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Page No 252:

Question 1:

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 No 252:

Question 2:

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 = a3c4d



Page No 253:

Question 3:

Express 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

(iii) 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36

(iv) 3125 = 5 × 5 × 5 × 5 × 5 = 55

Page No 253:

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

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 No 253:

Question 5:

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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Video Solution for Exponents and Powers (Page: 253 , Q.No.: 5)

NCERT Solution for Class 7 math - Exponents and Powers 253 , Question 5

Page No 253:

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

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 No 253:

Question 7:

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 No 253:

Question 8:

Compare the following numbers:

(i) 2.7 × 1012; 1.5 × 108

(ii) 4 × 1014; 3 × 1017

Answer:

(i) 2.7 × 1012; 1.5 × 108

2.7 × 1012 > 1.5 × 108

(ii) 4 × 1014; 3 × 1017

3 × 1017 > 4 × 1014
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Video Solution for Exponents and Powers (Page: 253 , Q.No.: 8)

NCERT Solution for Class 7 math - Exponents and Powers 253 , Question 8



Page No 260:

Question 1:

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) (x) 8t ÷ 82

Answer:

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

= 314

(ii) 615 ÷ 610 = (6)15 − 10 (am ÷ an = amn)

= 65

(iii) a3 × a2 = a(3 + 2) (am × an = am+n)

= a5

(iv) 7x + 72 = 7x + 2 (am × an = am+n)

(v) (52)3 ÷ 53

= 52 × 3 ÷ 53 (am)n = amn

= 56 ÷ 53

= 5(6 − 3) (am ÷ an = amn)

= 53

(vi) 25 × 55

= (2 × 5)5 [am × bm = (a × b)m]

= 105

(vii) a4 × b4

= (ab)4 [am × bm = (a × b)m]

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

(ix) (220 ÷ 215) × 23

= (220 − 15)× 23 (am ÷ an = amn)

= 25 × 23

= (25 + 3) (am × an = am+n)

= 28

(x) 8t ÷ 82 = 8(t − 2) (am ÷ an = amn)

Page No 260:

Question 2:

Simplify and express each of the following in exponential form:

(i) (ii) (iii)

(iv) (v) (vi) 20 + 30 + 40

(vii) 20 × 30 × 40 (viii) (30 + 20) × 50 (ix)

(x) (xi) (xii)

Answer:

(i)

(ii) [(52)3 × 54] ÷ 57

= [52 × 3 × 54] ÷ 57 (am)n = amn

= [56 × 54] ÷ 57

= [56 + 4] ÷ 57 (am × an = am+n)

= 510 ÷ 57

= 510 − 7 (am ÷ an = amn)

= 53

(iii) 254 ÷ 53 = (5 ×5)4 ÷ 53

= (52)4 ÷ 53

= 52 × 4 ÷ 53 (am)n = amn

= 58 ÷ 53

= 58 − 3 (am ÷ an = amn)

= 55

(iv)

= 1 × 7 × 115 = 7 × 115

(v)

(vi) 20 + 30 + 40 = 1 + 1 + 1 = 3

(vii) 20 × 30 × 40 = 1 × 1 × 1 = 1

(viii) (30 + 20) × 50 = (1 + 1) × 1 = 2

(ix)

(x)

(xi)

(xii) (23 × 2)2 = (am × an = am+n)

= (24)2 = 24 × 2 (am)n = amn

= 28

Page No 260:

Question 3:

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

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

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 No 261:

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

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
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(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

Video Solution for Exponents and Powers (Page: 261 , Q.No.: 4)

NCERT Solution for Class 7 math - Exponents and Powers 261 , Question 4

Page No 261:

Question 5:

Simplify:

(i) (ii) (iii)

Answer:

(i)

(ii)

(iii)



Page No 263:

Question 1:

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 No 263:

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

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 No 263:

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

Answer:

(i) 50000000 = 5 × 107

(ii) 7000000 = 7 × 106

(iii) 3186500000 = 3.1865 × 109

(iv) 390878 = 3.90878 × 105

(v) 39087.8 = 3.90878 × 104

(vi) 3908.78 = 3.90878 × 103

Page No 263:

Question 4:

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:

(a) 3.84 × 108 m

(b) 3 × 108 m/s

(c) 1.2756 × 107 m

(d) 1.4 × 109 m

(e) 1 × 1011 stars

(f) 1.2 × 1010 years

(g) 3 × 1020 m

(h) 6.023 × 1022

(i) 1.353 × 109 cubic km

(j) 1.027 × 109



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