -3 }$

(ii) $\left(-4\right)^{-2}$

(iii) $\frac{ 1 }{ 3^{-2}}$

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

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

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

(ii) $\left (-4 \right ) ^{ -2 } = \frac{ 1 }{ \left (-4 \right )^{ 2 } } = \frac{ 1 } { 16 }$

(iii) $\frac{ 1 }{ 3^{-2}} = 3^{2} = 9$

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

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

Question 2. Find the values of the following:

(i) $3^{ -1 } + 4^{ -1 }$

(ii) $\left ( 3^{0} + 4^{-1}\right ) \times 2^{2}$

(iii)  $\left ( 3^{ -1 } + 4^{ -1 } + 5^{ -1 } \right )^{ 0 }$

(iv) $\left (\left ( \frac{1}{3} \right )^{-1} – \left ( \frac{1}{4} \right )^{-1} \right ) ^{-1}$

(i) We know from the property of powers that for every natural number a, $a^{-1} = \frac{1}{a}$, Then:

$3^{ -1 } + 4^{ -1 } = \frac{ 1 }{ 3 } + \frac{ 1 }{ 4 }$

= $\frac{ 4 + 3 }{ 12 }$

= $\frac{ 7 }{ 12 }$

(ii) We know from the property of powers that for every natural number a, $a^{-1} = \frac{1}{a}$.

Moreover, $a^{0}$ is 1 for every natural number a not equal to 0. Then,

$\left ( 3^{0} + 4^{-1}\right ) \times 2^{2}$

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

= $\frac{5}{4} \times 4$

= 5

(iii) We know from the property of powers that for every natural number a, $a^{-1} = \frac{1}{a}$.

Moreover, $a^{0}$ is 1 for every natural number a not equal to 0. Then,

$\left ( 3^{ -1 } + 4^{ -1 } + 5^{ -1 } \right )^{ 0 }$ = 1     —> (Ignore the expression inside the bracket and use a0 = 1)

(iv) We know from the property of powers that for every natural number a, $a^{-1} = \frac{1}{a}$.

Then:

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

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

= -1

Question 3.Find the value of each of the following:

(i) $\left (\frac{1}{2} \right ) ^{-1} + \left (\frac{1}{3} \right ) ^{-1} + \left (\frac{1}{4} \right ) ^{-1}$

(ii) $\left (\frac{1}{2} \right ) ^{-2} + \left (\frac{1}{3} \right ) ^{-2} + \left (\frac{1}{4} \right ) ^{-2}$

(iii) $\left (2 ^{-1} \times 4 ^{-4} \right ) \div 2 ^{-2}$

(iv) $\left ( 5 ^{-1} \times 2 ^{-1} \right ) \div 6 ^{-1}$

(i) $\left (\frac{1}{2} \right ) ^{-1} + \left (\frac{1}{3} \right ) ^{-1} + \left (\frac{1}{4} \right ) ^{-1}$

= $\frac{1}{ \frac{1}{2}} + \frac{1}{ \frac{1}{3}} + \frac{1}{ \frac{1}{4}}$

= 2 + 3 + 4 = 12

(ii) $\left (\frac{1}{2} \right ) ^{-2} + \left (\frac{1}{3} \right ) ^{-2} + \left (\frac{1}{4} \right ) ^{-2}$

= $\frac{1}{ \left (\frac{1}{2} \right )^{2}} + \frac{1}{ \left (\frac{1}{3} \right )^{2}} + \frac{1}{\left ( \frac{1}{4} \right ) ^{2}}$

= $\frac{1}{ \frac{1}{4}} + \frac{1}{ \frac{1}{9}} + \frac{1}{ \frac{1}{16}}$

= 4 + 9 + 16 = 29

(iii) $\left (2 ^{-1} \times 4 ^{-4} \right ) \div 2 ^{-2}$

= $\frac{1}{2} \times \frac{1}{4} \div \frac{ 1 }{2 ^{2} }$

= $\frac{1} {8} \times 4$ = $\frac{1} {2}$

(iv) $\left ( 5 ^{-1} \times 2 ^{-1} \right ) \div 6 ^{-1}$

= $\left (\frac{1 } { 5} \times \frac{1} {2} \right ) \div \frac{1} {6}$

= $\frac{ 1} {10} \times 6$ = $\frac{3} {5}$

Question 4. Simplify:

