NCERT Solutions for Class 7 Maths Chapter 11 Exercise 11.2 Exponents and Powers in Hindi and English Medium for CBSE Session 2025-26. The solutions of ex. 11.2 class VII mathematics are revised according to new syllabus and latest textbooks issued for academic year 2025-26.

Class: 7Mathematics
Chapter: 11Exercise: 11.2
Topic Name:Exponents and Powers
Content:Exercise and Extra Questions
Academic Year:CBSE 2025-26
Medium:Hindi and English Medium

Class 7 Maths Chapter 11 Exercise 11.2 Solution

All the solutions and study material is updated according to latest NCERT Textbooks. If student feel any difficulty in PDF solutions, they may refer to video solution to understand properly. Videos related to each exercises are given separately on website as well as in app.

Class 7 Offline Maths App

Class 7 Maths Chapter 11 Exercise 11.2 Solution in Videos

Class 7 Maths App in Hindi Medium

Class 7 Maths Exercise 11.2 Important Questions

Simplify and write the answer in exponential form: (i) (62)4 (ii) (22)100 (iii) (750)2 (iv) (53)7

We have:
(i) (62)4 = 62 x 62 x 62 x 62 = 62+2+2+2 = 62×4 = 68
(ii) (22)100 = 22×100 = 2200
(iii) (750)2 = 750×2 = 7100
(iv) (53)7 = 53×7 = 521

Express the following terms in the exponential form: (i) (2 ร— 3)โต (ii) (2a)โด (iii) (โ€“ 4m)ยณ

We have:
(i) (2 ร— 3)โต = (2 ร— 3) ร— (2 ร— 3) ร— (2 ร— 3) ร— (2 ร— 3) ร— (2 ร— 3)
= (2 ร— 2 ร— 2 ร— 2 ร— 2) ร— (3 ร— 3ร— 3 ร— 3 ร— 3)
= 2โต ร— 3โต
(ii) (2a)โด = 2a ร— 2a ร— 2a ร— 2a = (2 ร— 2 ร— 2 ร— 2) ร— (a ร— a ร— a ร— a) = 2โด ร— aโด
(iii) (โ€“ 4m)ยณ = (โ€“ 4 ร— m)ยณ = (โ€“ 4 ร— m) ร— (โ€“ 4 ร— m) ร— (โ€“ 4 ร— m) = (โ€“ 4) ร— (โ€“ 4) ร— (โ€“ 4) ร— (m ร— m ร— m)
= (โ€“ 4)ยณ ร— (m)ยณ

Power of a Power

Consider the following
Simplify {(2)ยณ}ยฒ = (2)ยณ x (2)ยณ
Or, (2)ยณโบยณ = (2)โถ

Class 7 Maths Exercise 11.2 Important Questions

Simplify: (6โปยน โ€“ 8โปยน)โปยน + (2โปยน โ€“ 3โปยน)โปยน

We have:
(6โปยน โ€“ 8โปยน)โปยน + (2โปยน โ€“ 3โปยน)โปยน
= (1/6 โ€“ 1/8)โปยน + (1/2 โ€“ 1/3)โปยน
= {(4 โ€“ 3)/24}-ยน + {(3 โ€“ 2)/6}-ยน
= (1/24)โปยน + (1/6)โปยน
= (24/1) + (6/1)
= 24 + 6 = 30

By what number should we multiply 3โปโน so that the product is equal to 3?

Let the required number be x.
Then, 3โปโน X x = 3
Or, x = 3/3โปโน
Or, x = 3 X 3โน
X = 3ยนโบโน = 3ยนโฐ
Hence, the required number is 3ยนโฐ

By what number should we multiply (-8)โปยน to obtain a product equal to 10โปยน?

Let the required number be x. Then,
Then, (-8)โปยน X x = 10โปยน
Or, x = 10โปยน/(-8)โปยน
Or, x = -8 / 10
X = -4/5

Multiplying Powers with the Same Exponents

simplify 2ยณ ร— 3ยณ? Notice that here the two terms 2ยณ and 3ยณ have different bases, but the same exponents.
Now, 2ยณ ร— 3ยณ = (2 x 2 x 2) x (3 x 3 x 3)
Or, (2 x 3) x (2 x 3) x (2 x 3)
= 6 x 6 x 6 = 6ยณ (Observe 6 is the product of bases 2 and 3)

In general, for any non-zero integer a
Aแต ร— bแต = (ab)แต (where m is any whole number)

CBSe Class 7 Maths Exercise 11.2 solution guide
Class 7 Maths Ex. 11.2 solution
Class 7 Maths Exercise 11.2 solutions for new session
Class 7 Maths Exercise 11.2
Last Edited: April 21, 2023
Content Reviewed: April 21, 2023
Content Reviewer

Mayank Tiwari

I have completed my M. Tech. in Computer Science and Engineering, Specialization in Artificial Intelligence in Delhi. Since, then I am working for Tiwari Academy as quality manager in Tech and Content formation.