NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers

Question

Here you will find NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers, which is one of the most important study materials for students preparing for the CBSE board examination. Our NCERT Solutions are based on the most recent CBSE syllabus. These solutions, prepared and designed by subject matter experts, are to the point and explained in a point-by-point manner to make it easier for students to learn, study, and revise at the last minute. All of the shortcuts, equations, rules, theorems, and axioms have also been covered. We have explained all of the topics in detail in the NCERT Solution Class 7 that we have provided, which will undoubtedly help students clear all of the required concepts. Important topics covered here include:

• Introduction
• Exponents
• Laws of Exponents
• Example of Laws of Exponents
• Decimal Number System
• Expressing Larger Numbers in Standard Form

Our NCERT Solutions Class 7 Maths is the best resource for practicing and learning all of the key concepts. Regular practice of our solutions will assist students in developing subject-related problem-solving skills. We have answers to all types of questions, including exercise and in-text questions.

Answer 1

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers

1. Find the value of:

(i) 2⁶

(ii) 9³

(iii) 11²

(iv) 5⁴

Answer:

(i) 2⁶

The above value can be written as,

= 2 × 2 × 2 × 2 × 2 × 2

= 64

16

(ii) 9³

The above value can be written as,

= 9 × 9 × 9

= 729

(iii) 11²

The above value can be written as,

= 11 × 11

= 121

(iv) 5⁴

The above value can be written as,

= 5 × 5 × 5 × 5

= 625

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

The given question can be expressed in the exponential form as 6⁴.

(ii) t × t

The given question can be expressed in the exponential form as t².

(iii) b × b × b × b

The given question can be expressed in the exponential form as b⁴.

(iv) 5 × 5× 7 × 7 × 7

The given question can be expressed in the exponential form as 5² × 7³.

(v) 2 × 2 × a × a

The given question can be expressed in the exponential form as 2² × a².

(vi) a × a × a × c × c × c × c × d

The given question can be expressed in the exponential form as a³ × c⁴ × d.

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

(i) 512

(ii) 343

(iii) 729

(iv) 3125

Answer:

(i) 512

The factors of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

So it can be expressed in the exponential form as 2⁹.

(ii) 343

The factors of 343 = 7 × 7 × 7

So it can be expressed in the exponential form as 7³.

(iii) 729

The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

So it can be expressed in the exponential form as 3⁶.

(iv) 3125

The factors of 3125 = 5 × 5 × 5 × 5 × 5

So it can be expressed in the exponential form as 5⁵.

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 or 3⁴

The expansion of 4³ = 4 × 4 × 4 = 64

The expansion of 3⁴ = 3 × 3 × 3 × 3 = 81

Clearly,

64 < 81

So, 4³ < 3⁴

Hence, 3⁴ is the greater number.

(ii) 5³ or 3⁵

The expansion of 5³ = 5 × 5 × 5 = 125

The expansion of 3⁵ = 3 × 3 × 3 × 3 × 3= 243

Clearly,

125 < 243

So, 5³ < 3⁵

Hence, 3⁵ is the greater number.

(iii) 2⁸ or 8²

The expansion of 2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256

The expansion of 8² = 8 × 8= 64

Clearly,

256 > 64

So, 2⁸ > 8²

Hence, 2⁸ is the greater number.

(iv) 100² or 2¹⁰⁰

The expansion of 100² = 100 × 100 = 10000

The expansion of 2¹⁰⁰

2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

Then,

2¹⁰⁰ = 1024 × 1024 ×1024 × 1024 ×1024 × 1024 × 1024 × 1024 × 1024 × 1024 = (1024)¹⁰

Clearly,

100² < 2¹⁰⁰

Hence, 2¹⁰⁰ is the greater number.

(v) 2¹⁰ or 10²

The expansion of 2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

The expansion of 10² = 10 × 10= 100

Clearly,

1024 > 100

So, 2¹⁰ > 10²

Hence, 2¹⁰ is the greater number.

