NCERT Solutions for 7th Class Maths: Chapter 13-Exponents and Powers

Class 7: Maths Chapter 13 solutions. Complete Class 7 Maths Chapter 13 Notes.

NCERT Solutions for 7th Class Maths: Chapter 13-Exponents and Powers

NCERT 7th Maths Chapter 13, class 7 Maths Chapter 13 solutions

Exercise 13.1 

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

2. Express the following in exponential form: 

(ii) t × t

(i) 6 × 6 × 6 × 6  

(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 x d

3. Express each of the following numbers using exponential notation: 

(i) 512  

(ii) 343  

(iii) 729  

(iv) 3125

Answer

(i) 512

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁹

(ii) 343

343 = 7 × 7 × 7 = 7³

(iii) 729 

729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶

(iv) 3125

3125 = 5 × 5 × 5 × 5 × 5 = 5⁵

4. Identify the greater number, wherever possible, in each of the following: 

(i)  4³ and 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

Since 64 < 81

Thus, 34 is greater than 43.

(ii) 5³ = 5 × 5 × 5 = 125

3⁵ = 3 × 3 × 3 × 3 × 3 = 243

Since, 125 < 243

Thus, 34 is greater than 53.

(iii) 2⁸= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256

8² = 8×8 = 64

Since, 256 > 64

Thus, 28 is greater than 82.

(iv) 100² = 100 × 100 = 10,000

2¹⁰⁰ = 2 × 2 × 2 × 2 × 2 …….14 times ….× 2 = 16,384 ….× 2

Since, 10,000 < 16,384 ….×2

Thus, 2100 is greater than 1002.

(v)  210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1,024

10² = 10×10 = 100

Since, 1,024 > 100

Thus, 2¹⁰ >10²

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 = 2³ × 3⁴

(ii) 405 = 5 × 3⁴

(iii) 540 = 2² × 3³ × 5

(iv) 3,600 = 2⁴ × 3² × 5²

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  = 2,000

(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) 5³ × 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  =  90,000

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) = 225

(iv) (-2)³×(-10)³ =(-2)×(-2)×(-2)×(-10)×(-10)×(-10)

8. Compare the following numbers: 

(i) 2.7 × 10¹²; 1.5 × 10⁸ 

(ii) 4 × 10¹⁴; 3 × 10¹⁷ 

Answer

(i) 2.7 × 10¹² and 1.5 × 10⁸

On comparing the exponents of base 10,

2.7 × 10¹² > 1.5 × 10⁸

(ii) 4 × 10¹⁴ and 3 × 10¹⁷

On comparing the exponents of base 10,

4 × 10¹⁴ < 3 × 10¹⁷

NCERT 7th Maths Chapter 13, class 7 Maths Chapter 13 solutions

1. Using laws of exponents, simplify and write the answer in exponential form: 

(i) 3² × 3⁴ x 3⁸   

(ii) 6¹⁵ ÷ 6¹⁰

(iii) a³ × a²    

(iv) 7 × x 7²

(v) (5²)² ÷ 5³   

(vi) 2⁵ × 5⁵

(vii) a⁴ × b⁴

(viii) (3⁴)³

(ix) [2²⁰ ÷ 2¹⁵) × 2³

(x) 8^t × 8²

Answer

(i) 3² × 3⁴ × 3⁸ = 3^(2+4+8) = 3¹⁴ [∵ a^m× aⁿ = a^m+n]

(ii) 6¹⁵ ÷ 6¹⁰ = 6^15-10 = 6⁵ [∵ a^m ÷ aⁿ = a^m-n]

(iii) a³ × a² = a³⁺² = a⁵ [∵ am × an = a m+n]

(iv) 7^x × 7² = 7^x+2 [∵ a^m × aⁿ = a^m+n]

(v) (5²)² ÷ 5³ = 5^2×3 ÷ 5³ = 5⁶ ÷ 5³ [∵ (a^m)ⁿ = a^m×n]

= 5^6-3 = 5³ [∵ a^m ÷ aⁿ = a^m-n]

(vi) 2⁵ × 2⁵ = (2×5)⁵= 10⁵ [∵ a^m×b^m = (a×b)^m]

(vii) a⁴ × b⁴ = (a×b)⁴ [∵ a^m×b^m = (a×b)^m]

