NCERT Solutions for 8th Class Maths: Chapter 12-Exponents and Powers

NCERT Solutions for 8th Class Maths: Chapter 12-Exponents and Powers

Page No: 197

Exercise 12.1

1. Evaluate:
(i) 3^-2
(ii) (-4)^-2
(iii) (1/2)^-5 

Answer

(i) 3^-2
= 1/3² [ ∵ a^-m = 1/a^m ]
= 1/9

(ii) (-4)^-2
= 1/(-4)² [∵ a-m = 1/am ]
= 1/16

(iii) (1/2)^-5
= (2/1)⁵  [ ∵ a-m = 1/am ]
 = (2)⁵ = 32

2. Simplify and express the result in power notation with positive exponent:
(i) ( -4)⁵ ÷ (-4)⁸ 
(ii) (1/2³)² 
(iii) (-3)⁴ × (5/3)⁴ 
(iv) (3^-5 ÷ 3^-10) × 3^-5
(v) 2^-3 × (-7)^-3 

Answer

(i) ( -4)⁵ ÷ (-4)⁸
= (-4)^5-8 [ ∵ a^m ÷ aⁿ = a^m-n]
= (-4)^-3
= 1/(-4)³ [∵ a^-m = 1/a^m ]

(ii) (1/2³ )²
= 1²/(2³)² [∵ (a/b)^m = a^m/aⁿ]
= 1/2³×²
= 1/2⁶ [(a^m)ⁿ = a^m×ⁿ]

(iii) (-3)⁴ × (5/3)⁴
= (-3)⁴ × 5⁴/3⁴ [∵ (a/b)^m = a^m/aⁿ]
= {(-1)⁴ × 3⁴} × 5⁴/3⁴ [ ∵ (ab)^m = a^mb^m]
= 3^4-4 × 5⁴ [ ∵ a^m ÷ aⁿ = a^m-n]
= 3⁰ × 5⁴
= 5⁴ [a⁰ = 1]

(iv) (3^-7 ÷ 3^-10) × 3^-5
= 3^-7-(-10) × 3^-5 [∵ a^m ÷ aⁿ = a^m-n]
= 3^-7+10 × 3^-5
= 3³ × 3^-5 = 3^3+(-5) [∵ a^m × aⁿ = a^m+n]
=3^-2
= 1/3² [a^-m = 1/a^m]

(v) 2^-3 × (-7)^-3
= 1/2³ ×1/(-7)³ [∵ a^-m = 1/a^m]
= 1/{2 × (-7)}³
= 1/(-14)³ [∵ (ab)^m = a^mb^m]

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 

Answer 

(i) ( 3⁰ + 4^-1) × 2²
= (1 + 1/4) × 2² [∵ a^-m = 1/a^m]
= {(4 + 1)/4} × 2²
= 5/4 × 2²
= 5/2² × 2²
= 5 × 2^2-2 [∵ a^m ÷ aⁿ = a^m-n]
= 5 × 2⁰
= 5 × 1
= 5 [∵ a⁰ = 1]

(ii) ( 2^-1 × 4^-1) ÷ 2^-2 =[∵ a^-m = 1/a^m]
= [∵ a^m ×aⁿ = a^m+n]
=[∵ a^m ÷ aⁿ = a^m-n]
= 1/2 [∵ a^-m = 1/a^m]

(iii) (1/2)^-2 + (1/3)^-3 + (1/4)^-2
= (2^-1)^-2 + (3^-1)^-2 + (4^-1)^-2  [∵ a^-m = 1/a^m]
= 2^-1×(-2) + (3)^-1×(-2) + (4)^-1×(-2) [∵ (a^m)ⁿ = a^m×n]
= 2² + 3² + 4²
= 4 + 9 + 16
= 29

(iv) (3 + 4 + 5)⁰
= (1/3 + 1/4 + 1/5)⁰ [∵ a^-m = 1/a^m]
= {(20 + 15 + 12)/60}⁰
= (47/60)⁰
= 1 [∵ a⁰ = 1]

(v) [∵ (a^m)ⁿ = a^m×n]
=  [∵ a^-m = 1/a^m]
= 81/16

4. Evaluate:
(i) (8^-1 × 5³)/2^-4 
(ii) ( 5^-1 × 2^-1) × 6^-1 

Answer

(i) (8^-1 × 5³)/2^-4
= (8^-1 × 5³)/2^-4
= {(2³)^-1 × 5³}/2^-4
= (2^-3 × 5³)/2^-4  [∵ (a^m)ⁿ = a^m×n]
= 2^-3(-4) × 5³
= 2^-3+4 × 5³ [∵ a^m ÷ aⁿ = a^m-n]
= 2×125
= 250

(ii)( 5^-1 × 2^-1) 6^-1
= (1/5 × 1/2) × 1/6 [∵ a^-m = 1/a^m]
= 1/10 × 1/6 = 1/60

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

Answer

5^m ÷ 5^-3 = 5⁵
= 5^m-(-3) = 5⁵ [∵ a^m ÷ aⁿ = a^m-n]
⇒ 5^m+3 = 5⁵
Comparing exponents both sides, we get
⇒ m + 3 = 5
⇒ m = 5 - 3
⇒ m = 2

6. Evaluate:

(i) 
(ii) 

Answer

(i) 
= [∵ a^-m = 1/a^m]
={3 -4} = -1
(ii) 
= 
=  [∵ a^m ÷ aⁿ = a^m-n]
=  [∵ a^-m = 1/a^m]
= 512/125

7. Simplify:

(i) 
(ii) 

Answer

(i)
= 
=  
= 

(ii)
=

=
= 
= 
= 
= 1×1×3125 [∵ a⁰ = 1]
= 3125

Page No. 200

Exercise 12.2

1. Express the following numbers in standard form:
(i) 0.0000000000085
(ii) 0.00000000000942
(iii) 6020000000000000
(iv) 0.00000000837
(v) 31860000000

Answer

(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 

Answer

(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 × 10⁶
= 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 a bacteria is 0.0000005 m.
(iv) Size of a plant cell is 0.00001275 m.
(v) Thickness if a thick paper is 0.07 mm.

Answer

(i) 1 micron
= 1/1000000
= 1/10⁶ = 1 × 10^-6 m

(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

(iv) Size of a plant cell is 0.00001275 m
= (0.00001275 × 10⁵)/10⁵
= 1.275 × 10^-5 m

(v) Thickness of a thick paper = 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?

Answer

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
= 1.0008 × 10² mm

Chapterwise NCERT Solutions for 8th Class Maths

Chapter 1 – Rational Numbers
Chapter 2 – Linear Equations in One Variable
Chapter 3 – Understanding Quadrilaterals
Chapter 4 – Practical Geometry
Chapter 5 – Data Handling
Chapter 6 – Squares and Square Roots
Chapter 7 – Cubes and Cube Roots
Chapter 8 – Comparing Quantities
Chapter 9 – Algebraic Expressions and Identities
Chapter 10 – Visualizing Solid Shapes
Chapter 11 – Mensuration
Chapter 12 – Exponents and Powers
Chapter 13 – Direct and Inverse Proportions
Chapter 14 – Factorization
Chapter 15 – Introduction to Graphs
Chapter 16 – Playing with Numbers

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