## 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

**(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

**(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}.

**(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 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.

**(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 =

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 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}

**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

**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 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 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 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 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 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 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 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 same exponents = a^{m }× b^{m} = ab^{m}

Then,

= (a × b)^{4}

= ab^{4}

**(viii) (3 ^{4})^{3}**

**Solution:-**

By the rule of taking 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 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 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 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}

(5^{2})^{4} 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}

= 2a^{2} … [∵(a^{m})^{n }= a^{mn}]

**(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:**

**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})

**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 × 10^{6}) + (0 × 10^{5}) + (0 × 10^{4}) + (6 × 10^{3}) + (1 × 10^{2}) + (9 × 10^{1}) + (4 × 10^{0})

**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 × 10^{6}) + (8 × 10^{5}) + (0 × 10^{4}) + (6 × 10^{3}) + (1 × 10^{2}) + (9 × 10^{1}) + (6 × 10^{0})

**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})

**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 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 an 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.

**(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 sea water.**

**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}.

## Frequently Asked Questions on NCERT Solutions for Class 7 Maths Chapter 13

### Where to download NCERT Solutions for Class 7 Maths Chapter 13?

NCERT Solutions for Class 7 Maths Chapter 13 can be downloaded at BYJU’S website. It can be avail in free PDF. To download go to BYJU’S NCERT Solutions website/select the Class 7/opt the subject as Maths/click on to the desired chapter that is Chapter 13 Some Applications of Trigonometry.

### What are the applications of learning NCERT Solutions for Class 7 Maths Chapter 13?

Applications of learning NCERT Solutions for Class 7 Maths Chapter 13 are finding the

1. How are Expressions Formed

2. Terms of An Expression

3. Like and Unlike Terms

4. Monomials, Binomials, Trinomials and Polynomials

5. Addition and Subtraction of Algebraic Expressions

6. Finding The Value of An Expression

7. Using Algebraic Expressions – Formulas and Rules.

By learning these concepts students will be able to answer all the questions based on algebraic expressions as well as it may help in writing class tests and board exams.

### Is NCERT Solutions for Class 7 Maths Chapter 13 enough for board exams?

Yes, you have to learn all the questions thoroughly and practice them for more time. Once you are done with solving all the questions then you can refer to other reference books and questions provided NCERT Exemplar textbooks.