# Express each of the following decimals in the form

Question:

Express each of the following decimals in the form $\frac{p}{a}$, where $p, q$ are integers and $q \neq 0$.

(i) $0 . \overline{2}$

(ii) $0 . \overline{53}$

(iii) $2 . \overline{93}$

(iv) $18 . \overline{48}$

(v) $0 . \overline{235}$

(vi) $0.00 \overline{32}$

(vii) 1. $3 \overline{23}$

(viii) $0.3 \overline{178}$

(ix) $32.12 \overline{35}$

(x) $0.40 \overline{7}$

Solution:

(i) $0 . \overline{2}$

Let x = 0.222...                               .....(i)
Only one digit is repeated so, we multiply x by 10.
10x = 2.222...                                 .....(ii)
Subtracting (i) from (ii) we get

$9 x=2$

$\Rightarrow x=\frac{2}{9}$

(ii) $0 . \overline{53}$

Let x = 0.5353...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 53.5353...                                 .....(ii)
Subtracting (i) from (ii) we get

$99 x=53$

$\Rightarrow x=\frac{53}{99}$

(iii) $2 . \overline{93}$

Let x = 2.9393...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 293.9393...                                 .....(ii)
Subtracting (i) from (ii) we get

$99 x=291$

$\Rightarrow x=\frac{291}{99}=\frac{97}{33}$

(iv) $18 \cdot \overline{48}$

Let x = 18.4848...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 1848.4848...                                 .....(ii)
Subtracting (i) from (ii) we get

$99 x=1830$

$\Rightarrow x=\frac{1830}{99}=\frac{610}{33}$

(v) $0 . \overline{235}$

Let x = 0.235235...                               .....(i)
Three digits are repeated so, we multiply x by 1000.
1000x = 235.235235...                                 .....(ii)
Subtracting (i) from (ii) we get

$999 x=235$

$\Rightarrow x=\frac{235}{999}$

(vi) $0.00 \overline{32}$

Let x = 0.003232...                               .....(i)
we multiply x by 100.
100x = 0.3232...                                 .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 32.3232...                           .....(iii)
Subtracting (ii) from (iii) we get

$9900 x=32$

$\Rightarrow x=\frac{32}{9900}=\frac{8}{2475}$

(vii) $1.3 \overline{23}$

Let x = 1.32323...                               .....(i)
we multiply x by 10.
10x = 13.2323...                                 .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
1000x = 1323.2323...                           .....(iii)
Subtracting (ii) from (iii) we get

$990 x=1310$

$\Rightarrow x=\frac{131}{99}$

(viii) $0.3 \overline{178}$

Let x = 0.3178178...                               .....(i)
we multiply x by 10.
10x = 3.178178...                                 .....(ii)
Again multiplying by 1000 as there are 3 repeating numbers after decimals we get
10000x = 3178.178178...                           .....(iii)
Subtracting (ii) from (iii) we get

$9990 x=3175$

$\Rightarrow x=\frac{3175}{9990}=\frac{635}{1998}$

(ix) $32.12 \overline{35}$

Let x = 32.123535...                               .....(i)
we multiply x by 100.
100x = 3212.3535...                                 .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 321235.35...                           .....(iii)
Subtracting (ii) from (iii) we get

$9900 x=318023$

$\Rightarrow x=\frac{318023}{9900}$

(x) $0.40 \overline{7}$

Let x = 0.40777...                               .....(i)
we multiply x by 100.
100x = 40.7777...                                 .....(ii)
Again multiplying by 10 as there is 1 repeating number after decimals we get
1000x = 407.777...                           .....(iii)
Subtracting (ii) from (iii) we get

$900 x=367$

$\Rightarrow x=\frac{367}{900}$