NCERT Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.1, 11.2, 11.3 Solutions
In this page we have provided solutions of the Exercises 11.1, 11.2, 11.3 of NCERT Class 7 Maths Chapter 11 Exponents and Powers. These solutions are made by our team of expert teachers. Practice these solutions carefully for a better understanding of the topic which will help in scoring good marks in the examination.
Publishing Organisation |
NCERT |
Class |
7 |
Subject |
Mathematics |
Chapter |
11: Exponents and Powers |
Exercise |
11.1, 11.2, 11.3 |
Chapter – 11
Exercise – 11.1
(1) Find the value of:
(i) 26
Ans:- 2 × 2 × 2 × 2 × 2 × 2 = 64
(ii) 93
Ans:- 9 × 9 × 9 = 729
(iii) 112
Ans:- 11 × 11 = 121
(iv) 54
Ans:- 5 × 5 × 5 × 5 = 625
(2) Express the following in exponential form:
(i) 6 × 6 × 6 × 6
Ans:- 64
(ii) t × t
Ans:- t2
(iii) b × b × b × b
Ans:- b4
(iv) 5 × 5 × 7 × 7 × 7
Ans:- 52 × 73
(v) 2 × 2 × a × a
Ans:- 22 × a2
(vi) a × a × a × c × c × c × c × d
Ans:- a3c4d
(3) Express each of the following numbers using exponential notation:
(i) 512
Ans:- 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29
(ii) 343
Ans:- 7 × 7 × 7 = 73
(iii) 729
Ans:- 3 × 3 × 3 × 3 × 3 × 3 = 36
(iv) 3125
Ans:- 5 × 5 × 5 × 5 × 5 = 55
(4) Identify the greater number, wherever possible, in each of the following?
(i) 43 or 34
Ans:- 43 = 4 × 4 × 4 = 64
34 = 3 × 3 × 3 × 3 = 81
∵ 81 > 64
∴ 34 > 43
(ii) 53 or 35
Ans:- 53 = 5 × 5 × 5 = 125
35 = 3 × 3 × 3 × 3 × 3 = 243
∵ 243 > 125
∴ 35 > 53
(iii) 28 or 82
Ans:- 28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
82 = 8 × 8 = 64
∵ 256 > 64
∴ 28 > 82
(iv) 1002 or 2100
Ans:- 1002 = 100 × 100 = 10000
2100 = (210)10
= (2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2) = 210
= (1024)10 = [(1024)2]5
= (1024 × 1024)5 = (1048576)5
∵ 1048576 > 10000
∴ (1048576)5 > 10000
⇒ (210)10 > 10000
⇒ 2100 > (100)2
(v) 210 or 102
Ans:- 210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 1024
102 = 10 × 10 = 100
∵ 1024 > 100
∴ 210 > 102
(5) Express each of the following as product of powers of their prime factors:
(i) 648
Ans:- 2 × 2 × 2 × 3 × 3 × 3 × 3 = 23,34
(ii) 405
Ans:- 3 × 3 × 3 × 3 × 5 = 34.5
(iii) 540
Ans:- 2 × 2 × 3 × 3 × 3 × 5 = 22.33.5
(iv) 3,600
Ans:- 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 24.32.52
(6) Simplify:
(i) 2 × 103
Ans:- 2 × 10 × 10 × 10 = 2000
(ii) 72 × 22
Ans:- 7 × 7 × 2 × 2 = 49 × 4 = 196
(iii) 23 × 5
Ans:- 2 × 2 × 2 × 5 = 8 × 5 = 40
(iv) 3 × 44
Ans:- 3 × 4 × 4 × 4 × 4 = 12 × 64 = 768
(v) 0 × 102
Ans:- 0 × 10 × 10 = 0
(vi) 52 × 33
Ans:- 5 × 5 × 3 × 3 × 3 = 25 × 27 = 675
(vii) 24 × 32
Ans:- 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144
(viii) 32 × 104
Ans:- 3 × 3 × 10 × 10 × 10 × 10 = 90 × 1000 = 90000.
