NCERT Solutions for Class 8 Maths Chapter 10 Exponents and Powers

NCERT
SOLUTIONS
CLASS - 8TH

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

Class : 8th
Subject : Maths
Chapter : 12
Chapter Name : Exponents and Powers

Exercise 12.1

Q1 Evaluate
−5
(i) 3−2 (ii) (−4)−2 (iii) ( 12 )

(a−m = a1m )
−2
Answer. (i) 33 = 12 = 19
3

(a−m = a1m )
−2 1 1
(ii) (−4)(−4) = (−4) 2 = 16

−5
(iii) ( 12 ) = (2)1−5 = (2)5 = 2 × 2 × 2 × 2 × 2 = 32

Page : 197 , Block Name : EXERCISE 12.1

Q2 Simplify and express the result in power notation with positive exponent.
2
(i) (−4)5 ÷ (−4)8 (ii) ( 13 )
2
4
(ii) (−3)4 × ( 53 ) (iv) (3−7 ÷ 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 = a1m )

2
(ii) ( 13 ) = 1
= 16 ((am ) = aan )
n
2 3 2 2
(2 )

4
(iii) (−3)4 × ( 3 ) = (−1 × 3)4 × 54
4
5
3
4
= (−1)4 × 3 4
× 54 [(ab)m = am × bm ]
3
= (−1)4 × 54
4
[(−1)4 = 1]
=5

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

(iv) (3−7 ÷ 3−10 ) × 3−5 = (3−7−(−10) ) × 3−5 (am ÷ an = am−n )
= 33 × 3−5
= 33+(−5) (am × an = am+n )
= 3−2
= 12 (a−m = a1m )
3
1 1
×
(v)2−3 × (−7)−3 = 23 (−7)3

1
= [2×(−7)]3

1
= (−14) 3

Hint : (a−m = a1m )
Hint : [am × bm = (ab)m ]

Page : 197 , Block Name : EXERCISE 12.1

Q3 Find the value of.
(i) (3∘ + 4−1 ) × 22 (ii) (2−1 × 4−1 ) ÷ 2−2
−2 −2 −2
(iii) ( 12 ) + ( 13 ) + ( 14 )
−2 2
(iv) (3 −1
+4 −1
+5 ) −1 0
(v){( −2
3
) }

(30 + 4−1 ) × 22 = (1 + 14 ) × 22
Answer. (i)
= 54 × 4 = 5
Hint : (a0 = 1 and a−m = a1m )

(ii) (2−1 × 4−1 ) ÷ 2−2 = [2−1 × {(2)2 } ] ÷ 2−2
−1

= (2−1 × 2−2 ) ÷ 2−2 ((am ) = a−m )
n

= 2−1+(−2) ÷ 2−2 (am × an = am+n )
= 2−3 ÷ 2−2
= 2−3 − (−2) (am ÷ an = am−n )
−3+2 −1
(a−m = a1m )
=2 =2
= 12

−2 −2 −2 2 2 2
( 12 ) + ( 13 ) + ( 14 ) = ( 21 ) + ( 31 ) + ( 41 )
(iii) (∴ a−m = a1m )
= 22 + 32 + 42 = 4 + 9 + 16 = 29

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

0
(iv) (3−1 + 4−1 + 5−1 ) = ( 13 + 14 + 15 ) (a−m = a1m )
0

= 1 (a0 = 1)

−2 2 2 2
(v) {( 3 ) } = {( −2 ) } (a−m = a1m )
−2 3

2 m 2
= { (−2) 2 } [( b ) = abm ] = ( 94 ) = 81
2 m
3 a
16

Page : 197 , Block Name : EXERCISE 12.1

Q4 Evaluate
−1 3
(i) 8 −4
×5
( ii) (5−1 × 2−1 ) × 6−1
2

(a−m = a1n )
−1 3 4 3
Answer. (i) 8 −4
×5
= 2 ×5
1
2 8
4 3
= 2 ×53
=2 4−3 3
× 5 (am ÷ an = am−n )
2

(ii) (5−1 × 2−1 ) × 6−1 = ( 15 × 12 ) × 16 (a−m = a1m )
1
= 10 × 16 = 60
1

Page : 198 , Block Name : EXERCISE 12.1

Q5 Find the value of m for which 5m ÷ 5−3 = 55

5m ÷ 5−3 = 55
Answer. 5m−(−3) = 55 (am ÷ an = am−n )
5m+3 = 55
Since the powers have same bases on both sides, their respective exponents must
be equal.
m+3=5
m=5−3
m=2

