# RD Sharma Solutions for Class 8 Maths Chapter 2 - Powers Exercise 2.1

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## RD Sharma Solutions for Class 8 Maths Exercise 2.1 Chapter 2 Powers

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

(i) 2-3

(ii) (-4)-2

(iii) 1/(3)-2

(iv) (1/2)-5

(v) (2/3)-2

Solution:

(i) 2-3 = 1/23 = 1/2×2×2 = 1/8 (we know that a-n = 1/an)

(ii) (-4)-2 = 1/-42 = 1/-4×-4 = 1/16 (we know that a-n = 1/an)

(iii) 1/(3)-2 = 32 = 3×3 = 9 (we know that 1/a-n = an)

(iv) (1/2)-5 = 25 / 15 = 2×2×2×2×2 = 32 (we know that a-n = 1/an)

(v) (2/3)-2 = 32 / 22 = 3×3 / 2×2 = 9/4 (we know that a-n = 1/an)

2. Find the values of each of the following:

(i) 3-1 + 4-1

(ii) (30 + 4-1) × 22

(iii) (3-1 + 4-1 + 5-1)0

(iv) ((1/3)-1 – (1/4)-1)-1

Solution:

(i) 3-1 + 4-1

1/3 + 1/4 (we know that a-n = 1/an)

LCM of 3 and 4 is 12

(1×4 + 1×3)/12

(4+3)/12

7/12

(ii) (30 + 4-1) × 22

(1 + 1/4) × 4 (we know that a-n = 1/an, a0 = 1)

LCM of 1 and 4 is 4

(1×4 + 1×1)/4 × 4

(4+1)/4 × 4

5/4 × 4

5

(iii) (3-1 + 4-1 + 5-1)0

(We know that a0 = 1)

(3-1 + 4-1 + 5-1)0 = 1

(iv) ((1/3)-1 – (1/4)-1)-1

(31 – 41)-1 (we know that 1/a-n = an, a-n = 1/an)

(3-4)-1

(-1)-1

1/-1 = -1

3. Find the values of each of the following:

(i) (1/2)-1 + (1/3)-1 + (1/4)-1

(ii) (1/2)-2 + (1/3)-2 + (1/4)-2

(iii) (2-1 × 4-1) ÷ 2-2

(iv) (5-1 × 2-1) ÷ 6-1

Solution:

(i) (1/2)-1 + (1/3)-1 + (1/4)-1

21 + 31 + 41 (we know that 1/a-n = an)

2+3+4 = 9

(ii) (1/2)-2 + (1/3)-2 + (1/4)-2

22 + 32 + 42 (we know that 1/a-n = an)

2×2 + 3×3 + 4×4

4+9+16 = 29

(iii) (2-1 × 4-1) ÷ 2-2

(1/21 × 1/41) / (1/22) (we know that a-n = 1/an)

(1/2 × 1/4) × 4/1 (we know that 1/a ÷ 1/b = 1/a × b/1)

1/2

(iv) (5-1 × 2-1) ÷ 6-1

(1/51 × 1/21) / (1/61) (we know that a-n = 1/an)

(1/5 × 1/2) × 6/1 (we know that 1/a ÷ 1/b = 1/a × b/1)

3/5

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

Solution:

(i) (4-1 × 3-1)2 (we know that a-n = 1/an)

(1/4 × 1/3)2

(1/12)2

(1×1 / 12×12)

1/144

(ii) (5-1 ÷ 6-1)3

((1/5) / (1/6))3 (we know that a-n = 1/an)

((1/5) × 6)3 (we know that 1/a ÷ 1/b = 1/a × b/1)

(6/5)3

6×6×6 / 5×5×5

216/125

(iii) (2-1 + 3-1)-1

(1/2 + 1/3)-1 (we know that a-n = 1/an)

LCM of 2 and 3 is 6

((1×3 + 1×2)/6)-1

(5/6)-1

6/5

(iv) (3-1 × 4-1)-1 × 5-1

(1/3 × 1/4)-1 × 1/5 (we know that a-n = 1/an)

(1/12)-1 × 1/5

12/5

5. Simplify:

(i) (32 + 22) × (1/2)3

(ii) (32 – 22) × (2/3)-3

(iii) ((1/3)-3 – (1/2)-3) ÷ (1/4)-3

(iv) (22 + 32 – 42) ÷ (3/2)2

Solution:

(i) (32 + 22) × (1/2)3

(9 + 4) × 1/8 = 13/8

(ii) (32 – 22) × (2/3)-3

(9-4) × (3/2)3

5 × (27/8)

135/8

(iii) ((1/3)-3 – (1/2)-3) ÷ (1/4)-3

(33 – 23) ÷ 43 (we know that 1/a-n = an)

(27-8) ÷ 64

19 × 1/64 (we know that 1/a ÷ 1/b = 1/a × b/1)

19/64

(iv) (22 + 32 – 42) ÷ (3/2)2

(4 + 9 – 16) ÷ (9/4)

(-3) × 4/9 (we know that 1/a ÷ 1/b = 1/a × b/1)

-4/3

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

Solution:

Let us consider a number x

So, 5-1 × x = (-7)-1

1/5 × x = 1/-7

x = (-1/7) / (1/5)

= (-1/7) × (5/1)

= -5/7

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

Solution:

Let us consider a number x

So, (1/2)-1 × x = (-4/7)-1

1/(1/2) × x = 1/(-4/7)

x = (-7/4) / (2/1)

= (-7/4) × (1/2)

= -7/8

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

Solution:

Let us consider a number x

So, (-15)-1 ÷ x = (-5)-1

1/-15 × 1/x = 1/-5

1/x = (1×-15)/-5

1/x = 3

x = 1/3

## RD Sharma Solutions for Class 8 Maths Exercise 2.1 Chapter 2 Powers

Download the free RD Sharma Solutions Chapter 2 in PDF format, which provide answers to all the questions. Class 8 Maths Chapter 2 Powers Exercise 2.1 is based on problems which include negative exponents. These solutions are prepared by experienced faculty in accordance with the CBSE syllabus for the 8th Standard. The exercise-wise solutions are explained in a simple and easily understandable language to help students excel in the annual exam.