# NCERT Solutions for Class 7 Maths Chapter 11 Exponents and Powers

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers are given here in detail. Maths is a subject which requires understanding and reasoning skills accompanied by logic. Along with this, students should practise solving Maths problems regularly.

**Exercise 11.1**

**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

55

**(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}.

105

**(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.

**(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.

99

**(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

**(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

**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}

**NCERT Solutions for Class 7 Maths Chapter 10 Algebraic Expressions**

**Exercise 11.2 **

**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}

**(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.

**THE NATURE CONSERVANCY: NURTURING A SUSTAINABLE FUTURE**

**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

153

**Exercise 11.3 **

**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) The 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) The 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) The 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}.