# NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers

Contents

NCERT Solutions for Class 8 Chapter 12 Exponents and Powers will help the students aiming for high grade in their exam. Refer to the NCERT Class 8 Mathematics Exercise Solutions and practice the questions and answers as well examples based on the CBSE curriculum to understand the concepts covered in the chapter. As per our opinion an expression which represents repeated multiplication of same factor is called a power. Such as, in case of 62, the number 6 is called the base, and number 2 is the exponent. Exponent corresponds to the number of times the base is utilized as a factor in an expression.

## NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Exercise 12.1

1. Evaluate:

(i) 3-2 (ii) (-4)-2 (iii) (1/2)-5

Solution:

(i) 3-2 = (1/3)2

= 1/9

(ii) (-4)-2 = (1/-4)2

= 1/16

(iii) (1/2)-5 = (2/1)5

= 25

= 32

2.  Simplify and express the result in power notation with positive exponent:

(i) (-4)÷(-4)8

(ii) (1/23)2

(iii) -(3)4×(5/3)4

(iv) (3-7÷3-10)×3-5

(v) 2-3×(-7)-3

Solution:

(i)

= (-4)5/(-4)8

= (-4)5-8

= 1/(-4)3

(ii) (1/23)2

= 12/(23)2

= 1/23×2 = 1/26

(iii) -(3)4×(5/3)4

= (-1)4×34×(54/34 )

= 3(4-4)×54

= 30×54 = 54

(iv) (3-7/3-10)× 3-5

= 3-7 – (-10) × 3-5

= 3(-7+10)×3-5

= 33×3-5

= 3(3+-5)

= 3-2

=1/32

(v) 2-3×(-7) – 3

= (2×-7)-3

(Because am×bm = (ab)m)

= 1/(2×-7)3

= 1/(-14)3

3. Find the value of :

(i) (30+4-1)×22 (ii) (2-1×4-1)÷2 – 2

(iii) (1/2)-2+(1/3)-2+(1/4)-2 (iv) (3-1+4-1+5-1)0

(v) {(-2/3)-2}2

Solution:

(i)(30+4– 1)×22

= (1+(1/4))×22

= ((4+1)/4 )×22

= (5/4)×22

= (5/22)×22

= 5×2(2-2)

= 5×20

= 5×1 = 5

(ii)(2-1×4-1)÷2-2

= [(1/2)×(1/4)] ÷(1/4)

= (1/2×1/22 )÷ 1/4

= 1/23÷1/4

= (1/8)×(4)

= 1/2

(iii) (1/2)-2+(1/3)-2+(1/4)-2

= (2-1)-2+(3-1)-2+(4-1)-2

= 2(-1×-2)+3(-1×-2)+4(-1×-2)

= 22+32+42

= 4+9+16

=29

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

= 1

(v) {(-2/3)-2}2 = (-2/3)-2×2

= (-2/3)-4

= (-3/2)4

= 81/16

4. Evaluate

(i) (8-1×53)/2-4

(ii) (5-1×2-2)×6-1

Solution:

(i) (8-1×53)/2-4

=

= 2×125 = 250

(ii) (5-1×2-2)×6-1

= (1/10)×1/6

= 1/60

5. Find the value of m for which 5÷ 5-3 = 55

Solution:

5m ÷ 5-3 = 55

5(m-(-3) ) = 55

5m+3 =55

Comparing exponents both sides, we get

m+3 = 5

m = 5-3

m = 2

6. Evaluate

(i)

(ii)

Solution:

(i)

= 3-4

= -1

(ii)

=

=

=

=  512/125

7. Simplify.

(i)

(ii)

Solution:

(i)

=
=

(ii)

=

=

=
=

= 1×1×3125

= 3125

## NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Exercise 12.2

1. Express the following numbers in standard form.

(i) 0.0000000000085

(ii) 0.00000000000942

(iii) 6020000000000000

(iv) 0.00000000837

(v) 31860000000

Solution:

(i) 0.0000000000085

= 0.0000000000085×(1012/1012)

= 8.5 ×10-12

(ii) 0.00000000000942

= 0.00000000000942×(1012/1012)

= 9.42×10-12

(iii) 6020000000000000

= 6020000000000000×(1015/1015)

= 6.02×1015

(iv) 0.00000000837

= 0.00000000837×(109/109)

= 8.37×10-9

(v) 31860000000

= 31860000000×(1010/1010)

= 3.186×1010

2.Express the following numbers in usual form.

(i) 3.02×10-6

(ii) 4.5×104

(iii)3×10-8

(iv)1.0001×109

(v) 5.8×1012

(vi)3.61492×106

Solution:

(i) 3.02×10-6 = 3.02/106 = 0 .00000302

(ii) 4.5×104 = 4.5×10000 = 45000

(iii) 3×10-8 = 3/108 = 0.00000003

(iv) 1.0001×109 = 1000100000

(v) 5.8×1012 = 5.8×1000000000000 = 5800000000000

(vi) 3.61492×10= 3.61492×1000000 = 3614920

3. Express the number appearing in the following statements in standard form.

(i) 1 micron is equal to 1/1000000 m.
(ii) Charge of an electron is 0.000, 000, 000, 000, 000, 000, 16 coulomb.
(iii) Size of bacteria is 0.0000005 m
(iv)  Size of a plant cell is 0.00001275 m
(v) Thickness of a thick paper is 0.07 mm

Solution :

(i) 1 micron = 1/1000000

= 1/106

= 1×10-6

(ii) Charge of an electron is 0.00000000000000000016 coulombs.

= 0.00000000000000000016×1019/1019

= 1.6×10-19 coulomb

(iii) Size of bacteria = 0.0000005

=  5/10000000 = 5/107 = 5×10-7 m

(iv) Size of a plant cell is 0.00001275 m

= 0.00001275×105/105

= 1.275×10-5m

(v) Thickness of a thick paper = 0.07 mm

0.07 mm = 7/100 mm = 7/102 = 7×10-2 mm

4. In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?

Solution:

Thickness of one book = 20 mm

Thickness of 5 books = 20×5 = 100 mm

Thickness of one paper = 0.016 mm

Thickness of 5 papers = 0.016×5 = 0.08 mm

Total thickness of a stack = 100+0.08 = 100.08 mm

= 100.08×102/102 mm

mm

## NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers

NCERT Solutions for Class 8 Maths Chapter 12, Exponents and Powers is all about the law of exponents and powers as well their applications.

These NCERT Solutions will helps the students to gating a better understanding of the topic and learn the all concepts. This study material is designed by subject experts as per the CBSE current syllabus prescribed by the board. Class 8 CBSE chapter 12 of Maths, students will learn about the how to write large numbers more conveniently by using powers and exponents, express numbers in standard form, dealing with the negative exponents, various laws of exponents.

The main topics are covered in this chapter include:

12.1 Introduction
12.2 Powers with Negative Exponents
12.3 Laws of Exponents
12.4 Using Exponents to Express Small Numbers as Standard Form

## Characteristics of NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers.

• simple and easy language used.
• NCERT Solutions will assist students to increase their level of confidence.
• The subject matter experts have grouped all exercise questions into one place for your review.
• All solutions are structured in an easy and logical language for quick revisions.
• These NCERT Solutions are very helpful for competitive exams.

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