5. Express each of the following as a product of powers of their prime factors:

(i) 648

(ii) 405

(iii) 540

(iv) 3,600

Answer:

(i) 648

Factors of 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3

= 2³ × 3⁴

(ii) 405

Factors of 405 = 3 × 3 × 3 × 3 × 5

= 3⁴ × 5

(iii) 540

Factors of 540 = 2 × 2 × 3 × 3 × 3 × 5

= 2² × 3³ × 5

(iv) 3,600

Factors of 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

= 2⁴ × 3² × 5²

Answer 2

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

The above question can be written as,

= 2 × 10 × 10 × 10

= 2 × 1000

= 2000

(ii) 7² × 2²

The above question can be written as,

= 7 × 7 × 2 × 2

= 49 × 4

= 196

(iii) 2³ × 5

The above question can be written as,

= 2 × 2 × 2 × 5

= 8 × 5

= 40

(iv) 3 × 4⁴

The above question can be written as,

= 3 × 4 × 4 × 4 × 4

= 3 × 256

= 768

(v) 0 × 10²

The above question can be written as,

= 0 × 10 × 10

= 0 × 100

= 0

(vi) 5² × 3³

The above question can be written as,

= 5 × 5 × 3 × 3 × 3

= 25 × 27

= 675

(vii) 2⁴ × 3²

The above question can be written as,

= 2 × 2 × 2 × 2 × 3 × 3

= 16 × 9

= 144

(viii) 3² × 10⁴

The above question can be written as,

= 3 × 3 × 10 × 10 × 10 × 10

= 9 × 10000

= 90000

7. Simplify:

(i) (– 4)³

(ii) (–3) × (–2)³

(iii) (–3)² × (–5)²

(iv) (–2)³ × (–10)³

Answer:

(i) (– 4)³

The expansion of -4³

= – 4 × – 4 × – 4

= – 64

(ii) (–3) × (–2)³

The expansion of (-3) × (-2)³

= – 3 × – 2 × – 2 × – 2

= – 3 × – 8

= 24

(iii) (–3)² × (–5)²

The expansion of (-3)² × (-5)²

= – 3 × – 3 × – 5 × – 5

= 9 × 25

= 225

(iv) (–2)³ × (–10)³

The expansion of (-2)³ × (-10)³

= – 2 × – 2 × – 2 × – 10 × – 10 × – 10

= – 8 × – 1000

= 8000

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

By observing the question

Comparing the exponents of base 10,

Clearly,

2.7 × 10¹² > 1.5 × 10⁸

(ii) 4 × 10¹⁴ ; 3 × 10¹⁷

By observing the question

Comparing the exponents of base 10,

Clearly,

4 × 10¹⁴ < 3 × 10¹⁷

Answer 3

9. 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) (5²)³ ÷ 5³

(vi) 2⁵ × 5⁵

(vii) a⁴ × b⁴

(viii) (3⁴)³

(ix) (2²⁰ ÷ 2¹⁵) × 2³

(x) 8^t ÷ 8²

Answer:

(i) 3² × 3⁴ × 3⁸

By the rule of multiplying the powers with the same base = a^m × aⁿ = a^{m + n}

Then,

= (3)^{2 + 4 + 8}

= 3¹⁴

(ii) 6¹⁵ ÷ 6¹⁰

By the rule of dividing the powers with the same base = a^m ÷ aⁿ = a^{m – n}

Then,

= (6)^{15 – 10}

= 6⁵

(iii) a³ × a²

By the rule of multiplying the powers with the same base = a^m × aⁿ = a^{m + n}

Then,

= (a)^{3 + 2}

= a⁵

(iv) 7^x × 7²

By the rule of multiplying the powers with the same base = a^m × aⁿ = a^{m + n}

Then,

= (7)^{x + 2}

(v) (5²)³ ÷ 5³

By the rule of taking the power of as power = (a^m)ⁿ = a^mn

(5²)³ can be written as = (5)^{2 × 3}

= 5⁶

Now, 5⁶ ÷ 5³

By the rule of dividing the powers with the same base = a^m ÷ aⁿ = a^{m – n}

Then,

= (5)^{6 – 3}

= 5³

(vi) 2⁵ × 5⁵

By the rule of multiplying the powers with the same exponents = a^m × b^m = ab^m

Then,

= (2 × 5)⁵

= 10⁵

(vii) a⁴ × b⁴

By the rule of multiplying the powers with the same exponents = a^m × b^m = ab^m

Then,

= (a × b)⁴

= ab⁴

(viii) (3⁴)³

By the rule of taking the power of as power = (a^m)ⁿ = a^mn

(3⁴)³ can be written as = (3)^{4 × 3}

= 3¹²

(ix) (2²⁰ ÷ 2¹⁵) × 2³

By the rule of dividing the powers with the same base = a^m ÷ aⁿ = a^{m – n}

(2²⁰ ÷ 2¹⁵) can be simplified as,

= (2)^{20 – 15}

= 2⁵

Then,

By the rule of multiplying the powers with the same base = a^m × aⁿ = a^{m + n}

2⁵ × 2³ can be simplified as,

= (2)^{5 + 3}

= 2⁸

(x) 8^t ÷ 8²

By the rule of dividing the powers with the same base = a^m ÷ aⁿ = a^{m – n}

Then,

= (8)^{t – 2}

Answer 4

10. 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 × 11³)

