*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 11.

NCERT Solutions for Class 7 Maths Exercise 13.3 Chapter 13 Exponents and Powers in simple PDF are available here. This exercise of NCERT Solutions for Class 7 Maths Chapter 13 contains topics related to the decimal number system and expressing large numbers in the standard form. Students have to practise these solutions regularly to become an expert in Maths. To strengthen their knowledge of concepts and topics, students can download these NCERT Solutions for Class 7 Chapter 13 in PDF and practise them without any time constraints.

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

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Access Answers to NCERT Class 7 Maths Chapter 13 – Exponents and Powers Exercise 13.3

1. Write the following numbers in the expanded forms:

(a) 279404

Solution:-

The expanded form of the number 279404

= (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

Solution:-

The expanded form of the number 3006194

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

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

Solution:-

The expanded form of the number 2806196

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

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

Solution:-

The expanded form of the number 120719

= (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

Solution:-

The expanded form of the number 20068

= (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).

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

= (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

= (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

= (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 × 107

(ii) 70,00,000

Solution:-

The standard form of the given number is 7 × 106

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

Solution:-

The standard form of the given number is 3.1865 × 109

(iv) 3,90,878

Solution:-

The standard form of the given number is 3.90878 × 105

(v) 39087.8

Solution:-

The standard form of the given number is 3.90878 × 104

(vi) 3908.78

Solution:-

The standard form of the given number is 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.

Solution:-

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

(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 × 108m/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 × 107m.

(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 × 109m.

(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 × 1011.

(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 × 1010.

(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 × 1020m.

(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 × 1022.

(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 × 109 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 × 109.

Also, explore – 

NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 

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