NCERT Solutions for Chapter 12 Exponents and Powers Class 8 Mathematics

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/am ]
= 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 [ ∵ am ÷ an = am-n]
= (-4)-3
= 1/(-4)3 [∵ a-m = 1/am ]

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

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

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

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

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

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

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

(v) [∵ (am)n = am×n]
[∵ a-m = 1/am]
= 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  [∵ (am)n = am×n]
= 2-3(-4) × 53
= 2-3+4 × 53 [∵ am ÷ an = am-n]
= 2×125
= 250

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

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

Answer

5m ÷ 5-3 = 55
= 5m-(-3) = 55 [∵ am ÷ an = am-n]
⇒ 5m+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/am]
={3 -4} = -1
(ii) 

[∵ am ÷ an = am-n]
[∵ a-m = 1/am]
= 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

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