NCERT Solutions for Class 7 Maths Chapter 11 Exponents and Powers

NCERT
SOLUTIONS
CLASS - 7TH

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Book : Mathematics Ncert Solutions | Chapter-13 Maths

Class : 7th
Subject : Maths
Chapter : 13
Chapter Name : Exponents and Powers

Exercise 13.1

Q1 Find the value of:
(i) 26
(ii) 93
(iii) 112
(iv) 54

Answer. (i) 26 = 2 × 2 × 2 × 2 × 2 × 2 = 64
(ii) 93 = 9 × 9 × 9 = 729
(iii) 112 = 11 × 11 = 121
(iv) 54 = 5 × 5 × 5 × 5 = 625

Page : 252 , Block Name : Exercise 13.1

Q2 Express the following in exponential form:
(i) 6 × 6 × 6 × 6 (ii) t × t (iii) b × b × b × b (iv) 5 × 5× 7 × 7 × 7 (v) 2 × 2 × a × a (vi) a × a × a × c
×c×c×c×d

Answer. (i) 6 × 6 × 6 × 6 = 64
(ii) t × t = t2
(iii) b × b × b × b = b4
(iv) 5 × 5 × 7 × 7 × 7 = 52 × 73
(v) 2 × 2 × a × a = 22 × a2
(vi) a × a × a × c × c × c × c × d = a3 c4 d

Page : 252 , Block Name : Exercise 13.1

Q3 Express each of the following numbers using exponential notation: (i) 512 (ii) 343 (iii)
729 (iv) 3125

Answer. (i) 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29
(ii) 343 = 7 × 7 × 7 = 73

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Book : Mathematics Ncert Solutions | Chapter-13 Maths

(iii) 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36
(iv) 3125 = 5 × 5 × 5 × 5 × 5 = 55

Page : 253 , Block Name : Exercise 13.1

Q4 Identify the greater number, wherever possible, in each of the following?
(i) 43 or 34
(ii) 53 or 35
(iii) 28 or 82
(iv) 1002 or 2100
(v) 210 or 102

Answer. (i) 43 = 4 × 4 × 4 = 64
34 = 3 × 3 × 3 × 3 = 81
Therefore, 34 > 43
(ii) 53 = 5 × 5 × 5 = 125
35 = 3 × 3 × 3 × 3 × 3 = 243
Therefore, 35 > 53
(iii) 28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
82 = 8 × 8 = 64
Therefore, 28 > 82
(iv) 1002 or 2100
210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
2100 = 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024
1002 = 100 × 100 = 10000
Therefore, 2100 > 1002
(v) 210 and 102
210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
102 = 10 × 10 = 100
Therefore, 210 > 102

Page : 253 , Block Name : Exercise 13.1

Q5 Express each of the following as product of powers of their prime factors: (i) 648 (ii)
405 (iii) 540 (iv) 3,600

Answer. (i) 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 23 ⋅ 34
(ii) 405 = 3 × 3 × 3 × 3 × 5 = 34 ⋅ 5
(iii) 540 = 2 × 2 × 3 × 3 × 3 × 5 = 22 ⋅ 33 ⋅ 5
(iv) 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 24 ⋅ 32 ⋅ 52

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Book : Mathematics Ncert Solutions | Chapter-13 Maths

Page : 253 , Block Name : Exercise 13.1

Q6 Simplify:
(i) 2 × 103
(ii) 72 × 22
(iii) 23 × 5
(iv) 3 × 44
(v) 0 × 102
(vi) 52 × 33
(vii) 24 × 32
(viii) 32 × 104

Answer. (i) 2 × 103 = 2 × 10 × 10 × 10 = 2 × 1000 = 2000
(ii) 72 × 22 = 7 × 7 × 2 × 2 = 49 × 4 = 196
(iii) 23 × 5 = 2 × 2 × 2 × 5 = 8 × 5 = 40
(iv) 3 × 44 = 3 × 4 × 4 × 4 × 4 = 3 × 256 = 768
(v) 0 × 102 = 0 × 10 × 10 = 0
(vi) 52 × 33 = 5 × 5 × 3 × 3 × 3 = 25 × 27 = 675
(vii) 24 × 32 = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144
(viii) 32 × 104 = 3 × 3 × 10 × 10 × 10 × 10 = 9 × 10000 = 90000

