Home » Class 7 Maths » NCERT Solutions for Class 7 Maths Chapter 13 – Exponents and Powers

# NCERT Solutions for Class 7 Maths Chapter 13 – Exponents and Powers

Exercise 13.1 : Solutions of Questions on Page Number : 252

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

(i) 26 = 2 x 2 x 2 x 2 x 2 x 2 = 64

(ii) 93 = 9 x 9 x 9 = 729

(iii) 112 = 11 x 11 = 121

(iv)54 = 5 x 5 x 5 x 5 = 625

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

(i) 6 x 6 x 6 x 6 = 64

(ii) t x t= t2

(iii) b x b x b x b = b4

(iv) 5 x 5 x 7 x 7 x 7 = 52 x 73

(v) 2 x 2 x a x a = 22 x a2

(vi) a x a x a x c x c x c x c x d = a3 c4 d

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

(i) 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 29

(ii) 343 = 7 x 7 x 7 = 73

(iii) 729 = 3 x 3 x 3 x 3 x 3 x 3 = 36

(iv) 3125 = 5 x 5 x 5 x 5 x 5 = 55

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

(i) 43 = 4 x 4 x 4 = 64
34 = 3 x 3 x 3 x 3 = 81
Therefore, 34 > 43

(ii) 53 = 5 x 5 x 5 =125
35 = 3 x 3 x 3 x 3 x 3 = 243
Therefore, 35 > 53

(iii) 28 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256
82 = 8 x 8 = 64
Therefore, 28 > 82

(iv)1002 or 2100
210 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1024
2100 = 1024 x 1024 x 1024 x 1024 x 1024 x 1024 x 1024 x 1024 x 1024 x 1024
1002 = 100 x 100 = 10000
Therefore, 2100 > 1002

(v) 210 and 102
210 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1024
102 = 10 x 10 = 100
Therefore, 210 > 102

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

(i) 648 = 2 x 2 x 2 x 3 x 3 x 3 x 3 = 23. 34

(ii) 405 = 3 x 3 x 3 x 3 x 5 = 34 . 5

(iii) 540 = 2 x 2 x 3 x 3 x 3 x 5 = 22. 33. 5

(iv) 3600 = 2 x 2 x 2 x 2 x 3 x 3 x 5 x 5 = 24. 32. 52

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

(i) 2 x 103 = 2 x 10 x 10 x 10 = 2 x 1000 = 2000

(ii) 72 x 22 = 7 x 7 x 2 x 2 = 49 x 4 = 196

(iii) 23 x 5 = 2 x 2 x 2 x 5 = 8 x 5 = 40

(iv) 3 x 44 = 3 x 4 x 4 x 4 x 4 = 3 x 256 = 768

(v) 0 x 102 = 0 x 10 x 10 = 0

(vi) 52 x 33 = 5 x 5 x 3 x 3 x 3 = 25 x 27 = 675

(vii) 24 x 32 = 2 x 2 x 2 x 2 x 3 x 3 = 16 x 9 = 144

(viii) 32 x 104 = 3 x 3 x 10 x 10 x 10 x 10 = 9 x 10000 = 90000

Q7 :  Simplify:
(i) (- 4)3 (ii) (- 3) x (- 2)3
(iii) (- 3)2 x (- 5)2 (iv)(- 2)3 x (-10)3

(i) (-4)3 = (-4) x (-4) x (-4) = -64

(ii) (-3) x (-2)3 = (-3) x (-2) x (-2) x (-2) = 24

(iii) (-3)2 x (-5)2 = (-3) x (-3) x (-5) x (-5) = 9 x 25 = 225

(iv) (-2)3 x (-10)3 = (-2) x (-2) x (-2) x (-10) x (-10) x (-10)
= (-8) x (-1000) = 8000

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

(i) 2.7 x 1012; 1.5 x 108
2.7 x 1012 > 1.5 x 108

(ii) 4 x 1014; 3 x 1017
3 x 1017 > 4 x 1014

Exercise 13.2 : Solutions of Questions on Page Number : 260

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 (v) (vi) 25 × 55
(vii) a4 × b4 (viii) (34)3
(ix)(220÷215)x23
(x) 8t÷82