(v) 3⁷/ (3⁴ × 3³)

(vi) 2⁰ + 3⁰ + 4⁰

(vii) 2⁰ × 3⁰ × 4⁰

(viii) (3⁰ + 2⁰) × 5⁰

(ix) (2⁸ × a⁵)/ (4³ × a³)

(x) (a⁵/a³) × a⁸

(xi) (4⁵ × a⁸b³)/ (4⁵ × a⁵b²)

(xii) (2³ × 2)²

Answer:

(i) (2³ × 3⁴ × 4)/ (3 × 32)

Factors of 32 = 2 × 2 × 2 × 2 × 2

= 2⁵

Factors of 4 = 2 × 2

= 2²

Then,

= (2³ × 3⁴ × 2²)/ (3 × 2⁵)

= (2^{3 + 2} × 3⁴) / (3 × 2⁵) … [∵a^m × aⁿ = a^{m + n}]

= (2⁵ × 3⁴) / (3 × 2⁵)

= 2^{5 – 5} × 3^{4 – 1} … [∵a^m ÷ aⁿ = a^{m – n}]

= 2⁰ × 3³

= 1 × 3³

= 3³

(ii) ((5²)³ × 5⁴) ÷ 5⁷

(5²)³ can be written as = (5)^{2 × 3} … [∵(a^m)ⁿ = a^mn]

= 5⁶

Then,

= (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

(25)⁴ can be written as = (5 × 5)⁴

= (5²)⁴

(5²)⁴ can be written as = (5)^{2 × 4} … [∵(a^m)ⁿ = a^mn]

= 5⁸

Then,

= 5⁸ ÷ 5³

= 5^{8 – 3} … [∵a^m ÷ aⁿ = a^{m – n}]

= 5⁵

(iv) (3 × 7² × 11⁸)/ (21 × 11³)

Factors of 21 = 7 × 3

Then,

= (3 × 7² × 11⁸)/ (7 × 3 × 11³)

= 3^1-1 × 7^2-1 × 11^{8 – 3}

= 3⁰ × 7 × 11⁵

= 1 × 7 × 11⁵

= 7 × 11⁵

(v) 3⁷/ (3⁴ × 3³)

= 3⁷/ (3⁴⁺³) … [∵a^m × aⁿ = a^{m + n}]

= 3⁷/ 3⁷

= 3^{7 – 7} … [∵a^m ÷ aⁿ = a^{m – n}]

= 3⁰

= 1

(vi) 2⁰ + 3⁰ + 4⁰

= 1 + 1 + 1

= 3

(vii) 2⁰ × 3⁰ × 4⁰

= 1 × 1 × 1

= 1

(viii) (3⁰ + 2⁰) × 5⁰

= (1 + 1) × 1

= (2) × 1

= 2

(ix) (2⁸ × a⁵)/ (4³ × a³)

(4)³ can be written as = (2 × 2)³

= (2²)³

(2²)³ can be written as = (2)^{2 × 3} … [∵(a^m)ⁿ = a^mn]

= 2⁶

Then,

= (2⁸ × a⁵)/ (2⁶ × a³)

= 2^{8 – 6} × a^{5 – 3} … [∵a^m ÷ aⁿ = a^{m – n}]

= 2² × a² … [∵(a^m)ⁿ = a^mn]

= 2a²

(x) (a⁵/a³) × a⁸

= (a^{5 -3}) × a⁸ … [∵a^m ÷ aⁿ = a^{m – n}]

= a² × a⁸

= a^{2 + 8} … [∵a^m × aⁿ = a^{m + n}]

= a¹⁰

(xi) (4⁵ × a⁸b³)/ (4⁵ × a⁵b²)

= 4^{5 – 5} × (a^{8 – 5} × b^{3 – 2}) … [∵a^m ÷ aⁿ = a^{m – n}]

= 4⁰ × (a³b)

= 1 × a³b

= a³b

(xii) (2³ × 2)²

= (2^{3 + 1})² … [∵a^m × aⁿ = a^{m + n}]

= (2⁴)²

(2⁴)² can be written as = (2)^{4 × 2} … [∵(a^m)ⁿ = a^mn]

= 2⁸

Answer 5

11. 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

Let us consider Left Hand Side (LHS) = 10 × 10¹¹

= 10^{1 + 11} … [∵a^m × aⁿ = a^{m + n}]

= 10¹²

Now, consider Right Hand Side (RHS) = 100¹¹

= (10 × 10)¹¹

= (10^{1 + 1})¹¹

= (10²)¹¹

= (10)^{2 × 11} … [∵(a^m)ⁿ = a^mn]

= 10²²

By comparing LHS and RHS,

LHS ≠ RHS

Hence, the given statement is false.

