Exercise 2.1 - Powers RD Sharma Solutions | Mathematics (Maths) Class 8 PDF Download

Q.1. Express each of the following as a rational number of the form , where p and q are integers and q ≠ 0.

(i) 2⁻³

(ii) (−4)⁻²

(iii)

(iv)

(v)

Ans: We know that a⁻ⁿ = 1/aⁿ. Therefore,

(i)

(ii)

(iii)

(iv)

(v)

Q.2. Find the value of each of the following:

(i) 3⁻¹ + 4⁻¹

(ii) (3⁰ + 4⁻¹) × 2²

(iii) (3⁻¹ + 4⁻¹ + 5⁻¹)⁰

(iv) {(13)⁻¹−(14)⁻¹}⁻¹

Ans: 

(i) We know from the property of powers that for every natural number a, a⁻¹ = 1/a. Then:

---> (a⁻¹ = 1/a)

(ii) We know from the property of powers that for every natural number a, a⁻¹ = 1/a.

Moreover, a0 is 1 for every natural number a not equal to 0. Then:

(3⁰+4⁻¹)×2²

=

=
= 5

(iii) We know from the property of powers that for every natural number a, a⁻¹ = 1/a.

Moreover, a⁰ is 1 for every natural number a not equal to 0. Then:

(3⁻¹+4⁻¹+5⁻¹) = 1          ---> (Ignore the expression inside the bracket and use a⁰ = 1 immediately.)

(iv) We know from the property of powers that for every natural number a, a⁻¹ = 1/a. Then:

 ---> (a−1 = 1/a)

= 1

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

(i)

(ii)

(iii) (2⁻¹ × 4⁻¹) ÷ 2⁻²

(iv) (5⁻¹ × 2⁻¹) ÷ 6⁻¹

Ans: 

(i)

=  ---> (a−1 = 1/a)

= 2 + 3 + 4

= 12

(ii)

=

= 4 + 9 +16

= 29

(iii) 

(2⁻¹ × 4⁻¹) ÷ 2⁻² =

= 2

(iv) 

(5⁻¹ × 2⁻¹) ÷ 6⁻¹ =

=

=

Q.4. Simplify:

(i) (4⁻¹ × 3⁻¹)²

(ii) (5⁻¹ ÷ 6⁻¹)³

(iii) (2⁻¹ + 3⁻¹)⁻¹

(iv) (3⁻¹ × 4⁻¹)⁻¹ × 5⁻¹

Ans: 

(i) (4⁻¹ × 3⁻¹)² 

=

=

=

=

(ii) 

(5⁻¹ ÷ 6⁻¹)³ 

=

=

=

=

(iii) 

(2⁻¹ + 3⁻¹)⁻¹ 

=

=

=

=

(iv) 

(3⁻¹ × 4⁻¹)⁻¹ × 5⁻¹

=

=

=

Q.5. Simplify:

(i)

(ii)

(iii)

(iv)

Ans:  

(i) 

=

(ii)

=

=

=

(iii)

=

= (27−8)÷64

=19 × 1/64

 =19/64

(iv)

=

=

= 

Q.6. By what number should 5−¹ be multiplied so that the product may be equal to (−7)⁻¹?

Ans: Using the property a⁻¹ = 1/a for every natural number a, we have 5⁻¹ = 1/5 and (−7)⁻¹ = −1/7. We have to find a number x such that

Multiplying both sides by 5, we get:

Hence, the required number is −5/7.

Q.7. By what number should (1/2)⁻¹ be multiplied so that the product may be equal to (−4/7)⁻¹?

Ans: Using the property a⁻¹ = 1/a for every natural number a, we have (1/2)⁻¹ = 2 and (−4/7)⁻¹ = −7/4. We have to find a number x such that

Dividing both sides by 2, we get:

Hence, the required number is −7/8.

Q.8. By what number should (−15)⁻¹ be divided so that the quotient may be equal to (−5)⁻¹?

Ans: Using the property a⁻¹ = 1/a for every natural number a, we have (−15)⁻¹ = −1/15 and (−5)⁻¹ = −1/5. We have to find a number x such that

or

or

Hence, (−15)⁻¹ should be divided by 1/3 to obtain (−5)⁻¹.