Class 7 Maths Chapter 13 Exponents and Powers NCERT Solutions

Class 7 Maths Chapter 13 Exponents and Powers NCERT Solutions

• Class 7 Maths Exponents and Powers Exercise 13.1 
• Class 7 Maths Exponents and Powers Exercise 13.2
• Class 7 Maths Exponents and Powers Exercise 13.3

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.1
Ex 13.1 Class 7 Maths Question 1.
Find the value of
(i) 26
(ii) 93
(iii) 112
(iv) 54
Solution:
(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

Ex 13.1 Class 7 Maths Question 2.
Exress 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
Solution:
(i) 6 × 6 × 6 × 6 = 63
(ii) t × t = t2
(iii) b × b × b × b = b4
(iv) 5 × 5× 7 × 7 × 7 = 52 × 73 = 52 · 73
(v) 2 × 2 × a × a = 22 × a2 = 22 · a2
(vi) a × a ×a × c × c × c × c × d = a3 × c4 × d = a3 · c4 · d

Ex 13.1 Class 7 Maths Question 3.
Express each of the following numbers using exponential notation:
(i) 512
(ii) 343
(iii) 729
(iv) 3125
Solution:

Ex 13.1 Class 7 Maths Question 4.
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
Solution:
(i) 43 or 34
43 = 4 × 4 × 4 = 64,
34 = 3 × 3 × 3 × 3 = 81
Since 81 > 64
∴ 34 is greater than 43.

(ii) 53 or 35
53 = 5 × 5 × 5 = 125
35 = 3 × 3 × 3 × 3 × 3 = 243
Since 243 > 125
∴ 35 is greater than 53.

(iii) 28 or 82
28 =2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
82 = 8 × 8 = 64
Since 256 > 64
∴ 28 is greater than 28.

(iv) 1002 or 2100
1002 = 100 × 100 = 10000
2100 = 2 × 2 × 2 × … 100 times
Here 2 × 2 × 2 ×2 × 2 × 2 × 2 ×2 × 2 × 2 × 2 × 2 × 2 × 2 = 214 = 16384
Since 16384 > 10,000
∴ 2100 is greater than 1002.

(v) 210 or 102
210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
102 = 10 × 10 = 100
Since 1024 > 100
∴ 210 is greater than 102.

Ex 13.1 Class 7 Maths Question 5.
Express each of the following as the product of powers of their prime
(i) 648
(ii) 405
(iii) 540
(iv) 3600
Solution:

Ex 13.1 Class 7 Maths Question 6.
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
Solution:
(i) 2 × 103 = 2 × 10 × 10 × 10 = = 2000
(ii) 72 × 22 = = 7 × 7 × 2 × 2 = 196
(iii) 23 × 5 = 2 × 2 × 2 × 5 = 40
(iv) 3 × 44 = 3 × 4 × 4 × 4 × 4 = 768
(v) 0 × 102 = 0 × 10 × 10 = = 0
(vi) 52 × 33 = 5 × 5 × 3 × 3 × 3 = 675
(vii) 24 × 32 = 2 × 2 × 2 × 2 × 3 × 3 = 144
(viii) 32 × 104 = 3 × 3 × 10 × 10 × 10 × 10 = 90000

Ex 13.1 Class 7 Maths Question 7.
Simplify:
(i) (-4)3
(ii) (-3) × (-2)3
(iii) (-3)2 × (-5)2
(iv) (-2)3 × (-10)3
Solution:
(i) (-4)2 = (-4) × (-4) × (-4) = -64 [∵ (-a)odd number = -aodd number]
(ii) (-3) × (-2)3 = (-3) × (-2) × (-2) × (-2)
= (-3) × (-8) = 24
(iii) (-3)2 × (-5)2 = [(-3) × (-5)]2
= 152 = 225 [∵ am × bm = (ab)m)
(iv) (-2)3 × (-10)3 = [(-2) × (-10)]3
= 202 = 8000 [∵ am × bm = (ab)m]