(ii) 2³ > 5²

Let us consider LHS = 2³

Expansion of 2³ = 2 × 2 × 2

= 8

Now, consider RHS = 5²

Expansion of 5² = 5 × 5

= 25

By comparing LHS and RHS,

LHS < RHS

2³ < 5²

Hence, the given statement is false.

(iii) 2³ × 3² = 6⁵

Let us consider LHS = 2³ × 3²

Expansion of 2³ × 3²= 2 × 2 × 2 × 3 × 3

= 72

Now, consider RHS = 6⁵

Expansion of 6⁵ = 6 × 6 × 6 × 6 × 6

= 7776

By comparing LHS and RHS,

72 ≠ 7776

LHS ≠ RHS

Hence, the given statement is false.

(iv) 3⁰ = (1000)⁰

Let us consider LHS = 3⁰

= 1

Now, consider RHS = 1000⁰

= 1

By comparing LHS and RHS,

LHS = RHS

3⁰ = 1000⁰

Hence, the given statement is true.

12. 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

The factors of 108 = 2 × 2 × 3 × 3 × 3

= 2² × 3³

The factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

= 2⁶ × 3

Then,

= (2² × 3³) × (2⁶ × 3)

= 2^{2 + 6} × 3^{3 + 1} … [∵a^m × aⁿ = a^{m + n}]

= 2⁸ × 3⁴

(ii) 270

The factors of 270 = 2 × 3 × 3 × 3 × 5

= 2 × 3³ × 5

(iii) 729 × 64

The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

= 3⁶

The factors of 64 = 2 × 2 × 2 × 2 × 2 × 2

= 2⁶

Then,

= (3⁶ × 2⁶)

= 3⁶ × 2⁶

(iv) 768

The factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

= 2⁸ × 3

13. 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)

8³ can be written as = (2 × 2 × 2)³

= (2³)³

We have,

= ((2⁵)² × 7³)/ ((2³)³ × 7)

= (2^{5 × 2} × 7³)/ ((2^{3 × 3} × 7) … [∵(a^m)ⁿ = a^mn]

= (2¹⁰ × 7³)/ (2⁹ × 7)

= (2^{10 – 9} × 7^{3 – 1}) … [∵a^m ÷ aⁿ = a^{m – n}]

= 2 × 7²

= 2 × 7 × 7

= 98

(ii) (25 × 5² × t⁸)/ (10³ × t⁴)

25 can be written as = 5 × 5

= 5²

10³ can be written as = 10³

= (5 × 2)³

= 5³ × 2³

We have,

= (5² × 5² × t⁸)/ (5³ × 2³ × t⁴)

= (5^{2 + 2} × t⁸)/ (5³ × 2³ × t⁴) … [∵a^m × aⁿ = a^{m + n}]

= (5⁴ × t⁸)/ (5³ × 2³ × t⁴)

= (5^{4 – 3} × t^{8 – 4})/ 2³ … [∵a^m ÷ aⁿ = a^{m – n}]

= (5 × t⁴)/ (2 × 2 × 2)

= (5t⁴)/ 8

(iii) (3⁵ × 10⁵ × 25)/ (5⁷ × 6⁵)

10⁵ can be written as = (5 × 2)⁵

= 5⁵ × 2⁵

25 can be written as = 5 × 5

= 5²

6⁵ can be written as = (2 × 3)⁵

= 2⁵ × 3⁵

Then we have,

= (3⁵ × 5⁵ × 2⁵ × 5²)/ (5⁷ × 2⁵ × 3⁵)

= (3⁵ × 5^{5 + 2} × 2⁵)/ (5⁷ × 2⁵ × 3⁵) … [∵a^m × aⁿ = a^{m + n}]

= (3⁵ × 5⁷ × 2⁵)/ (5⁷ × 2⁵ × 3⁵)

= (3^{5 – 5} × 5^{7 – 7} × 2^{5 – 5})

= (3⁰ × 5⁰ × 2⁰) … [∵a^m ÷ aⁿ = a^{m – n}]