(i) $\left (4^{-1} \times 3^{-1} \right ) ^{2}$

(ii) $\left ( 5^{-1} \div 6^{-1} \right )^{3}$

(iii) $\left ( 2^{-1} + 3^{-1} \right ) ^{-1}$

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

(i) $\left (4^{-1} \times 3^{-1} \right ) ^{2}$

= $\left (\frac{1}{4} \times \frac{1}{3} \right ) ^{2}$

= $\left ( \frac{1}{12} \right )^{2}$

= $\left ( \frac{1 ^{2} }{12 ^{2} } \right )$ = $\left ( \frac{1 }{24 } \right )$

(ii) $\left ( 5^{-1} \div 6^{-1} \right )^{3}$

= $\left ( \frac{1}{5} \div \frac{1}{6} \right )^{3}$

= $\left ( \frac{1}{5} \times 6 \right )^{3}$

= $\left ( \frac{6}{5} \right )^{3}$ = $\frac{216} {125}$

(iii) $\left ( 2^{-1} + 3^{-1} \right ) ^{-1}$

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

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

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

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

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

= $\left ( \frac{1} {12} \right )^{-1} \times \frac{1}{5}$ = $\frac{12}{5}$

Question 5. Simplify:

(i) $\left (3^{2} + 2^{2} \right ) \times \left ( \frac{1}{2} \right )^{3}$

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

(iii) $\left ( \left ( \frac{1}{3} ^{-3} \right ) – \left ( \frac{1}{2} \right ) ^{-3} \right ) \div \left ( \frac{1}{4} \right ) ^{-3}$

(iv) $\left (2^{2} + 3^{2} – 4^{2} \right ) \div \left ( \frac{3}{2} \right )^{2}$

(i) $\left (3^{2} + 2^{2} \right ) \times \left ( \frac{1}{2} \right )^{3}$

= $\left ( 9 + 4 \right ) \times \frac{1}{8}$ = $\frac{13} {8}$

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

= $\left ( 9 – 4 \right ) \times \frac{1} { \left (2/3 \right )^{3}}$

= $5 \times \frac{1} { \left (8/27 \right )}$ = $\frac{135} {8}$

(iii) $\left ( \left ( \frac{1}{3} ^{-3} \right ) – \left ( \frac{1}{2} \right ) ^{-3} \right ) \div \left ( \frac{1}{4} \right ) ^{-3}$

= $\left ( 3^{3} – 2^{3} \right ) \div 4^{3}$

= $\left ( 27 – 8 \right ) \div 64$

= $19 \times \frac{1}{64}$ = $\frac{19}{64}$

(iv) $\left (2^{2} + 3^{2} – 4^{2} \right ) \div \left ( \frac{3}{2} \right )^{2}$

= $\left (4 + 9 – 16 \right ) \div \left ( \frac{9}{4} \right )$

= $-3 \times \frac{4}{9}$ = $-\frac{4}{3}$

Question 6. By what number should $5^{-1}$ be multiplies so that the product may be equal to $-7^{-1}$?

Using the property $a^{-1} = \frac{1}{a}$ for every natural number a, we have $5^{-1} = \frac{1}{5}$ and $(-7)^{-1} = -\frac{1}{7}$. We have to find a number x such that

$\frac{1}{5} \times x = \frac{-1}{7}$

Multiply bith sides by 5, we get

$x = \frac{-5}{7}$

Hence, the required number is  $\frac{-5}{7}$

Question 7. By what number should $\left(\frac{1}{2}\right)^{-1}$  be multiplies so that the product may be equal to $\left (-\frac{4}{7} \right )^{-1}$?

Using the property $a^{-1} = \frac{1}{a}$ for every natural number a, we have $\left (\frac{1}{2} \right )^{-1} = 2 \: and \: \left (\frac{-4}{7} \right )^{-1} = \frac{-7}{4}$. We have to find the number x such that

$2x = \frac{-7}{4}$

Dividing both sides by 2, we get

$x = \frac{-7}{8}$

Hence, the required number is $\frac{-7}{8}$

Question 8. By what number should $\left ( -15 \right )^{-1}$ be multiplies so that the product may be equal to $\left ( -5 \right )^{-1}$

Using the property $a^{-1} = \frac{1}{a}$ for every natural number a, we have $\left ( -15 \right )^{-1} = -\frac{1}{15}$ and $\left ( -5 \right )^{-1} = -\frac{1}{5}$. We have to find a number x such that
$\frac{- \frac{1}{15}}{\frac{x}{1}} = \frac{-1}{5}$
Or $\frac{1}{15} \times \frac{1}{x} = – \frac{-1}{5}$
Or $x = \frac{1}{3}$
Hence, $\left ( -15 \right )^{-1}$ should be divided by $\frac{1}{3}$ to obtain $\left ( -5 \right )^{-1}$..