(i) 32 x 34 x 38 = (3)2 + 4 + 8 (am x an = am+n)
= 314

(ii) 615 ÷ 610 = (6)15 – 10 (am ÷ an = am-n)
= 65
(iii) a3 x a2 = a(3 + 2) (am x an = am+n)
= a5
(iv) 7x + 72 = 7x + 2 (am x an = am+n)

(v) (52)3 ÷ 53
= 52 x 3 ÷ 53 (am)n = amn
= 56 ÷ 53
= 5(6 – 3) (am ÷ an = am-n)
= 53

(vi) 25 x 55
= (2 x 5)5 [am x bm = (a x b)m]
= 105

(vii) a4 x b4
= (ab)4 [am x bm = (a x b)m]

(viii) (34)3 = 34 x 3 = 312 (am)n = amn

(ix) (220÷215) x 23
= (220 – 15) x 23 (am ÷ an = am-n)
= 25 x 23
= (25 + 3) (am x an = am+n)
= 28

(x) 8t ÷ 82 = 8(t – 2) (am ÷ an = am-n)    (i) 10 x 1011 = 10011 (ii) 23 > 52
(iii) 23 x 32 = 65 (iv) 30 = (1000)0

(i) 10 x 1011 = 10011
L.H.S. = 10 x 1011 = 1011 + 1 (am x an = am+n)
= 1012
R.H.S. = 10011 = (10 x 10)11= (102)11
= 102 x 11 = 1022 (am)n = amn
As L.H.S. ≠ R.H.S.,
Therefore, the given statement is false.

(ii) 23 > 52
L.H.S. = 23 = 2 x 2 x 2 = 8
R.H.S. = 52 = 5 x 5 = 25
As 25 > 8,
Therefore, the given statement is false.

(iii) 23 x 32 = 65
L.H.S. = 23 x 32 = 2 x 2 x 2 x 3 x 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.

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

(i) 108 x 192
= (2 x 2 x 3 x 3 x 3) x (2 x 2 x 2 x 2 x 2 x 2 x 3)
= (22 x 33) x (26 x 3)
= 26 + 2 x 33 + 1 (am x an = am+n)
= 28 x 34

(ii) 270 = 2 x 3 x 3 x 3 x 5 = 2 x 33 x 5

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

(iv) 768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 = 28 x 3  Exercise 13.3 : Solutions of Questions on Page Number : 263

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

279404 = 2 x 105 + 7 x 104 + 9 x 103 + 4 x 102 + 0 x 101 + 4 x 100

3006194 = 3 x 106 + 0 x 105 + 0 x 104 + 6 x 103 + 1 x 102 + 9 x 101 + 4 x 100

2806196 = 2 x 106 + 8 x 105 + 0 x 104 + 6 x 103 + 1 x 102 + 9 x 101 + 6 x 100

120719 = 1 x 105 + 2 x 104 + 0 x 103 + 7 x 102 + 1 x 101 + 9 x 100

20068 = 2 x 104 + 0 x 103 + 0 x 102 + 6 x 101 + 8 x 100

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

(a) 8 x 104 + 6 x 103 + 0 x 102 + 4 x 101 + 5 x 100
= 86045

(b) 4 x 105 + 5 x 103 + 3 x 102 + 2 x 100
= 405302

(c) 3 x 104 + 7 x 102 + 5 x 100
= 30705

(d) 9 x 105 + 2 x 102 + 3 x 101
= 900230

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

(i) 50000000 = 5 x 107

(ii) 7000000 = 7 x 106

(iii) 3186500000 = 3.1865 x 109

(iv) 390878 = 3.90878 x 105

(v) 39087.8 = 3.90878 x 104

(vi) 3908.78 = 3.90878 x 103

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.

(a) 3.84 x 108m

(b) 3 x 108m/s

(c) 1.2756 x 107m

(d) 1.4 x 109m

(e) 1 x 1011stars

(f) 1.2 x 1010years

(g) 3 x 1020m

(h) 6.023 x 1022

(i) 1.353 x 109 cubic km

(j) 1.027 x 109