In this chapter, we provide RD Sharma Solutions for Class 8 Maths Exercise 2.1 Chapter 2 Powers for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 8 Maths Exercise 2.1 Chapter 2 Powers pdf, free RD Sharma Solutions for Class 8 Maths Exercise 2.1 Chapter 2 Powers book pdf download. Now you will get step by step solution to each question.
Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 2.1 Chapter 2 Powers
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
All Chapter RD Sharma Solutions For Class 8 Maths
*************************************************
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.
If these solutions have helped you, you can also share ncertsolutionsfor.com to your friends.