Page : 198 , Block Name : EXERCISE 12.1

Q6 Evaluate

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

−1 −1 −1 −7 −4
(i) {( 13 ) − ( 14 ) } (ii) ( 58 ) × ( 85 )

−1 −1 −1 1 1 −1
Answer. (i) {( 13 ) − ( 14 ) } = {( 31 ) − ( 1 ) } (a−m = a1m )
4

1
= {3 − 4}−1 = (−1)−1 = −1 = −1

−7 −4 m
(ii) ( 8 ) × ( 85 ) = 5−7 × 8−4 [( ab ) = abm ]
−7 −4
5 m

8 5

= 87 × 54 (a−m = a1m )
7 4

5 8
87−4
= 7−4 (am ÷ an = am−n )
5
3
= 83 = 512
125
5

Page : 198 , Block Name : EXERCISE 12.1

Q7 Simplify
25×t−4
(i) −3 (t ≠ 0)
5 ×10×t−8

−5
×10−5 ×125
(ii) 3
5−7 ×6−5

2
25×t−4 −4
Answer. (i) −3−8
= −3 5 ×t −8
5 ×10×t 5 ×5×2×t
52 ×t−4
= −3+1 (a × a = am+n )
m n
5 ×2×t−5
2 −4
= −25 ×t −8
5 ×2×t
2−(−2) −4−(−8)
= 5 t
2
(am ÷ an = am−n )
4 4 4
= 5 2t = 625t 2

−5
×10−5 ×125 3−5 ×(2×5)−5 ×53
(ii) 3 =
5−7 ×6−5 5−7 ×(2×3)−5
−5 −5 −5 3
= 3 −7×2 ×5 ×5
[(a × b)m = am × bm ]
5 ×2−5 ×3−5
−5−(−5) −5−(−5) −5+3−(−7)
=3 ×2 ×5 m
(a ÷ an = am−a )
0 0 5 0
= 3 × 2 × 5 (a = 1)
= 55

Page : 198 , Block Name : EXERCISE 12.1

Exercise 12.2

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

Q1 Express the following numbers in standard form.
(i) 0.0000000000085
(ii) 0.00000000000942
(iii) 6020000000000000
(iv) 0.00000000837
(v) 31860000000

Answer.
(i) 0.0000000000085 = 8.5 × 10−12
(ii) 0.00000000000942 = 9.42 × 10−12
(iii) 6020000000000000 = 6.02 × 10−9
(iv) 0.00000000837 = 8.37 × 10−9
(v) 31860000000 = 3.186 × 1010

Page : 200 , Block Name : EXERCISE 12.2

Q2 Express the following numbers in usual form
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 = 0.00000302
(ii) 4.5 × 104 = 45000
(iii) 3 × 10−8 = 0.00000003
(iv) 1.0001 × 109 = 1000100000
(iv) 5.8 × 1012 = 5800000000000
(vi) 3.61492 × 106 = 3614920

Page : 200 , Block Name : EXERCISE 12.2

Q3 Express the number appearing in the following statements in standard form.
1
(i) 1 micron is equal to 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 of a thick paper is 0.07 mm

Answer. (i)

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

1
1000000
= 1 × 10−6
(ii) 0.000, 000, 000, 000, 000, 000, 16 = 1.6 × 10−19
(iii) 0.0000005 = 5 × 10−7
(iv) 0.00001275 = 1.275 × 10−5
(v) 0.07 = 7 × 10−2

Page : 200 , Block Name : EXERCISE 12.2

Q4 In a stack there are 5 books each of thickness 20mm and 5 paper sheets each of
thickness 0.016 mm. What is the total thickness of the stack.

Answer. Thickness Of each book = 20 mm
Hence, thickness of 5 books = (5 x 20) mm = 100 mm
Thickness Of each paper sheet = 0.016 mm
Hence, thickness Of 5 paper sheets = (5 x 0.016) mm = 0.080 mm
Total thickness Of the stack = Thickness Of 5 books + Thickness Of 5 paper sheets
= (100 + 0.080) mm
= 100.08 mm
= 1.0008 x 102 mm

Page : 200 , Block Name : EXERCISE 12.2

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