Page : 253 , Block Name : Exercise 13.1

Q7 Simplify:
(i) (−4)3
(ii) (−3) × (−2)3
(iii) (−3)2 × (−5)2
(iv) (−2)3 × (−10)3

Answer. (i) (−4)3 = (−4) × (−4) × (−4) = −64
(ii) (−3) × (−2)3 = (−3) × (−2) × (−2) × (−2) = 24
(iii) (−3)2 × (−5)2 = (−3) × (−3) × (−5) × (−5) = 9 × 25 = 225
(iv) (−2)3 × (−10)3 = (−2) × (−2) × (−2) × (−10) × (−10) × (−10)
= (−8) × (−1000) = 8000

Page : 253 , Block Name : Exercise 13.1

Q8 Compare the following numbers:
(i) 2.7 × 1012 ; 1.5 × 108
(ii) 4 × 1014 ; 3 × 1017

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Book : Mathematics Ncert Solutions | Chapter-13 Maths

Answer. (i) 2.7 × 1012 ; 1.5 × 108
2.7 × 1012 > 1.5 × 108
(ii) 4 × 1014 ; 3 × 1017
3 × 1017 > 4 × 1014

Page : 253 , Block Name : Exercise 13.1

Exercise 13.2

Q1 Using laws of exponents, simplify and write the answer in exponential form:
(i) 32 × 34 × 38
(ii) 615 ÷ 610
(iii) a3 × a2
(iv) 7x × 72
3
(v) (52 ) ÷ 53
(vi) 25 × 55
(vii) a4 × b4
3
(viii) (34 )
(ix) (220 ÷ 215 ) × 23
(x) 8t ÷ 82

(i)32 × 34 × 38 = (3)2+4+8 (am × an = am+n )
Answer.
= 314
(i1)615 ÷ 610 = (6)15−10 (am ÷ an = am−n )
= 65
(ili) a3 × a(3+2) (am × an = am+n )
= a5
( iv) 7x + 72 = 7x+2 (am × an = am+n )
( vi) 25 × 55
= (2 × 5)5 [am × bm = (a × b)m ]
= 105
(vii) a4 × b4
= (ab)4 [am × bm = (a × b)m ]

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Book : Mathematics Ncert Solutions | Chapter-13 Maths

3
(34 ) = 34×3 = 312 (am ) = amn
n

(1x) (220 ÷ 215 ) × 23
= (220−15 ) × 23 (am ÷ an = am−n )
= 25 × 23
= (25+3 ) (am × an = am+n )
= 28
(x)gt ÷ 82 = g(t−2) (am ÷ an = am−n )

Page : 260 , Block Name : Exercise 13.2

Q2 Simplify and express each of the following in exponential form:
3 4
(i) 2 3×32
×3 ×4

(ii) ((52 ) × 54 ) ÷ 5?
3

(iii) 254 ÷ 53
2 8
(iv) 3×7 ×113
21×11
37
(v) 4 3
3 ×3
(vi) 2 + 3 + 40
0 0

(vii) 20 × 30 × 40
(viii) (30 + 20 ) × 50
8 5
(ix) 23 ×a3
4 ×a
(x) ( aa3 ) × a8
5

5 8 3
(xi) 45 ×a5 b2
4 ×a b
3 2
(xii) (2 × 2)

3 4 3 4 3 4 2
Answer. (i) 2 3×32
×3 ×4 2 ×3 ×2×2
= 3×2×2×2×2×2 = 2 ×3 ×2
5
3×2
3+2 4
= 2 ×35
3×2
(a × a = am+a )
m n
5 4
= 2 ×35
3×2
= 25−5 × 34−1
(am + an = am−n )
= 20 33 = 1 × 33 = 33
(ii) [(52 ) × 54 ] ÷ 57
3

= [52×3 × 54 ] ÷ 57 (am ) = amn
n

= [56 × 54 ] ÷ 57

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Book : Mathematics Ncert Solutions | Chapter-13 Maths

= [56+4 ] ÷ 57 (am × an = am+n )
= 510 ÷ 57
= 510−7 (am ÷ an = am−n )
= 53
(iii) 254 ÷ 53 = (5 × 5)4 ÷ 53
4
= (52 ) ÷ 53
= 52×4 ÷ 53 (am ) = amn
n