(7) Simplify:
(i) (-4)3
Ans:- (-4) × (-4) × (-4) = -64
(ii) (-3) × (-2)3
Ans:- (-3) × (-2) × (-2) × (-2) = 24
(iii) (-3)2 × (-5)2
Ans:- (-3) × (-3) × (-5) × (-5) = 9 × 25 = 225
(iv) (-2)3 × (-10)3
Ans:- (-2) × (-2) × (-2) × (-10) × (-10) × (-10) = (-8) × (-1000)
= 8000
(8) Compare the following numbers:
(i) 2.7 × 1012; 1.5 × 108
Ans:- = 15 × 107 (Contains 9 digits)
Clearly, 2.7 × 1012 > 1.5 × 108
(ii) 4 × 1014; 3 × 1017
Ans:- 3 × 1017 (Contains 18 digits)
4 × 1014 (Contains 15 digits)
Clearly, 3 × 1017 > 4 × 1014
Exercise – 11.2
(1) Using laws of exponents, simplify and write the answer in exponential form:
(i) 32 × 34 × 38
Ans:- (3)2 + 4 + 8 (am × an = am + n)
= 314
(ii) 615 ÷ 610
Ans:- (6)15 – 10 (am ÷ an = am-n)
= 65
(iii) a3 × a2
Ans:- a (3 + 2) (am × an = am + n)
(iv) 7x × 72
Ans:- 7x + 2 (am × a2 = am + n)
(v) (52)3 ÷ 53
Ans:- (52)3 ÷ 53
= 52 × 3 ÷ 53 (am)n = amn
= 56 ÷ 53
= 5 (6 – 3) (am ÷ an = am – n)
= 53
(vi) 25 × 55
Ans:- (2 × 5)3 [am × bm = (a × b)m]
= 105
(vii) a4 × b4
Ans:- (ab)4 [am × bm = (a × b)m]
(viii) (34)3
Ans:- 34 × 3
= 312 (am)n
= amn
(ix) (220 ÷ 215) × 23
Ans:- (220 – 15) × 23 (am ÷ an = am – n)
= 25 × 23
= (25 + 3) (am × an = am + n)
= 28
(x) 8t ÷ 82
Ans:- 8(t – 2) (am ÷ an = am – n)
(2) Simplify and express each of the following in exponential form:
(i) 23 × 34 × 4/3 × 32
Ans:- 23 × 34 × 22/3 × 25
∵ 4 = 2 ∵ 2 = 22
32 = 2 × 2 × 2 × 2 × 2 = 25
= 23+2 × 34/25 × 3 1 am × an = am + n
= 25 × 34/25 × 3 = 34/31
= 34-1 = 33
1 am + an = am – n
(ii) ( (52)3 × 54 ) ÷ 57
Ans:- (52 ×3 × 54) ÷ 57
1 (am)n = amn
= (56 × 54) ÷ 57
= 56+4 ÷ 57
= 510 ÷ 57
= 510-7
= 53
(iii) 254 ÷ 53
Ans:- (52)4 ÷ 53
= 52×4 ÷ 53 (am)n = amn
= 58 ÷ 53
= 58-3 (am ÷ an = am – n)
= 55
(iv) 3 × 72 × 118/21 × 113
Ans:- 3 × 72 × 118/3 × 7 × 113
= 72 × 118/7 × 113
= 72-1 × 118-3
1 am + an = am – n
71 × 115
7 × 115
(v) 37/34 × 33
Ans:- 37/34+3
= 37/37
= 37-7
= 30 = 1
(vi) 20 + 30 + 40
Ans:- 1 + 1 + 1 = 3
1a2 = 1
(vii) 20 × 30 × 40
Ans:- 1 × 1 × 1
= 1
(viii) (30 + 20) × 50
Ans:- (1 + 1) × 1
= 2 × 1 = 2
(ix) 28 × a5/43 × a3
Ans:- 28 × a5/(22)3 × a3
= 28 × a5/22×3 × a3
= 28 × a5/26 × a3
= 28.6 × a5-3
= 22 × a2
= (2 × a)
= (2a)2
(x) (a5/a3) × a8
Ans:- a5-3 × a8
= a2 × a3 = a2 + 8
= a10
∵ am/an = am-n
(xi) 45 × a8b3/45 × a5b2
Ans:- a8b3/a5b2
= a8-5 b8-2
= a3b1
= a3b
(xii) (23 × 2)2
Ans:- (23 × 21)2
= (23-1)2
= (24)2
= 24×2
= 28
(3) Say true or false and justify your answer:
(i) 10 × 1011 = 10011
Ans:- L.H.S = 10 × 1011 = 1011+1 (am × an = am+n)
= 1012
R.H.S = 10011 = (10 × 10)11 = (102)11
= 102×11 = 1022 (am)n = amn
As L.H.S ≠ R.H.S,
Therefore, the given Statement is false.