Ex 13.1 Class 7 Maths Question 8.
Compare the following:
(i) 2.7 × 1012; 1.5 × 108
(ii) 4 × 1014; 3 × 1014
Solution:
(i) 2.7 × 1012; 1.5 × 108
Here, 1012 > 108
∴ 2.7 × 1012> 1.5 × 108
(ii) 4 × 1014; 3 × 1017
Here, 1017 > 1014
∴ 4 × 1014 < 3 × 1017

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.2
Ex 13.2 Class 7 Maths Question 1.
Using laws of e×ponents, simplify and write the answer in e×ponential form:
(i) 32 × 34 × 38
(ii) 615 ÷ 610
(iii) a3 × a2
(iv) 7x × 72
(v) (52)3 ÷ 53
(vi) 25 × 55
(vii) a4 × b4
(viii) (34)3
(ix) (220 ÷ 215) × 23
(x) 8t ÷ 82
Solution:
(i) 32 × 34 × 38 = 32+4+8 = 314 [am ÷ an = am+n]
(ii) 615 ÷ 610 = 615-10 = 65 [am ÷ an = am-n]
(iii) a3 × a2 = a3+2 = a5 [am × an = am+n]
(iv) 7x × 72 = 7x+2 [am × an = am+n]
(v) (52)3 ÷ 53 = 52×3 ÷ 53 = 56 ÷ 53 = 56-3 = 53 [(a3)n = amn, am ÷ an = am-n]
(vi) 25 × 55 = (2 × 5)5 = 105 [am × bm = (ab)m]
(vii) a4 × b4 = (ab)4 [am × bm = (ab)4]
(ix) (220 ÷ 215) × 23 = 220-15 × 23
=25 × 23 = 25+3 = 28
(x) 8t ÷ 82 = 8t-2 [am ÷ an = am-n]

Ex 13.2 Class 7 Maths Question 2.
Simplify and express each of the following in exponential form:

Solution:

Ex 13.2 Class 7 Maths Question 3.
Say true or false and justify your answer:
(i) 10 × 1011 = 10011
(ii) 23 > 52
(iii) 23 × 32 = 65
(iv) 320 = (1000)0
Solution:
(i) 10 × 1011 = 101+11 = 1012
RHS = 10011 = (102)11 = 1022
1012 ≠ 1022
∴ Statement is false.

(ii) 23 > 52
LHS = 23 = 8
RHS = 522 = 25
8 < 25
∴ 23 < 52
Thus, the statement is false.

(iii) 23 × 32 = 65
LHS = 233 × 32 = 8 × 9 = 72
RHS = 65 = 6 × 6 × 6 × 6 × 6 = 7776
∴ 72 ≠ 7776
∴ The statement is false.

(iv) 30 = (1000)0
⇒ 1 = 1 True [∵ a0 = 1]

Ex 13.2 Class 7 Maths Question 4.
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
Solution:
(i) 108 × 192 = 2 × 2 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
=28 × 34

(iii) 729 × 64 = 3 × 3 × 3 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2
=36 × 26

Ex 13.2 Class 7 Maths Question 5.
Simplify:

Solution:

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.3
Ex 13.3 Class 7 Maths Question 1.
Write the following numbers in the e×panded forms:
279404, 3006194, 2806196, 120719, 20068
Solution:
(i) 279404 = 2 × 100000 + 7 × 10000 + 9 × 1000 + 4 × 100 + 0 × 10 + 4
= 2 × 105 + 7 × 104 + 9 × 1032 + 4 × 102 + 0 × 101 + 4 × 100
(ii) 3006194 = 3 × 1000000 + 0 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 4
= 3 × 106 + 0 × 105 + 0 × 104 + 6 × 103 + 1 × 102 + 9 × 101 + 4 × 100
(iii) 2806196 = 2 × 1000000 + 8 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 6
= 2 × 106 + 8 × 105 + 0 × 104 + 6 × 103 + 1 × 102 + 9 × 101 + 6 × 100
(iv) 120719 = 1 × 100000 + 2 × 10000 + 0 × 1000 + 7 × 100 + 1 × 10 + 9
= 1 × 105 + 2 × 104 + 0 × 103 + 7 × 102 + 1 × 101 + 9 × 100
(v) 20068 = 2 × 10000 + 0 × 1000 + 0 × 100 + 6 × 10 + 8
= 2 × 104 + 0 × 103 + 0 × 102 + 6 × 101 + 8 × 100

