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

NCERT Solutions for Class 7 Maths Chapter 13 – Exponents and Powers. Furthermore, here we’ve provided you with the latest solution for Class 7 Maths Chapter 13 – Exponents and Powers. As a result here you’ll find solutions to all the exercises. This NCERT Class 7 solution will help you to score good marks in your exam.

Students can refer to our solution for NCERT Class 7 Maths Chapter 13 – Exponents and Powers. The Chapter 13 Solution of NCERT will help students prepare for the exams and easily crack the exam. Below we’ve provided you with the exercise-wise latest solution.

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

NCERT Solutions for Class 7 Maths Chapter 13 – Exponents and Powers Exercise Wise Solution

Exercise 13.1 – Page 252 of NCERT
Exercise 13.2 – Page 260 of NCERT
Exercise 13.3 – Page 263 of NCERT

NCERT Solutions for Class 7 Maths Chapter 13 – Exponents and Powers Exercise 13.1 Solution

Here you’ll find NCERT Chapter 13 – Exponents and Powers Exercise 13.1 Solution.
Exercise 13.1: Solutions of Questions on Page Number: 252

Q1: Find the value of:

(i) 2⁶(ii) 9³

(iii) 11² (iv)5⁴

Answer:

(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

Answer:

(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 = 5²x 7³

(v) 2 x 2 x a x a = 2²x a²

(vi) a x a x a x c x c x c x c x d = a³c⁴d

Q3: Express the following numbers using exponential notation:

(i) 512 (ii) 343

(iii) 729 (iv) 3125

Answer :

(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) 4³or 3⁴(ii) 5³or 3⁵

(iii) 2⁸or 8²(iv) 100²or 2¹⁰⁰(v) 2¹⁰or 10²

Answer:

(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)100²or 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) 2¹⁰and 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

Answer:

(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) 7²x 2²(iii) 2³x 5 (iv) 3 x 4⁴

(v) 0 x 10²(vi) 5²x 3³

(vii) 2⁴x 3²(viii) 3²x 10⁴

Answer:

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

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

(iii) 2³x 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) 5²x 3³= 5 x 5 x 3 x 3 x 3 = 25 x 27 = 675

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

(viii) 3²x 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)²x (- 5)²(iv)(- 2)³x (-10)³

Answer:

(i) (-4)³= (-4) x (-4) x (-4) = -64

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

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

(iv) (-2)³x (-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

Answer:

(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

NCERT Solutions for Class 7 Maths Chapter 13 – Exponents and Powers Exercise 13.2 Solution

Here you’ll find NCERT Chapter 13 – Exponents and Powers Exercise 13.2 Solution.
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) 7^x× 7²(v)

(vii) a⁴× b⁴(viii) (3⁴)³

(vi) 2⁵× 5⁵
(ix) (x) 8^t÷ 8²

Answer:

(i) 3²× 3⁴× 3⁸ = (3)²⁺⁴⁺⁸ (a^m x aⁿ = a^m+n)

= 314

(ii) 6¹⁵÷ 6¹⁰= (6)^{15 – 10}(a^m÷ aⁿ= a^m^–ⁿ)

= 6⁵

(iii) a³x a²= a^{(3 + 2)}(a^mx aⁿ= a^m⁺ⁿ)

= a⁵

(iv) 7^x+ 7²= 7^x^{+ 2}(a^mx aⁿ= a^m⁺ⁿ)

(v) (5²)³÷ 5³

= 5^2×3 ÷ 5³ (a^m)ⁿ = a^mn

= 5⁶÷ 5³

= 5^{6 – 3}(a^m ÷ aⁿ = a^m–n)

= 5³

(vi) 2⁵x 5⁵

= (2 x 5)⁵[a^mx b^m= (a x b)^m]

= 10⁵

(vii) a⁴x b⁴

= (ab)⁴[a^mx b^m= (a x b)^m]

(viii) (3⁴)³= 3^{4 x 3}= 3¹²(a^m)ⁿ= a^mn

(ix) (2²⁰ ÷ 2¹⁵) x 2³

= (2^{20 – 15}) x 23 (a^m ÷ aⁿ = a^m–n)

= 2⁵x 2³

= (2⁵⁺³) (a^m x aⁿ = a^m+n)

= 2⁸

(x) 8^t÷ 8²= 8⁽^t^{– 2)}(a^m÷ aⁿ= a^m^–ⁿ)

Q2: 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

= [5^{2 × 3}× 5⁴] ÷ 5⁷(a^m)ⁿ= a^mn=

[5⁶× 5⁴] ÷ 5⁷

= [5^{6 + 4}] ÷ 5⁷(a^m× aⁿ= a^m⁺ⁿ)

= 5¹⁰÷ 5⁷

= 5^10-7 (a^m ÷ aⁿ = a^{m – n})

= 5³

(iii) 25⁴ ÷ 5³= (5 ×5)⁴÷ 5³

= (5²)^4÷ 5³

= 52 × 4 ÷ 53 (a^m)ⁿ = a^mn = 5⁸÷ 5³

= 58 – 3 (a^m ÷ aⁿ = a^{m – n})

= 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= (a^m× aⁿ= a^m⁺ⁿ)

= (2⁴)² = 2^{4 × 2} (a^m)ⁿ = a^mn

= 2⁸

Q3: Say true or false and justify your answer:

(i) 10 x 10¹¹= 100¹¹(ii) 2³> 5²

(iii) 2³x 3²= 6⁵(iv) 3⁰= (1000)⁰

Answer:

(i) 10 x 10¹¹= 100¹¹

L.H.S. = 10 x 10¹¹= 10^{11 + 1}(a^mx aⁿ= a^m⁺ⁿ)

= 10¹²

R.H.S. = 100¹¹= (10 x 10)¹¹= (10²)¹¹

= 10^{2 x 11} = 10²² (a^m)ⁿ = a^mn

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) 2³x 3²= 6⁵

L.H.S. = 2³x 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

Answer :

(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)

= (2²x 3³) x (2⁶x 3)

= 2^{6 + 2} x 3^{3 + 1} (a^m x aⁿ = a^m+n)

= 2⁸x 3⁴

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

= 3⁶x 2⁶

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

Q5: Simplify:

(i) (ii) (iii)

Answer :

(i)

(ii)

(iii)

NCERT Solutions for Class 7 Maths Chapter 13 – Exponents and Powers Exercise 13.3 Solution

Here you’ll find NCERT Chapter 13 – Exponents and Powers Exercise 13.3 Solution.
Exercise 13.3: Solutions of Questions on Page Number: 263

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

Answer:

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

Answer:

(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

= 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

Answer:

(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.

The distance between Earth and Moon is 384,000,000 m.
Speed of light in vacuum is 300,000,000 m/s.
Diameter of the Earth is 1,27,56, 000 m.
Diameter of the Sun is 1,400,000,000 m.
In a galaxy there are on an average 100,000,000,000 stars.
The universe is estimated to be about 12,000,000,000 years old.
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.
60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.
The earth has 1,353,000,000 cubic km of sea water.
The population of India was about 1,027,000,000 in March, 2001.

Answer :

(a) 3.84 x 10⁸m

(b) 3 x 10⁸m/s

(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 10⁹cubic km

(j) 1.027 x 10⁹