= 58 ÷ 53
= 58−3 (am ÷ an = am−n )
= 55
2 8 2 8
(iv) 3×7 ×113 = 3×7 ×113
21×11 3×7×11
1−1 2−1
=3 ×7 × 118−3
(a + a = a )
m n n−n

= 30 × 71 × 115
= 1 × 7 × 115 = 7 × 115
7 7
(v) 43 3 = 34+3
3 ×3 3
(am × an = am+n )
7
= 37 = 37−7
3
(a × an = am−n )
m

= 3∘ = 1
(vi) 20 + 30 + 40 = 1 + 1 + 1 = 3
(vii) 20 × 30 × 40 = 1 × 1 × 1 = 1
(viii) (30 + 20 ) × 50 = (1 + 1) × 1 = 2
8 5
28 ×a5 28 ×a5
(ix) 23 ×a3 = 3 3 =
4 ×a (2×2) ×a 3
(22 ) ×a3
23 ×a5
=
(22×3 )×a3
m n
[(a ) = a ] m
8 5
= 26 ×a3
2 ×a
= 28−6 × a5−3
(am × an = am−n )
= 22 × a2 = (2 × a)2
[am × bm = (a × b)m ]
= (2a)2
(x) ( aa3 ) × a8 = a5−3 × a8
5

(am ÷ an = am−n )
= a2 × a8
= a2+8 = a10
(am × an = am+n )
5 8 3
(xi) 45 ×a5 b2 = 45−5 × a8−5 × b3−2
4 ×a b

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Book : Mathematics Ncert Solutions | Chapter-13 Maths

(am ÷ an = am−n )
= 40 × a3 × b1 = 1 × a3 × b = a3 b
2 2
(xii) (23 × 2) = (23+1 ) (am × an = am+n )
2
= (24 ) = 24×2 (am ) = amn
n

= 28

Page : 260 , Block Name : Exercise 13.2

Q3 Say true or false and justify your answer:
(i) 10 × 1011 = 10011
(ii) 23 > 52
(iii) 23 × 32 = 65
(iv) 30 = (1000)0

Answer. (i) 10 × 1011 = 10011
L.H.S = 10 × 1011 = 1011+1 (am × an = am+n )
= 1012
11
R.H.S = 10011 = (10 × 10)11 = (102 )
= 102×11 = 1022 (am ) = amn
n

As L.H.S ≠ R.H.S.,
Therefore, the given statement is false.
(ii) 23 > 52
L.H.S = 23 = 2 × 2 × 2 = 8
R.H.S = 52 = 5 × 5 = 25
As 25 > 8
Therefore, the given statement is false.
(iii) 23 × 32 = 65
L.H.S = 23 × 32 = 2 × 2 × 2 × 3 × 3 = 72
R.H.S = 65 = 7776
As L.H.S ≠ R.H.S.,
Therefore, the given statement is false.
(iv) 30 = (1000)0
L.H.S = = 30 = 1
R.H.S = (1000)0 = 1 = L.H.S
Therefore, the given statement is true.

Page : 260 , Block Name : Exercise 13.2

Q4 Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192 (ii) 270 (iii) 729 × 64 (iv) 768

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Book : Mathematics Ncert Solutions | Chapter-13 Maths

Answer. (i) 108 × 192
= (2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2 × 3)
= (22 × 33 ) × (26 × 3)
= 26+2 × 33+1 (am × an = am+n )
= 28 × 34
(ii) 270 = 2 × 3 × 3 × 3 × 5 = 2 × 33 × 5
(iii) 729 × 64 = (3 × 3 × 3 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2)
= 36 × 26
(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 28 × 3

Page : 261 , Block Name : Exercise 13.2

Q5 Simplify:
2
(25 ) ×73
(i)
83 ×7
2
×t8
(ii) 25×53
10 ×t4
5 5
(iii) 3 ×10 ×25
5 ×65
7

2
(25 ) ×73 5⋅2
2 ×7 3 n
Answer. (i) 3 = (2×2×2) 3 ×7 [(a∗ ) = a−m ]
8 ×7
210 ×73
10 3
= 3 3 = 23×3×7
n
[(am ) = amn ]
(2 ) ×7 2 ×7
10 3
= 2 9×7 = 210−9 × 73−1 (am ÷ an = am−n )
2 ×7
1 2
= 2 × 7 = 2 × 7 × 7 = 98
2 2
×t8 ×t8
(ii) 25×5
3 = 5×5×5 (a × b)m = (am × bm )
10 ×t 4 (5×2) ×t4
3

