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

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 and thus become an expert in Maths. To strengthen the fundamentals we advise students to practise more of NCERT Solutions for Class 7 Chapter 13.

## Download the PDF of NCERT Solutions For Class 7 Maths Chapter 13 Exponents and Powers â€“ Exercise 13.3

Â

Â

### Access Other Exercises of NCERT Solutions For Class 7 Maths Chapter 13 â€“ Exponents and Powers

Exercise 13.1 Solutions

Exercise 13.2 Solutions

### Access Answers to NCERT Class 7 Maths Chapter 13 â€“ Exponents and Powers Exercise 13.3

1. Write the following numbers in the expanded forms:

279404

Solution:-

The expanded form of the number 279404 is,

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

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

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

3006194

Solution:-

The expanded form of the number 3006194 is,

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

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

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

2806196

Solution:-

The expanded form of the number 2806196 is,

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

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

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

120719

Solution:-

The expanded form of the number 120719 is,

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

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

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

20068

Solution:-

The expanded form of the number 20068 is,

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

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

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

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

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

Solution:-

The expanded form is,

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

= 80000 + 6000 + 0 + 40 + 5

= 86045

(b) (4 Ã—10)5 + (5 Ã— 10)3 + (3 Ã— 10)2 + (2 Ã— 10)0

Solution:-

The expanded form is,

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

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

= 405302

(c) (3 Ã—10)4 + (7 Ã— 10)2 + (5 Ã— 10)0

Solution:-

The expanded form is,

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

= 30000 + 0 + 700 + 0 + 5

= 30705

(d) (9 Ã—10)5 + (2 Ã— 10)2 + (3 Ã— 10)1

Solution:-

The expanded form is,

= (9 Ã— 100000) + (0 Ã— 10000) + (0 Ã— 1000) + (2 Ã— 100) + (3 Ã— 10) + (0 Ã— 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 Ã— 107

(ii) 70,00,000

Solution:-

The standard form of the given number is 7 Ã— 106

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

Solution:-

The standard form of the given number is 3.1865 Ã— 109

(iv) 3,90,878

Solution:-

The standard form of the given number is 3.90878 Ã— 105

(v) 39087.8

Solution:-

The standard form of the given number is 3.90878 Ã— 104

(vi) 3908.78

Solution:-

The standard form of the given number is 3.90878 Ã— 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 Ã— 108m.

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

Solution:-

The standard form of the number appearing in the given statement is 3 Ã— 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 Ã— 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 Ã— 109m.

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

Solution:-

The standard form of the number appearing in the given statement is 1 Ã— 1011 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 Ã— 1010 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.

Solution:-

The standard form of the number appearing in the given statement is 3 Ã— 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 Ã— 1022 molecules.

(i) The earth has 1,353,000,000 cubic km of sea water.

Solution:-

The standard form of the number appearing in the given statement is 1.353 Ã— 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 Ã— 109.

Â