RD Sharma Solutions for Class 8 Maths Chapter 2 – Powers Updated for 2025-26

RD Sharma Solutions for Class 8 Maths Chapter 2 - Powers are designed to assist students in their exam preparation and help them score well in Maths. Our expert solutions provide a detailed approach to understanding and solving all the exercise problems, with practical examples and easy-to-follow steps. These RD Sharma solutions, created by our subject specialists, include helpful tips and shortcuts that simplify complex concepts for better comprehension.

This chapter focuses on the laws of integral exponents (both positive and negative) of rational numbers. To make the learning process smoother, Home-Tution provides clear and step-by-step solutions for all the questions in Chapter 2. By practicing these solutions, students can strengthen their understanding and build a solid foundation in the subject.

For convenient access, students can download the PDF of RD Sharma Class 8 Maths Chapter 2 from the provided links.

RD Sharma Solutions for Class 8 Maths Chapter 2 – Powers

RD Sharma Class 8 Chapter 2 is part of the 27 chapters in the book. This chapter focuses on Powers and includes 3 practical exercises to help students test their understanding with quality questions.

Access answers to RD Sharma Solutions for Class 8 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

(i) 2 - 3

2 - 3 = -1

As a rational number, this is: -1 = −1/1

(ii) (-4) - 2

-4 - 2 = -6

As a rational number, this is: -6 = −6/1

(iii) 1/3 - 2

1/3 - 2 = 1/3 - 6/3 = (1 - 6)/3 = −5/3

So, the rational number is: −5/3

(iv) 1/2 - 5

1/2 - 5 = 1/2 - 10/2 = (1 - 10)/2 = −9/2

So, the rational number is: −9/2

(v) 2/3 - 2

2/3 - 2 = 2/3 - 6/3 = (2 - 6)/3 = −4/3

So, the rational number is: −4/3

2. 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

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

To solve this:

• (1/2)^-1 = 2
• (1/3)^-1 = 3
• (1/4)^-1 = 4

The expression becomes: 2 + 3 + 4 = 9

Answer: 9

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

To solve this:

• (1/2)^-2 = 4
• (1/3)^-2 = 9
• (1/4)^-2 = 16

The expression becomes: 4 + 9 + 16 = 29

Answer: 29

(iii) (2^-1 × 4^-1) ÷ 2^-2

To solve this:

• 2^-1 = 1/2
• 4^-1 = 1/4
• 2^-2 = 1/4

Now, calculate: (1/2 × 1/4) ÷ (1/4) = (1/8) ÷ (1/4) = 1/8 × 4/1 = 4/8 = 1/2

Answer: 1/2

(iv) (5^-1 × 2^-1) ÷ 6^-1

To solve this:

• 5^-1 = 1/5
• 2^-1 = 1/2
• 6^-1 = 1/6

Now, calculate: (1/5 × 1/2) ÷ (1/6) = (1/10) ÷ (1/6) = 1/10 × 6/1 = 6/10 = 3/5

Answer: 3/5

3. 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

(i) (4^-1 × 3^-1)²

4^-1 = 1/4

3^-1 = 1/3

(1/4) × (1/3) = 1/12

(1/12)² = 1/144

Answer: 1/144

(ii) (5^-1 ÷ 6^-1)³

5^-1 = 1/5

6^-1 = 1/6

(1/5) ÷ (1/6) = (1/5) × 6 = 6/5

(6/5)³ = 216/125

Answer: 216/125

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

2^-1 = 1/2

3^-1 = 1/3

(1/2) + (1/3) = (3+2)/6 = 5/6

(5/6)^-1 = 6/5

Answer: 6/5

(iv) (3^-1 × 4^-1)^-1 × 5^-1

3^-1 = 1/3

4^-1 = 1/4

(1/3) × (1/4) = 1/12

(1/12)^-1 = 12

12 × (1/5) = 12/5

Answer: 12/5

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

We are given the equation:

5^-1 × x = (-7)^-1

Step 1: Simplify the expressions involving exponents.

5^-1 = 1/5

(-7)^-1 = 1/-7

Step 2: The equation becomes:

1/5 × x = 1/-7

Step 3: To find x, multiply both sides of the equation by 5 to isolate x:

x = 1/-7 × 5

x = 5/-7

The number that 5^-1 should be multiplied by to get (-7)^-1 is:

x = -5/7

Benefits of using RD Sharma for Class 8 Maths Chapter 2

Here are some benefits of using RD Sharma for Class 8 Maths Chapter 2 (Powers):

1. Clear Conceptual Understanding: RD Sharma offers clear explanations of the fundamental concepts like exponents, integral exponents, and their laws, making it easier for students to understand how to work with powers and their properties.
2. Step-by-Step Solutions: Each exercise in RD Sharma comes with detailed solutions that guide students through each step of the problem-solving process. This ensures they can follow the logic and apply similar steps to solve other problems.
3. Variety of Practice Questions: The chapter contains a wide range of practice questions, from basic to complex, allowing students to test their understanding and improve their problem-solving skills.
4. Develops Problem-Solving Skills: By practicing the problems in the book, students learn how to approach different kinds of questions and develop critical thinking and analytical skills.
5. Enhanced Exam Preparation: RD Sharma clear structure and gradual difficulty levels in each exercise help students build their knowledge progressively, which is crucial for strong exam performance.
6. Comprehensive Coverage: This chapter covers all key topics related to powers, ensuring students gain in-depth knowledge and are prepared for any variations in exam questions.
7. Accessible Language: The language used in RD Sharma is simple and easy to follow, which helps students grasp complex concepts quickly without feeling overwhelmed.
8. Helps Build Foundation for Higher Classes: The concepts learned in this chapter serve as a solid foundation for future studies in mathematics, especially when dealing with algebra and number theory.
9. Useful for Competitive Exams: The comprehensive nature of RD Sharma makes it an excellent resource not only for school exams but also for competitive exams where students need to be well-prepared in mathematics.