# Class 7 Math NCERT Solutions for Chapter – 13 Exponents and Powers Ex – 13.3

## Exponents and Powers

1. Write the following numbers in the expanded forms:

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

3006194

Solution:-

The expanded form of the number 3006194 is,

= (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)

2806196

Solution:-

The expanded form of the number 2806196 is,

= (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)

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

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 × 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 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 × 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 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 an average 100,000,000,000 stars.

Solution:-

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