51+1+2 ×t8
= 3 3 4 (am × an = am+n )
5 ×2 ×t
4 8 4−3 8−4
= 35 ×t3 4 = 5 ×t 3
(am ÷ an = am−n )
5 ×2 ×t 2
51 ×t4 5t4
= 2×2×2 = 8
35 ×105 ×25 35 ×(2×5)5 ×5×5
(iii) =
57 ×65 57 ×25 ×35
3 ×25 ×55 ×52
5
= 7 5 5 (a × b)m = (am × bm )
5 ×2 ×3
5 5
×55+2
= 3 ×2 (am × an = am+n )
5 ×25 ×35
7
5 5 7
= 37 ×25 ×55
5 ×2 ×3
5−5 5−5 7−7
=3 ×2 ×5 (a ÷ a = a m n m−n
)
0 0 0
=3 ×2 ×5 =1×1×1=1

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Book : Mathematics Ncert Solutions | Chapter-13 Maths

Page : 261 , Block Name : Exercise 13.2

Exercise 13.3

Q1 Write the following numbers in the expanded forms:
279404, 3006194, 2806196, 120719, 20068

Answer. 279404 = 2 × 105 + 7 × 104 + 9 × 103 + 4 × 102 + 0 × 101 + 4 × 100
3006194 = 3 × 106 + 0 × 105 + 0 × 104 + 6 × 103 + 1 × 102 + 9 × 101 + 4 × 100
2806196 = 2 × 106 + 8 × 105 + 0 × 104 + 6 × 103 + 1 × 102 + 9 × 101 + 6 × 100
120719 = 1 × 105 + 2 × 104 + 0 × 103 + 7 × 102 + 1 × 101 + 9 × 100
20068 = 2 × 104 + 0 × 103 + 0 × 102 + 6 × 101 + 8 × 100

Page : 263 , Block Name : Exercise 13.3

Q2 Find the number from each of the following expanded forms:
(a) 8 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
(b) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100
(c) 3 × 104 + 7 × 102 + 5 × 100
(d) 9 × 105 + 2 × 102 + 3 × 101

Answer. (a) 8 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
= 86045
(b) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100
= 405302
(c) 3 × 104 + 7 × 102 + 5 × 100
= 30705
(d) 9 × 105 + 2 × 102 + 3 × 101
= 900230

Page : 263 , Block Name : Exercise 13.3

Q3 Express the following numbers in standard form:
(i) 5,00,00,000 (ii) 70,00,000 (iii) 3,18,65,00,000 (iv) 3,90,878 (v) 39087.8 (vi) 3908.78

Answer. (i) 50000000 = 5 × 107
(ii) 7000000 = 7 × 106
(iii) 3186500000 = 3.1865 × 109

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Book : Mathematics Ncert Solutions | Chapter-13 Maths

(iv) 390878 = 3.90878 × 105
(v) 39087.8 = 3.90878 × 104
(vi) 3908.78 = 3.90878 × 103

Page : 263 , Block Name : Exercise 13.3

Q4 Express the number appearing in the following statements in standard form.
(a) The distance between Earth and Moon is 384,000,000 m. (b) Speed of light in vacuum
is 300,000,000 m/s. (c) Diameter of the Earth is 1,27,56,000 m. (d) Diameter of the Sun is
1,400,000,000 m. (e) In a galaxy there are on an average 100,000,000,000 stars. (f) The
universe is estimated to be about 12,000,000,000 years old. (g) The distance of the Sun
from the centre of the Milky Way Galaxy is estimated to be 300,000,000,000,000,000,000
m. (h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water
weighing 1.8 gm. (i) The earth has 1,353,000,000 cubic km of sea water. (j) The
population of India was about 1,027,000,000 in March, 2001.

Answer. (i) 3.84 × 108 m
(ii) 3 × 108 m/s
(iii) 1.2756 × 107 m
(iv) 1.4 × 109 m
(v) 1 × 1011
(vi) 1.2 × 1010
(vii) 3 × 1020 m
(viii) 6.023 × 1022
(ix) 1.353 × 109 cubic km
(x) 1.027 × 109

Page : 263 , Block Name : Exercise 13.3

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