RD Sharma Solutions Class 8 Powers Exercise 2.1

RD Sharma Solutions Class 8 Chapter 2 Exercise 2.1

RD Sharma Class 8 Solutions Chapter 2 Ex 2.1 PDF Free Download

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

 

Answer:

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

 

Answer:

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

 

Answer:

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

Answer:

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

 

Answer:

(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 \( -7^{-1} \)?

Answer:

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 \( \left (-\frac{4}{7} \right )^{-1}\)?

Answer:

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 \( \left ( -5 \right )^{-1} \)

Answer:

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 \(\left ( -5 \right )^{-1}\)..


Practise This Question

If a 3 digit number abc is reversed, then the difference of original and reversed number is not divisible by 9.