(viii) (3⁴)³ = 3^4×3 = 3¹² [∵ (a^m)ⁿ = a^m×n]

(ix) (2²⁰ ÷ 2¹⁵)×2³ = (2^20-15)×2³ [∵ a^m ÷ aⁿ = a^m-n]

= 2⁵ × 2³ = 2⁵⁺³ = 2⁸ [∵ a^m × aⁿ = a^m+n]

(x) 8^t ÷ 8² = 8^t-2 [∵ a^m ÷ aⁿ = a^m-n]

2. Simplify and express each of the following in exponential form:

(i) 2³ × 3⁴ × 4/3×32

(ii) [(5²)³ ×5⁴] ÷ 5⁷

(iii) 25⁴ ÷ 5³

(iv) 3 × 7² × 11⁸ / 21 × 11

(v) 3⁷ / 3⁴ × 3³

(vi) 2⁰ + 3⁰ + 4⁰

(vii) 2⁰ × 3⁰ × 4⁰

(viii) (3⁰ + 2⁰) × 5⁰

(ix) 2⁸ × a ⁵ / 4³ × a³

(x) (a⁵ / a³) × a⁸

(xi) 4⁵ × a⁸b³ / 4⁵ × a⁵b²

(xii) (2³ × 2)³

Answer

(i) 2³ × 3⁴ × 4 / 3× 32 = 2³ × 3⁴ × 2² / 3 × 2⁵ = 2³⁺² × 3⁴ / 3×2⁵ [∵ a^m× aⁿ = a^m+n]

= 2⁵ × 3⁴ / 3 × 2⁵ = 2^5-5 × 3^4-3 [∵ a^m ÷ aⁿ = a^m-n]

= 2⁰ × 3³ = 1× 3³ = 3³

(ii) [(5²)³ × 5⁴] ÷ 5⁷  [∵ (a^m)ⁿ = a^m×n]

= [5⁶⁺⁴] ÷ 5⁷ = 5¹⁰ ÷ 5⁷  [∵ a^m× aⁿ = a^m+n]

= 5^10-7 = 5³ [∵ a^m ÷ aⁿ = a^m-n]

(iii) 25⁴ ÷ 5³ = (5²)⁴ ÷ 5³ = 5⁸ ÷ 5³ [∵ (a^m)ⁿ = a^m×n]

= 5^8-3 = 5⁵ [∵ a^m ÷ aⁿ = a^m-n]

(iv) 3 × 7² × 11⁸ / 21 × 11² = 3 × 72 × 11⁸ / 3 × 7 × 11³ = 3^1-1 × 7^2-1 × 11^8-3   [∵ a^m ÷ aⁿ = a^m-n]

= 3⁰ × 7¹ × 11⁵ = 7 × 11⁵

(v) 3⁷ / 3⁴ × 3³ = 3⁷ / × 3⁴⁺³ = 3⁷ / 3⁷ [∵ a^m× aⁿ = a^m+n]

= 37-7 = 30 = 1 [∵ a^m× aⁿ = a^m+n]

(vi) 2⁰ + 3⁰ + 4⁰ + 1+1+1 = 3  [∵ a⁰ = 1]

(vii) 2⁰ × 3⁰ × 4⁰ = 1×1×1 = 1 [∵ a⁰ = 1]

(viii) (3⁰ + 2⁰) × 5⁰ = (1+1)×1 = 2×1 = 2 [∵ a⁰ = 1]

(ix) 2⁸ × a⁵  / 4³ × a³ = 2⁸ × a⁵/ (2²)³ × a³ = 2⁸×a⁵ / 2⁶ × a³ [∵ (a^m)ⁿ = a^m×n]

= 2^8-6 × a^5-2 = 2² × a² [∵ a^m ÷ aⁿ = a^m-n]

= (2a)² [∵ a^m×b^m = (a×b)^m]

(x) (a⁵ / a³) × a⁸ = (a ^5-3) × a⁸ = a² × a⁸ [∵ a^m ÷ aⁿ = a^m-n]

= a²⁺⁸ = a¹⁰ [∵ a^m× aⁿ = a^m+n]

(xi) 4⁵ × a⁸b³ / 4⁵ × a⁵b² = 4^5-5 × a^8-5 × b^3-2 = 4⁰ × a³ × b [∵ a^m ÷ aⁿ = a^m-n]