(ii) 23 > 52
Ans:- 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
Ans:- 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
Ans:- L.H.S = 30 = 1
R.H.S = 30 = 1
R.H.S = (1000)0 = 1 = L.H.S
Therefore, the given statement is true.
(4) Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
Ans:- (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
Ans:- 2 × 3 × 3 × 3 × 5 = 2 × 33 × 5
(iii) 729 × 64
Ans:- (3 × 3 × 3 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2)
= 36 × 26
(iv) 768
Ans:- 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 28 × 3
(5) Simplify:
(i) (25)2 × 73/83 × 7
Ans:- 25 ×2 × 73/(23)3 × 7
= (am)n = amn
= 210 × 73/23×3×7
= 210 × 73/29 × 7 / am/an = am-n
= 2010-9 × 73-1
= 21 × 72
= 2 × 7 × 7
= 98
(ii) 25 × 52 × t8/103 × t4
Ans:- 52 × 52 × t8/ (2 × 5)3 × t4
= 52 + 2 × t8/23 × 53 × t4
= 54 × t8/23 × 53 × t4
= 54-3 × t8-4/23
= 51 × t4/23
= 5t4/8
(iii) 35 × 105 × 25/57 × 65
Ans:- 105 = 2 × 5, 25 = 52, 6 = 2 × 3
= 35 × (2 × 5)5 × 52/57 × (2 × 3)5
= 35 × 25 × 55 × 52/57 × 25 × 35 / (ab)m = ambm
= 35 × 25 × 55 + 2/25 × 35 × 57
= 25 × 35 × 57/25 × 35 × 57
= 25-5 × 35-5 × 57-7
= 20 × 30 × 50
= 1 × 1 × 1 = 1
1a0 = 1
Exercise – 11.3
(1) Write the following numbers in the expanded forms: 279404, 3006194, 2806196, 120719, 20068.
Ans:-
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 × 103 + 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
(2) Find the number from each of the following expanded forms:
(a) 8 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
Ans:- 86045
(b) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100
Ans:- 405302
(c) 3 × 104 + 7 × 102 + 5 × 100
Ans:- 30705
(d) 9 × 105 + 2 × 102 + 3 × 101
Ans:- 900230
(3) Express the following numbers in standard form:
(i) 5,00,00,000
Ans:- 5 × 107
(ii) 70,00,000
Ans:- 7 × 106
(iii) 3,18,65,00,000
Ans:- 3.1865 × 109
(iv) 3,90,878
Ans:- 3.90878 × 105
(v) 39087.8
Ans:- 3.90878 × 104
(vi) 3908.78
Ans:- 3.90878 × 103
(4) Express the number appearing in the following statements in standard form.
(a) The distance between Earth and Moon is 384,000,000 m.
Ans:- 3.84 × 108m
(b) Speed of light in vacuum is 300,000,000 m/s.
Ans:- 3 × 108 m/s
(c) Diameter of the Earth is 1,27,56,000 m.
Ans:- 1.2756 × 107m
(d) Diameter of the Sun is 1,400,000,000 m.
Ans:- 1.4 × 109 m
(e) In a galaxy there are on an average 100,000,000,000 stars.
Ans:- 1 × 1011 Stars
(f) The universe is estimated to be about 12,000,000,000 years old.
Ans:- 1.2 × 1010 Years
(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.
Ans:- 3 × 1020 m
(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.
Ans:- 6.023 × 1022
(i) The earth has 1,353,000,000 cubic km of sea water.
Ans:- 1.353 × 109 Cubic Km
(j) The population of India was about 1,027,000,000 in March, 2001.
Ans:- 1.027 × 109
Chapter 10 Algebraic Expressions