NCERT solution class 7 chapter 13 Exponents and powers 13.1 mathematics

EXERCISE 13.1

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"Exponents and Powers" Chapter 13 - Introduction - NCERT Class 7th Maths Solutions

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Introduction - Laws of Exponents - NCERT Class 7th Maths Solutions

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Q 1, Ex 13.1 - Exponents and Powers - Chapter 13 - Maths Class 7th - NCERT

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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

Q 2, Ex 13.1 - Exponents and Powers - Chapter 13 - Maths Class 7th - NCERT

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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 = a³ c⁴ d

Q 3, Ex 13.1 - Exponents and Powers - Chapter 13 - Maths Class 7th - NCERT

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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

Q 4, Ex 13.1 - Exponents and Powers - Chapter 13 - Maths Class 7th - NCERT

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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

Q 5, Ex 13.1 - Exponents and Powers - Chapter 13 - Maths Class 7th - NCERT

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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

Q 6, Ex 13.1 - Exponents and Powers - Chapter 13 - Maths Class 7th - NCERT

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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

Q 7, Ex 13.1 - Exponents and Powers - Chapter 13 - Maths Class 7th - NCERT

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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

Q 8, Ex 13.1 - Exponents and Powers - Chapter 13 - Maths Class 7th - NCERT

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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