Ex 13.3 Class 7 Maths Question 2.
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
Solution:
(a) 8 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
= 8 × 10000 + 6 × 1000 + 0 × 100 + 4 × 10 + 5 × 1
= 80000 + 6000 + 0 + 40 + 5 = 86045

(b) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100
= 4 × 100000 + 5 × 1000 + 3 × 100 + 2 × 1
= 400000 + 5000 + 300 + 2 = 405302

(c) 3 × 104 + 7 × 102 + 5 × 100
= 3 × 10000 + 7 × 100 + 5 × 1
= 30000 + 700 + 5 = 30705

(d) 9 × 105 + 2 × 102 + 3 × 101
= 9 × 100000 + 2 × 100 + 3 × 10
= 900000 + 200 + 30 = 900230

Ex 13.3 Class 7 Maths Question 3.
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
Solution:
(i) 5,00,00,000 = 5 × 1077
(ii) 70,00,000 = 7 × 106
(iii) 3,18,65,00,000 = 3.1865 × 109
(iv) 3,90,878 = 3.90878 × 105
(v) 39087.8 = 3.90878 × 104
(vi) 3908.7 8 = 3.90878 × 103

Ex 13.3 Class 7 Maths Question 4.
Express the number appearing in the following statements in standard form:
(a) The distance between Earth and Moon is 384.0. 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 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.
Solution:
(a) 384,000,000 m = 3.84 × 108 m
(b) 300,000,000 m/s = 3 × 108 m/s
(c) 1,27,56,000 m = 1.2756 × 1072 m
(d) 1,400,000,000 m = 1.4 × 109 m
(e) 100,000,000,000 stars = 1 × 1011 stars
(f) 12,000,000,000 years old = 1.2 × 1010 years old
(g) 300,000,000,000,000,000,000 m = 3 × 1020 m
(h) 60, 230, 000, 000, 000, 000, 000, 000 molecules = 6.023 × 1022 molecules
(i) 1,353,000,000 cubic km = 1.353 × 109 cubic km
(j) 1,0,27,000,000 = 1.027 × 109

Exponents and Powers Class 7 Extra Questions Very Short Answer Type

Question 1.
Express 343 as a power of 7.
Solution:
We have 343 = 7 × 7 × 7 = 73
Thus, 343 = 73

Question 2.
Which is greater 32 or 23?
Solution:
We have 32 = 3 × 3 = 9
23 = 2 × 2 × 2 = 8
Since 9 > 8
Thus, 32 > 23

Question 3.
Express the following number as a powers of prime factors:
(i) 144
(ii) 225
Solution:
(i) We have
144 = 2 × 2 × 2 × 2 × 3 × 3 = 24 × 32
Thus, 144 = 24 × 32

(ii) We have
225 = 3 × 3 × 5 × 5 = 32 × 52
Thus, 225 = 32 × 52

Question 4.
Find the value of:
(i) (-1)1000
(ii) (1)250
(iii) (-1)121
(iv) (10000)0
Solution:
(i) (-1)1000 = 1 [∵ (-1)even number = 1]
(ii) (1)250 = 1 [∵ (1)even number = 1]
(iii) (-1)121 = -1 [∵ (-1)odd number = -1]
(iv) (10000)0 = 1 [∵ a0 = 1]

Question 5.
Express the following in exponential form:
(i) 5 × 5 × 5 × 5 × 5
(ii) 4 × 4 × 4 × 5 × 5 × 5
(iii) (-1) × (-1) × (-1) × (-1) × (-1)
(iv) a × a × a × b × c × c × c × d × d
Solution:
(i) 5 × 5 × 5 × 5 × 5 = (5)5
(ii) 4 × 4 × 4 × 5 × 5 × 5 = 43 × 53
(iii) (-1) × (-1) × (-1) × (-1) × (-1) = (-1)5
(iv) a × a × a × b × c × c × c × d × d = a3b1c3d2