= 1 × a³ × b = a³b [∵ a⁰ = 1]

(xii) (2³ × 2)² = (2³⁺¹)² = (2⁴)² [∵ a^m× aⁿ = a^m+n]

= 2^4×2 = 2⁸

NCERT 7th Maths Chapter 13, class 7 Maths Chapter 13 solutions

3. Say true or false and justify your answer: 

(i) 10 x 10¹¹ = 10011   

(ii)  2³ > 5²

(iii) 2³× 3² = 6s    

(iv) 3⁰ = (1000)⁰

Answer 

(i) 10×10¹¹ = 100¹¹

L.H.S. 10¹⁺¹¹ = 10¹²    and R.H.S. (10²)¹¹=10²²

Since, L.H.S. ≠ R.H.S.

Therefore, it is false.

(ii) 2³ > 5²

L.H.S 2³ = 8    and R.H.S. 5² = 25

Since, L.H,S. is not greater than R.H.S.

Therefore, it is false,

(iii) 2³ ×3² = 6⁵

L.H.S. 2³ × 3² = 8 x 9 = 72    and R.H.S. 6⁵ =7.776

Since, L.H.S. ≠ R.H.S.

Therefore, it is false.

(iv) 3⁰ = (1000)⁰

L.H.S. 3⁰ = 1    and R.H.S. (1000)⁰ = 1

Since, L.H.S. = R.H.S.

Therefore, it is true.

NCERT 7th Maths Chapter 13, class 7 Maths Chapter 13 solutions

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 (a^m × aⁿ = a^m+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)

= 3⁶ × 2⁵

(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁸ × 3

5. Simplify:

(i) (2⁵)² × 7³ / 8³ × 7

(ii) 25 × 5² × t⁸ / 10³ × t⁴

(iii) 3⁵ × 10⁵ × 25 × 5⁷ × 6⁵

Answer

(i) (2⁵)² × 7³ / 8³ × 7 = 2^5×2 × 7³ / (2³)³ × 7

= 2¹⁰ × 7³ / 2⁹ × 7

= 2^10-9 × 7^3-1

= 2 × 7²

= 2 × 49

= 98

(ii) 25 × 5² ×t⁸ / 10³ – t⁴ = 5² × 5² × t⁸ / (5×2)³ × t⁴

= 5²⁺² × t^8-4/2³×3³

= 5⁴ × t⁴ / 2³ × 5³

= 5^4-3 × t⁴ / 2³

= 5t⁴ / 8

(iii) 3⁵ × 10⁵ × 25 / 5⁷ × 6⁵= 3⁵ ×(2×5)⁵ × 5² / 5⁷ × (2×3)⁵

= 3⁵ × 2⁵ × 5⁵ × 5² / 5⁷ × 2⁵ × 3⁵

= 3⁵ × 2⁵ × 5⁵⁺² / 5⁷ × 2⁵ × 3⁵

= 3⁵ × 2⁵ × 5⁷ / 5⁷ × 2⁵ × 3⁵

= 2^5-5 × 3^5-5 × 5^5-5

= 2⁰ × 3⁰ × 5⁰

= 1 × 1 × 1

= 1

NCERT 7th Maths Chapter 13, class 7 Maths Chapter 13 solutions

1. Write the following numbers in the expanded form: 

(i) 279404

(ii) 3006194

(iii) 2806196

(iv) 120719

(v) 20068

Answer

(i) 2,79,404

= 2,00,000 + 70,000 + 9,000 + 400 + 00 + 4 

= (2×100000) + (7 × 10000) + (9×1000) + (4 × 100) + (0× 10) + (4×1)

= (2×10⁵) + (7×10⁴) + (9×10³) + (4 ×10²) + (0×10¹) + (4×10⁰)

(ii) 3006194

= 3000000 + 0 + 0 + 6000 + 100 + 90 + 4

= (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) 28,06,196

= 20,00,000 + 8,00,000 + 0 + 6,000 + 100 + 90 + 6
= 2×1000000 + 8× 100000+ 0 × 10000 +6 × 1000 + 1 × 100 + 9 × 10 + 6 × 1
= 2 ×10⁶+8×10⁵+0×10⁴ + 6×10³+1×10²+9×10+6×10⁰