= 1 × 1 × 1

= 1

Answer 6

14. Write the following numbers in the expanded forms:

(a) 279404

(b) 3006194

(c) 2806196

(d) 120719

(e) 20068

Answer:

(a) 279404

The expanded form of the number 279404 is,

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

Now we can express it using powers of 10 in the exponent form,

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

(b) 3006194

The expanded form of the number 3006194 is,

= (3 × 1000000) + (0 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + (4 × 1)

Now we can express it using powers of 10 in the exponent form,

= (3 × 10⁶) + (0 × 10⁵) + (0 × 10⁴) + (6 × 10³) + (1 × 10²) + (9 × 10¹) + (4 × 10⁰)

(c) 2806196

The expanded form of the number 2806196 is,

= (2 × 1000000) + (8 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + (6 × 1)

Now we can express it using powers of 10 in the exponent form,

= (2 × 10⁶) + (8 × 10⁵) + (0 × 10⁴) + (6 × 10³) + (1 × 10²) + (9 × 10¹) + (6 × 10⁰)

(d) 120719

The expanded form of the number 120719 is,

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

Now we can express it using powers of 10 in the exponent form,

= (1 × 10⁵) + (2 × 10⁴) + (0 × 10³) + (7 × 10²) + (1 × 10¹) + (9 × 10⁰)

(e) 20068

The expanded form of the number 20068 is,

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

Now we can express it using powers of 10 in the exponent form,

= (2 × 10⁴) + (0 × 10³) + (0 × 10²) + (6 × 10¹) + (8 × 10⁰)

15. Find the number from each of the following expanded forms:

(a) (8 × 10)⁴ + (6 × 10)³ + (0 × 10)² + (4 × 10)¹ + (5 × 10)⁰

(b) (4 × 10)⁵ + (5 × 10)³ + (3 × 10)² + (2 × 10)⁰

(c) (3 × 10)⁴ + (7 × 10)² + (5 × 10)⁰

(d) (9 × 10)⁵ + (2 × 10)² + (3 × 10)¹

Answer:

(a) (8 × 10)⁴ + (6 × 10)³ + (0 × 10)² + (4 × 10)¹ + (5 × 10)⁰

The expanded form is,

= (8 × 10000) + (6 × 1000) + (0 × 100) + (4 × 10) + (5 × 1)

= 80000 + 6000 + 0 + 40 + 5

= 86045

(b) (4 × 10)⁵ + (5 × 10)³ + (3 × 10)² + (2 × 10)⁰

The expanded form is,

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

= 400000 + 0 + 5000 + 300 + 0 + 2

= 405302

(c) (3 × 10)⁴ + (7 × 10)² + (5 × 10)⁰

The expanded form is,

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

= 30000 + 0 + 700 + 0 + 5

= 30705

(d) (9 × 10)⁵ + (2 × 10)² + (3 × 10)¹

The expanded form is,

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

= 900000 + 0 + 0 + 200 + 30 + 0

= 900230

16. Express the following numbers in standard form:

(i) 5,00,00,000

(iii) 3,18,65,00,000

(iv) 3,90,878

(v) 39087.8

(vi) 3908.78

Answer:

(i) 5,00,00,000

The standard form of the given number is 5 × 10⁷

(ii) 70,00,000

The standard form of the given number is 7 × 10⁶

(iii) 3,18,65,00,000

The standard form of the given number is 3.1865 × 10⁹

(iv) 3,90,878

The standard form of the given number is 3.90878 × 10⁵

(v) 39087.8

The standard form of the given number is 3.90878 × 10⁴

(vi) 3908.78

The standard form of the given number is 3.90878 × 10³

Answer 7

17. 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 a 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 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 seawater.

(j) The population of India was about 1,027,000,000 in March 2001.

Answer:

(a) The standard form of the number appearing in the given statement is 3.84 × 10⁸m.

(b) The standard form of the number appearing in the given statement is 3 × 10⁸m/s.

(c) The standard form of the number appearing in the given statement is 1.2756 × 10⁷m.

(d) The standard form of the number appearing in the given statement is 1.4 × 10⁹m.

(e) The standard form of the number appearing in the given statement is 1 × 10¹¹ stars.

(f) The standard form of the number appearing in the given statement is 1.2 × 10¹⁰ years old.

(g) The standard form of the number appearing in the given statement is 3 × 10²⁰m.

(h) The standard form of the number appearing in the given statement is 6.023 × 10²² molecules.

(i) The standard form of the number appearing in the given statement is 1.353 × 10⁹ cubic km.

(j) The standard form of the number appearing in the given statement is 1.027 × 10⁹.