Question 6.
Express each of the following as product of powers of their prime factors:
(i) 405
(ii) 504
(iii) 500
Solution:
(i) We have
405 = 3 × 3 × 3 × 3 × 5 = 34 × 51
Thus, 405 = 34 × 51

(ii) We have
504 = 2 × 2 × 2 × 3 × 3 × 7 = 23 × 32 × 71
Thus, 504 = 23 × 32 × 71

(iii) We have
500 = 2 × 2 × 5 × 5 × 5 = 22 × 53
Thus, 500 = 22 × 53

Question 7.
Simplify the following and write in exponential form:
(i) (52)3
(ii) (23)3
(iii) (ab)c
(iv) [(5)2]2
Solution:
(i) (52)3 = 52×3 = 56
(ii) (23)3 = 23×3 = 29
(iii) (ab)c = ab×c = abc
(iv) [(5)2]2 = 52×2 = 54

Question 8.
Verify the following:

Solution:

Question 9.
Simplify:

Solution:

Question 10.
Simplify and write in exponential form:

Solution:

Exponents and Powers Class 7 Extra Questions Short Answer Type

Question 11.
Express each of the following as a product of prime factors is the exponential form:
(i) 729 × 125
(ii) 384 × 147
Solution:
(i) 729 × 125 = 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5 = 36 × 53
Thus, 729 × 125 = 36 × 53

(ii) 384 × 147 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7 = 27 × 32 × 72
Thus, 384 × 147 = 27 × 32 × 72

Question 12.
Simplify the following:
(i) 103 × 90 + 33 × 2 + 70
(ii) 63 × 70 + (-3)4 – 90
Solution:
(i) 103 × 90 + 33 × 2 + 70
= 1000 + 54 + 1
= 1055
(ii) 63 × 70 + (-3)4 – 90
= 216 × 1 + 81 – 1
= 216 + 80
= 296

Question 13.
Write the following in expanded form:
(i) 70,824
(ii) 1,69,835
Solution:
(i) 70,824
= 7 × 10000 + 0 × 1000 + 8 × 100 + 2 × 10 + 4 × 100
= 7 × 104 + 8 × 102 + 2 × 101 + 4 × 100
(ii) 1,69,835
= 1 × 100000 + 6 × 10000 + 9 × 1000 + 8 × 100 + 3 × 10 + 5 × 100
= 1 × 105 + 6 × 104 + 9 × 103 + 8 × 102 + 3 × 101 + 5 × 100

Question 14.
Find the number from each of the expanded form:
(i) 7 × 108 + 3 × 105 + 7 × 102 + 6 × 101 + 9
(ii) 4 × 107 + 6 × 103 + 5
Solution:
(i) 7 × 108 + 3 × 105 + 7 × 102 + 6 × 101 + 9
= 7 × 100000000 + 3 × 100000 + 7 × 100 + 6 × 10 + 9
= 700000000 + 300000 + 700 + 60 + 9
= 700300769
(ii) 4 × 107 + 6 × 103 + 5
= 4 × 10000000 + 6 × 1000 + 5
= 40000000 + 6000 + 5
= 40006005

Question 15.
Find the value of k in each of the following:

Solution:

Question 16.
Find the value of
(a) 30 ÷ 40
(b) (80 – 20) ÷ (80 + 20)
(c) (20 + 30 + 40) – (40 – 30 – 20)
Solution:
(a) We have 30 ÷ 40 = 1 ÷ 1 = 1 [∵ a0 = 1]
(6) (80 – 20) ÷ (80 + 20) = (1 – 1) ÷ (1 + 1) = 0 ÷ 2 = 0
(c) (20 + 30 + 40) – (40 – 30 – 20)
= (1 + 1 + 1) – (1 – 1 – 1) [∵ a0 = 1]
= 3 – 1
= 2

Question 17.
Express the following in standard form:
(i) 8,19,00,000
(ii) 5,94,00,00,00,000
(iii) 6892.25
Solution:
(i) 8,19,00,000 = 8.19 × 107
(ii) 5,94,00,00,00,000 = 5.94 × 1011
(iii) 6892.25 = 6.89225 × 103

Question 18.
Evaluate:

Solution:

Exponents and Powers Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 19.
Find the value of x, if

Solution:

Question 20.

Solution: