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NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 

Here, we have provided NCERT Solutions for Class 7 Maths Chapter 13. Students can refer to these solutions before exams to gain a better understanding of the chapter's concepts and themes.
authorImageAnanya Gupta2 Mar, 2024
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NCERT Solutions for Class 7 Maths Chapter 13

NCERT Solutions for Class 7 Maths Chapter 13

NCERT Solutions for Class 7 Maths Chapter 13: NCERT Solutions for Class 7 Maths Chapter 13 provide detailed explanations and answers to the exercises in the chapter on Exponents and Powers. With these solutions students can strengthen their understanding of the concepts and improve their problem-solving skills. Practicing with NCERT Solutions for Chapter 13 will help students prepare thoroughly for exams and build a strong foundation in mathematics.

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NCERT Solutions for Class 7 Maths Chapter 13 PDF

Below is the PDF link for NCERT Solutions for Class 7 Maths Chapter 13. This PDF provides detailed solutions to the exercises and problems presented in the chapter. By using this resource students can enhance their understanding of Exponents and Powers and practice solving various types of questions. Utilizing the PDF will aid students in strengthening their mathematical skills and preparing effectively for examinations.

NCERT Solutions for Class 7 Maths Chapter 13 PDF

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers

Exercise 13.1 Page: 252

1. Find the value of:

(i) 2 6

Solution:-

The above value can be written as, = 2 × 2 × 2 × 2 × 2 × 2 = 64
220

(ii) 9 3

Solution:-

The above value can be written as, = 9 × 9 × 9 = 729

(iii) 11 2

Solution:-

The above value can be written as, = 11 × 11 = 121
88

(iv) 5 4

Solution:-

The above value can be written as, = 5 × 5 × 5 × 5 = 625

2. Express the following in exponential form:

(i) 6 × 6 × 6 × 6

Solution:-

The given question can be expressed in the exponential form as 6 4 .
157

(ii) t × t

Solution:-

The given question can be expressed in the exponential form as t 2 .

(iii) b × b × b × b

Solution:-

The given question can be expressed in the exponential form as b 4 .

(iv) 5 × 5× 7 × 7 × 7

Solution:-

The given question can be expressed in the exponential form as 5 2 × 7 3 .

(v) 2 × 2 × a × a

Solution:-

The given question can be expressed in the exponential form as 2 2 × a 2 .

(vi) a × a × a × c × c × c × c × d

Solution:-

The given question can be expressed in the exponential form as a 3 × c 4 × d.

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

(i) 512

Solution:-

The factors of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 So it can be expressed in the exponential form as 2 9 .

(ii) 343

Solution:-

The factors of 343 = 7 × 7 × 7 So it can be expressed in the exponential form as 7 3 .

(iii) 729

Solution:-

The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3 So it can be expressed in the exponential form as 3 6 .

(iv) 3125

Solution:-

The factors of 3125 = 5 × 5 × 5 × 5 × 5 So it can be expressed in the exponential form as 5 5 .

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

(i) 4 3 or 3 4

Solution:-

The expansion of 4 3 = 4 × 4 × 4 = 64 The expansion of 3 4 = 3 × 3 × 3 × 3 = 81 Clearly, 64 < 81 So, 4 3 < 3 4 Hence, 3 4 is the greater number.

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(ii) 5 3 or 3 5

Solution:-

The expansion of 5 3 = 5 × 5 × 5 = 125 The expansion of 3 5 = 3 × 3 × 3 × 3 × 3= 243 Clearly, 125 < 243 So, 5 3 < 3 5 Hence, 3 5 is the greater number.
137

(iii) 2 8 or 8 2

Solution:-

The expansion of 2 8 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256 The expansion of 8 2 = 8 × 8= 64 Clearly, 256 > 64 So, 2 8 > 8 2 Hence, 2 8 is the greater number.

(iv) 100 2 or 2 100

Solution:-

The expansion of 100 2 = 100 × 100 = 10000 The expansion of 2 100 2 10 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024 Then, 2 100 = 1024 × 1024 ×1024 × 1024 ×1024 × 1024 × 1024 × 1024 × 1024 × 1024 = (1024) 10 Clearly, 100 2 < 2 100 Hence, 2 100 is the greater number.

(v) 2 10 or 10 2

Solution:-

The expansion of 2 10 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024 The expansion of 10 2 = 10 × 10= 100 Clearly, 1024 > 100 So, 2 10 > 10 2 Hence, 2 10 is the greater number.

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

(i) 648

Solution:-

Factors of 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 2 3 × 3 4

(ii) 405

Solution:-

Factors of 405 = 3 × 3 × 3 × 3 × 5 = 3 4 × 5

(iii) 540

Solution:-

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

(iv) 3,600

Solution:-

Factors of 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 2 4 × 3 2 × 5 2
127

6. Simplify:

(i) 2 × 10 3

Solution:-

The above question can be written as, = 2 × 10 × 10 × 10 = 2 × 1000 = 2000

(ii) 7 2 × 2 2

Solution:-

The above question can be written as, = 7 × 7 × 2 × 2 = 49 × 4 = 196
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(iii) 2 3 × 5

Solution:-

The above question can be written as, = 2 × 2 × 2 × 5 = 8 × 5 = 40

(iv) 3 × 4 4

Solution:-

The above question can be written as, = 3 × 4 × 4 × 4 × 4 = 3 × 256 = 768

(v) 0 × 10 2

Solution:-

The above question can be written as, = 0 × 10 × 10 = 0 × 100 = 0

(vi) 5 2 × 3 3

Solution:-

The above question can be written as, = 5 × 5 × 3 × 3 × 3 = 25 × 27 = 675

(vii) 2 4 × 3 2

Solution:-

The above question can be written as, = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144

(viii) 3 2 × 10 4

Solution:-

The above question can be written as, = 3 × 3 × 10 × 10 × 10 × 10 = 9 × 10000 = 90000

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7. Simplify:

(i) (– 4) 3

Solution:-

The expansion of -4 3 = – 4 × – 4 × – 4 = – 64

(ii) (–3) × (–2) 3

Solution:-

The expansion of (-3) × (-2) 3 = – 3 × – 2 × – 2 × – 2 = – 3 × – 8 = 24

(iii) (–3) 2 × (–5) 2

Solution:-

The expansion of (-3) 2 × (-5) 2 = – 3 × – 3 × – 5 × – 5 = 9 × 25 = 225

(iv) (–2) 3 × (–10) 3

Solution:-

The expansion of (-2) 3 × (-10) 3 = – 2 × – 2 × – 2 × – 10 × – 10 × – 10 = – 8 × – 1000 = 8000
614

8. Compare the following numbers:

(i) 2.7 × 10 12 ; 1.5 × 10 8

Solution:-

By observing the question Comparing the exponents of base 10, Clearly, 2.7 × 10 12 > 1.5 × 10 8

(ii) 4 × 10 14 ; 3 × 10 17

Solution:-

By observing the question Comparing the exponents of base 10, Clearly, 4 × 10 14 < 3 × 10 17

Exercise 13.2 Page: 260

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

(i) 3 2 × 3 4 × 3 8

Solution:-

By the rule of multiplying the powers with the same base = a m × a n = a m + n Then, = (3) 2 + 4 + 8 = 3 14

(ii) 6 15 ÷ 6 10

Solution:-

By the rule of dividing the powers with the same base = a m ÷ a n = a m – n Then, = (6) 15 – 10 = 6 5

(iii) a 3 × a 2

Solution:-

By the rule of multiplying the powers with the same base = a m × a n = a m + n Then, = (a) 3 + 2 = a 5

(iv) 7 x × 7 2

Solution:-

By the rule of multiplying the powers with the same base = a m × a n = a m + n Then, = (7) x + 2

(v) (5 2 ) 3 ÷ 5 3

Solution:-

By the rule of taking the power of as power = (a m ) n = a mn (5 2 ) 3 can be written as = (5) 2 × 3 = 5 6 Now, 5 6 ÷ 5 3 By the rule of dividing the powers with the same base = a m ÷ a n = a m – n Then, = (5) 6 – 3 = 5 3

(vi) 2 5 × 5 5

Solution:-

By the rule of multiplying the powers with the same exponents = a m × b m = ab m Then, = (2 × 5) 5 = 10 5

(vii) a 4 × b 4

Solution:-

By the rule of multiplying the powers with the same exponents = a m × b m = ab m Then, = (a × b) 4 = ab 4

(viii) (3 4 ) 3

Solution:-

By the rule of taking the power of as power = (a m ) n = a mn (3 4 ) 3 can be written as = (3) 4 × 3 = 3 12

(ix) (2 20 ÷ 2 15 ) × 2 3

Solution:-

By the rule of dividing the powers with the same base = a m ÷ a n = a m – n (2 20 ÷ 2 15 ) can be simplified as, = (2) 20 – 15 = 2 5 Then, By the rule of multiplying the powers with the same base = a m × a n = a m + n 2 5 × 2 3 can be simplified as, = (2) 5 + 3 = 2 8

(x) 8 t ÷ 8 2

Solution:-

By the rule of dividing the powers with the same base = a m ÷ a n = a m – n Then, = (8) t – 2

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

(i) (2 3 × 3 4 × 4)/ (3 × 32)

Solution:-

Factors of 32 = 2 × 2 × 2 × 2 × 2 = 2 5 Factors of 4 = 2 × 2 = 2 2 Then, = (2 3 × 3 4 × 2 2 )/ (3 × 2 5 ) = (2 3 + 2 × 3 4 ) / (3 × 2 5 ) … [∵a m × a n = a m + n ] = (2 5 × 3 4 ) / (3 × 2 5 ) = 2 5 – 5 × 3 4 – 1 … [∵a m ÷ a n = a m – n ] = 2 0 × 3 3 = 1 × 3 3 = 3 3

(ii) ((5 2 ) 3 × 5 4 ) ÷ 5 7

Solution:-

(5 2 ) 3 can be written as = (5) 2 × 3 … [∵(a m ) n = a mn ] = 5 6 Then, = (5 6 × 5 4 ) ÷ 5 7 = (5 6 + 4 ) ÷ 5 7 … [∵a m × a n = a m + n ] = 5 10 ÷ 5 7 = 5 10 – 7 … [∵a m ÷ a n = a m – n ] = 5 3

Related Links -

NCERT Solutions for Class 7 Maths Chapter 1 NCERT Solutions for Class 7 Maths Chapter 7
NCERT Solutions for Class 7 Maths Chapter 2 NCERT Solutions for Class 7 Maths Chapter 10
NCERT Solutions for Class 7 maths Chapter 3 NCERT Solutions for Class 7 Maths Chapter 11
NCERT Solutions for Class 7 Maths Chapter 4 NCERT Solutions for Class 7 Maths Chapter 9
NCERT Solutions for Class 7 Maths NCERT Solutions for Class 7 Maths Chapter 13
NCERT Solutions for Class 7 Maths Chapter 5 NCERT Solutions for Class 7 Maths Chapter 14
NCERT Solutions for Class 7 Maths Chapter 6 NCERT Solutions for Class 7 Maths Chapter 15
NCERT Solutions for Class 7 Maths Chapter 8

(iii) 25 4 ÷ 5 3

Solution:-

(25) 4 can be written as = (5 × 5) 4 = (5 2 ) 4 (5 2 ) 4 can be written as = (5) 2 × 4 … [∵(a m ) n = a mn ] = 5 8 Then, = 5 8 ÷ 5 3 = 5 8 – 3 … [∵a m ÷ a n = a m – n ] = 5 5

(iv) (3 × 7 2 × 11 8 )/ (21 × 11 3 )

Solution:-

Factors of 21 = 7 × 3 Then, = (3 × 7 2 × 11 8 )/ (7 × 3 × 11 3 ) = 3 1-1 × 7 2-1 × 11 8 – 3 = 3 0 × 7 × 11 5 = 1 × 7 × 11 5 = 7 × 11 5

(v) 3 7 / (3 4 × 3 3 )

Solution:-

= 3 7 / (3 4+3 ) … [∵a m × a n = a m + n ] = 3 7 / 3 7 = 3 7 – 7 … [∵a m ÷ a n = a m – n ] = 3 0 = 1

(vi) 2 0 + 3 0 + 4 0

Solution:-

= 1 + 1 + 1 = 3

(vii) 2 0 × 3 0 × 4 0

Solution:-

= 1 × 1 × 1 = 1

(viii) (3 0 + 2 0 ) × 5 0

Solution:-

= (1 + 1) × 1 = (2) × 1 = 2

(ix) (2 8 × a 5 )/ (4 3 × a 3 )

Solution:-

(4) 3 can be written as = (2 × 2) 3 = (2 2 ) 3 (2 2 ) 3 can be written as = (2) 2 × 3 … [∵(a m ) n = a mn ] = 2 6 Then, = (2 8 × a 5 )/ (2 6 × a 3 ) = 2 8 – 6 × a 5 – 3 … [∵a m ÷ a n = a m – n ] = 2 2 × a 2 … [∵(a m ) n = a mn ] = 2a 2

(x) (a 5 /a 3 ) × a 8

Solution:-

= (a 5 -3 ) × a 8 … [∵a m ÷ a n = a m – n ] = a 2 × a 8 = a 2 + 8 … [∵a m × a n = a m + n ] = a 10

(xi) (4 5 × a 8 b 3 )/ (4 5 × a 5 b 2 )

Solution:-

= 4 5 – 5 × (a 8 – 5 × b 3 – 2 ) … [∵a m ÷ a n = a m – n ] = 4 0 × (a 3 b) = 1 × a 3 b = a 3 b

(xii) (2 3 × 2) 2

Solution:-

= (2 3 + 1 ) 2 … [∵a m × a n = a m + n ] = (2 4 ) 2 (2 4 ) 2 can be written as = (2) 4 × 2 … [∵(a m ) n = a mn ] = 2 8

3. Say true or false and justify your answer:

(i) 10 × 10 11 = 100 11

Solution:-

Let us consider Left Hand Side (LHS) = 10 × 10 11 = 10 1 + 11 … [∵a m × a n = a m + n ] = 10 12 Now, consider Right Hand Side (RHS) = 100 11 = (10 × 10) 11 = (10 1 + 1 ) 11 = (10 2 ) 11 = (10) 2 × 11 … [∵(a m ) n = a mn ] = 10 22 By comparing LHS and RHS, LHS ≠ RHS Hence, the given statement is false.

(ii) 2 3 > 5 2

Solution:-

Let us consider LHS = 2 3 Expansion of 2 3 = 2 × 2 × 2 = 8 Now, consider RHS = 5 2 Expansion of 5 2 = 5 × 5 = 25 By comparing LHS and RHS, LHS < RHS 2 3 < 5 2 Hence, the given statement is false.

(iii) 2 3 × 3 2 = 6 5

Solution:-

Let us consider LHS = 2 3 × 3 2 Expansion of 2 3 × 3 2 = 2 × 2 × 2 × 3 × 3 = 72 Now, consider RHS = 6 5 Expansion of 6 5 = 6 × 6 × 6 × 6 × 6 = 7776 By comparing LHS and RHS, 72 ≠ 7776 LHS ≠ RHS Hence, the given statement is false.

(iv) 3 0 = (1000) 0

Solution:-

Let us consider LHS = 3 0 = 1 Now, consider RHS = 1000 0 = 1 By comparing LHS and RHS, LHS = RHS 3 0 = 1000 0 Hence, the given statement is true.

4. Express each of the following as a product of prime factors only in exponential form:

(i) 108 × 192

Solution:-

The factors of 108 = 2 × 2 × 3 × 3 × 3 = 2 2 × 3 3 The factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2 6 × 3 Then, = (2 2 × 3 3 ) × (2 6 × 3) = 2 2 + 6 × 3 3 + 1 … [∵a m × a n = a m + n ] = 2 8 × 3 4

(ii) 270

Solution:-

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

(iii) 729 × 64

The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3 6 The factors of 64 = 2 × 2 × 2 × 2 × 2 × 2 = 2 6 Then, = (3 6 × 2 6 ) = 3 6 × 2 6

(iv) 768

Solution:-

The factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2 8 × 3

5. Simplify:

(i) ((2 5 ) 2 × 7 3 )/ (8 3 × 7)

Solution:-

8 3 can be written as = (2 × 2 × 2) 3 = (2 3 ) 3 We have, = ((2 5 ) 2 × 7 3 )/ ((2 3 ) 3 × 7) = (2 5 × 2 × 7 3 )/ ((2 3 × 3 × 7) … [∵(a m ) n = a mn ] = (2 10 × 7 3 )/ (2 9 × 7) = (2 10 – 9 × 7 3 – 1 ) … [∵a m ÷ a n = a m – n ] = 2 × 7 2 = 2 × 7 × 7 = 98

(ii) (25 × 5 2 × t 8 )/ (10 3 × t 4 )

Solution:-

25 can be written as = 5 × 5 = 5 2 10 3 can be written as = 10 3 = (5 × 2) 3 = 5 3 × 2 3 We have, = (5 2 × 5 2 × t 8 )/ (5 3 × 2 3 × t 4 ) = (5 2 + 2 × t 8 )/ (5 3 × 2 3 × t 4 ) … [∵a m × a n = a m + n ] = (5 4 × t 8 )/ (5 3 × 2 3 × t 4 ) = (5 4 – 3 × t 8 – 4 )/ 2 3 … [∵a m ÷ a n = a m – n ] = (5 × t 4 )/ (2 × 2 × 2) = (5t 4 )/ 8

(iii) (3 5 × 10 5 × 25)/ (5 7 × 6 5 )

Solution:-

10 5 can be written as = (5 × 2) 5 = 5 5 × 2 5 25 can be written as = 5 × 5 = 5 2 6 5 can be written as = (2 × 3) 5 = 2 5 × 3 5 Then we have, = (3 5 × 5 5 × 2 5 × 5 2 )/ (5 7 × 2 5 × 3 5 ) = (3 5 × 5 5 + 2 × 2 5 )/ (5 7 × 2 5 × 3 5 ) … [∵a m × a n = a m + n ] = (3 5 × 5 7 × 2 5 )/ (5 7 × 2 5 × 3 5 ) = (3 5 – 5 × 5 7 – 7 × 2 5 – 5 ) = (3 0 × 5 0 × 2 0 ) … [∵a m ÷ a n = a m – n ] = 1 × 1 × 1 = 1 Exercise 13.3 Page: 263

1. Write the following numbers in the expanded forms:

(a) 279404

Solution:-

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 × 10 5 ) + (7 × 10 4 ) + (9 × 10 3 ) + (4 × 10 2 ) + (0 × 10 1 ) + (4 × 10 0 )

(b) 3006194

Solution:-

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 × 10 6 ) + (0 × 10 5 ) + (0 × 10 4 ) + (6 × 10 3 ) + (1 × 10 2 ) + (9 × 10 1 ) + (4 × 10 0 )

(c) 2806196

Solution:-

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 × 10 6 ) + (8 × 10 5 ) + (0 × 10 4 ) + (6 × 10 3 ) + (1 × 10 2 ) + (9 × 10 1 ) + (6 × 10 0 )

(d) 120719

Solution:-

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 × 10 5 ) + (2 × 10 4 ) + (0 × 10 3 ) + (7 × 10 2 ) + (1 × 10 1 ) + (9 × 10 0 )

(e) 20068

Solution:-

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

2. 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

Solution:-

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

Solution:-

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

Solution:-

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

Solution:-

The expanded form is, = (9 × 100000) + (0 × 10000) + (0 × 1000) + (2 × 100) + (3 × 10) + (0 × 1) = 900000 + 0 + 0 + 200 + 30 + 0 = 900230

3. Express the following numbers in standard form:

(i) 5,00,00,000

Solution:-

The standard form of the given number is 5 × 10 7

(ii) 70,00,000

Solution:-

The standard form of the given number is 7 × 10 6

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

Solution:-

The standard form of the given number is 3.1865 × 10 9

(iv) 3,90,878

Solution:-

The standard form of the given number is 3.90878 × 10 5

(v) 39087.8

Solution:-

The standard form of the given number is 3.90878 × 10 4

(vi) 3908.78

Solution:-

The standard form of the given number is 3.90878 × 10 3

4. Express the number appearing in the following statements in standard form.

(a) The distance between Earth and Moon is 384,000,000 m.

Solution:-

The standard form of the number appearing in the given statement is 3.84 × 10 8 m.

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

Solution:-

The standard form of the number appearing in the given statement is 3 × 10 8 m/s.

(c) Diameter of the Earth is 1,27,56,000 m.

Solution:-

The standard form of the number appearing in the given statement is 1.2756 × 10 7 m.

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

Solution:-

The standard form of the number appearing in the given statement is 1.4 × 10 9 m.

(e) In a galaxy, there are, on average, 100,000,000,000 stars.

Solution:-

The standard form of the number appearing in the given statement is 1 × 10 11 stars.

(f) The universe is estimated to be about 12,000,000,000 years old.

Solution:-

The standard form of the number appearing in the given statement is 1.2 × 10 10 years old.
192

(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.

Solution:-

The standard form of the number appearing in the given statement is 3 × 10 20 m.

(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.

Solution:-

The standard form of the number appearing in the given statement is 6.023 × 10 22 molecules.

(i) The Earth has 1,353,000,000 cubic km of seawater.

Solution:-

The standard form of the number appearing in the given statement is 1.353 × 10 9 cubic km.

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

Solution:-

The standard form of the number appearing in the given statement is 1.027 × 10 9 .

NCERT Solutions for Class 7 Maths Chapter 13 FAQs

What are exponents?

Exponents are shorthand notation used to represent repeated multiplication of a number by itself. For example, 2^3 means "2 raised to the power of 3," which equals 2 × 2 × 2 = 8.

What is the difference between exponent and power?

Exponent refers to the small raised number indicating how many times a base number is multiplied by itself, while power refers to the result of raising a base number to an exponent.

How do you read expressions with exponents?

Expressions with exponents are read as "the base raised to the power of the exponent." For example, 3^2 is read as "three raised to the power of two."

What are some properties of exponents?

Some properties of exponents include the product of powers property (a^m × a^n = a^(m+n)), the power of a power property ((a^m)^n = a^(m×n)), and the power of a product property ((ab)^n = a^n × b^n).
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