(iv) 1,20,719 = 1,00,000 + 20,000 + 0 + 700 + 10 + 9
= 1 × 100000 + 2 × 10000 + 0× 1000 + 7×100 + 1× 10+9×1
= 1 × 10⁵ + 2 × 10⁴ + 0 x 10³ + 7 × 10² + 1 x 101 + 9 × 10⁰ 

(v) 20,068= 20,000 + 00 + 00 + 60 + 8

= 2 × 10000 + 0 × 1000 + 0 × 100 + 6 × 10 + 8 × 1

= 2×10⁴ + 0 × 10³ + 0 × 10² + 6 x 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 

(b) 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 100 

(c) 3 × 10³ + 7 × 10² + 5 × 100

(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹ 

Answer

(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 

= 86,045

(b) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100

= 4 × 100000 + 0 × 10000 + 5 × 1000 + 3 × 100 + 0 × 10 + 2 × 1

= 400000 + 0 + 5000 + 3000 + 0 + 2 

= 4,05,302

(c) 3 × 10⁴ + 7 × 10² + 5 × 10⁰

= 3 × 10000 + 0 × 1000 + 7 × 100 + 0 × 10 + 5 × 1

= 30000 + 0 + 700 + 0 + 5 

= 30,705

(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹ 

= 9 × 100000 + 0 × 10000 + 0 × 1000 + 2 × 100 + 3 × 10 + 0 × 1

= 900000 + 0 + 0 + 200 + 30 + 0 

= 9,00,230

NCERT 7th Maths Chapter 13, class 7 Maths Chapter 13 solutions

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) 5,00,00,000 = 5 × 1,00,00,000 = 5 × 10⁷

(ii) 70,00,000 = 7 × 10,00,000 = 7 × 10⁶

(iii) 3,18,65,00,000 = 31865 × 100000

= 3.1865 × 10000 × 100000 = 3.1865 × 10⁹

(iv) 3,90,878 = 3.90878 × 100000 = 3.90878 × 10⁵

(v) 39087.8 = 3.90878 × 10000 = 3.90878 × 10⁴

(vi) 3908.78 = 3.90878 × 1000 = 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. 

(b) Speed of light in vacuum is 300,000,000 m/s. 

(c) Diameter of Earth id 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 100,000,000,0000 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) The distance between Earth and Moon = 384,000,000 m

= 384 × 1000000 m

= 3.84 × 100 × 1000000

= 3.84 × 10⁸ m

(b) Speed of light in vacuum = 300,000,000 m/s

= 3 × 100000000 m/s

= 3×10⁸ m/s

(c) Diameter of the Earth = 1,27,50,000 m

= 12756 × 1000 m

= 1.2756 × 10000 × 1000 m

= 1 2756×10⁷ m

(d) Diameter of the Sun = 1,400,000,000 m

= 14 × 100,000,000 m

= 1.4 × 10 × 100,000,000 m

= 1.4 × 10⁹ m

(e) Average of Stars = 100, 000, 000,000

= 1 × 100,000,000,000

= 1×10¹¹

(f) Years of Universe = 12,000,000,000 years

= 12 × 1000,000,000 years

= 1.2 × 10 × 1000,000,000 years

= 1.2 × 10¹⁰ years

(g) Distance of the Sun from the centre of the Milky Way Galaxy

= 300,000,000,000,000,000,000 m

= 3 × 100,000,000,000,000,000,000 m

= 3 × 10²⁰ m

(h) Number of molecules in a drop of water weighing 1.8 gm

= 60,230,000,000,000,000,000,000

= 6023 × 10,000,000,000,000,000,000

= 6,023 × 1000 × 10,000,000,000,000,000,000

= 6.023 × 1022

(i) The Earth has Sea water = 1,353,000,000 km³

= 1,353 × 1000000 km3

= 1,353 × 1000 × 1000,000 km³

= 1.353 × 10⁹ km³

(j) The population of India = 1,027,000,000

= 1027 × 1000000 = 1,027 × 1000 × 1000000

= 1.027×10⁹.

NCERT Solutions for 7th Class Maths: Chapter 13: Download PDF

NCERT Solutions for 7th Class Maths: Chapter 13-Exponents and Powers

Download PDF: NCERT Solutions for 7th Class Maths: Chapter 13-Exponents and Powers PDF

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