Convert the following decimal numbers in the form of p/q. Class 9

Convert the following decimal numbers in the form of p/q. Class 9

Question 1.

Convert the following decimal numbers in the form of $\frac{p}{q}$. Class 9

i. 12.6

ii. 0.0120

iii. $3.0 \overline{52}$

iv. $1.2 \overline{35}$

v. $0 . \overline{23}$

vi. $2.0 \overline{5}$

vii. $2.12 \overline{5}$

viii. $3.12 \overline{5}$

ix. $2 . \overline{1625}$

Solution:

i. 12.6

12.6 = $\frac{126}{10}=\frac{63}{5}$

∴ 12.6 = $\frac{63}{5}$

ii. 0.0120

∴ 0.0120 = 0.012

= $\frac{12}{1000}$

∴ 0.0120 = $\frac{3}{250}$

iii. $3.0 \overline{52}$

Let x = $3.0 \overline{52}$

Multiplying both sides by 10, we get

10x = $3.0 \overline{52}$ ...(i)

Multiplying by 100 on both sides, we get

1000x = $3052 . \overline{52}$ ...(ii)

Subtracting (i) from (ii), we get

1000x - 10x = $3052 . \overline{52}$ - $3.0 \overline{52}$

∴ 990x = 3022

∴ x = $\frac{3022}{990}=\frac{1511}{495}$

∴ $3.0 \overline{52}$ = $\frac{1511}{495}$

iv. $1.2 \overline{35}$

Let x = $1.2 \overline{35}$

Multiplying by 10 on both sides, we get

10x = $12 . \overline{35}$ ...(i)

Now multiplying by 100 on both sides, we get

1000x = $1235 . \overline{35}$ ...(ii)

Subtracting (i) from (ii), we get

1000x - 10x = $1235 . \overline{35}$ - $1.2 \overline{35}$

∴ 990x = 1223

∴ x = $\frac{1223}{990}$

∴ $1.2 \overline{35}=\frac{1223}{990}$

v. $0 . \overline{23}$

Let x = $0 . \overline{23}$ ...(i)

Multiplying by 100 on both sides, we get

100x = $23 . \overline{23}$ ...(ii)

Subtracting (i) from (ii), we get

100x - x = $23 . \overline{23}$ - $0 . \overline{23}$

∴ 99x = 23

∴ x = $\frac{23}{99}$

∴ $0 . \overline{23}=\frac{23}{99}$

vi. $2.0 \overline{5}$

Let x = $2.0 \overline{5}$

Multiplying by 10 on both sides, we get

10x = $20 . \overline{5}$ ...(i)

Multiplying again by 10 on both sides, we get

100x = $205 . \overline{5}$ ...(ii)

Subtracting (i) from (ii), we get

100x - 10x = $205 . \overline{5}$ - $2.0 \overline{5}$

∴ 90x = 185

∴ x = $\frac{185}{90}=\frac{37}{18}$

∴ $2.0 \overline{5}=\frac{37}{18}$

vii. $2.12 \overline{5}$

Let x = $2.12 \overline{5}$

Multiplying by 100 on both sides, we get

100x = $212 . \overline{5}$ ...(i)

Multiplying by 10 on both sides, we get

1000x = $2125 . \overline{5}$ ...(ii)

Subtracting (i) from (ii), we get

1000x - 100x = $2125 . \overline{5}$ - $212 . \overline{5}$

∴ 900x = 1913

∴ x = $\frac{1913}{900}$

∴ $2.12 \overline{5}=\frac{1913}{900}$

viii. $3.12 \overline{5}$

Let x = $3.12 \overline{5}$

Multiplying by 100 on both sides, we get

100x = $312 . \overline{5}$ ...(i)

Multiplying by 10 on both sides, we get

1000x = $3125 . \overline{5}$ ...(ii)

Subtracting (i) from (ii), we get

1000x - 100x = $3125 . \overline{5}$ - $312 . \overline{5}$

∴ 900x = 2813

∴ x = $\frac{2813}{900}$

∴ $3.12 \overline{5}=\frac{2813}{900}$

ix. $2 . \overline{1625}$

Let x = $2 . \overline{1625}$ ...(i)

There are 4 repeating digits.

Multiplying by 10⁴ = 10000, we get

10000x = $21625 . \overline{1625}$ ...(ii)

Subtracting (ii) from (i), we get

10000x - x = $21625 . \overline{1625}$ - 2.1625 9999x = $2 . \overline{1625}$

∴ 9999x = 21623

∴ x = $\frac{21623}{9999}$

∴ $2 . \overline{1625}=\frac{21623}{9999}$