# Express the following in the form

Question:

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

(i) $0.2$

(ii) $0.888 \ldots$

(iii) $5 . \overline{2}$

(iv) $0 . \overline{001}$

(v) $0.2555 \ldots$

(vi) $0.1 \overline{34}$

(vii) $0.00323232 \ldots \quad$

(viii) $0.404040 \ldots$

Solution:

(i) Let            $x=0.2=\frac{2}{10}=\frac{1}{5}$

(ii) Let            $x=0.888 \ldots \ldots \ldots$              $\ldots$ (i)

On multiplying both sides of Eq.(i) by 10 , we get

$10 x=8.888$                  ... (ii)

On subtracting Eq. (i) from Eq. (ii), we get

$10 x-x=(8.88)-(0.888)$

$\Rightarrow$                      $9 x=8$

$\therefore$                        $x=\frac{8}{9}$

(iii) Let                     $x=5 . \overline{2}=5.222 \ldots$ $\ldots$ (i)

On multiplying both sides of Eq. (i) by 10 , we get

$10 x=52222 \ldots \ldots \ldots$$\ldots$ (ii)

On subtracting Eq. (i) from Eq. (ii), we get

$10 x-x=(52.222 \ldots)-(5.222 \ldots)$

$\Rightarrow$               $9 x=47$

$\therefore$                    $x=\frac{47}{9}$

(iv) Let $x=0 . \overline{001}$

$\Rightarrow$             $x=0 . \overline{001}=0.001001$           $\ldots$ (i)

On multiplying both sides of Eq. (i) by 1000 , we get

$1000 x=001.001 \ldots \ldots . .$                      .......(ii)

On multiplying both sides of Eq. (ii) by 100 , we get

$1000 x=134.3434 \ldots \ldots \ldots$             ...(iii)

On subtracting Eq. (ii) from Eq. (iii), we get

$1000 x-10 x=134.34 \ldots-(1.3434 \ldots)$

$\Rightarrow$             $990 x=133$

$\therefore$                 $x=\frac{133}{990}$

(vii) Let                    $x=0.00323232$           $\ldots($ i)

On multiplying both sides of Eq. (i) by 100 we get

$100 x=0.3232 \ldots \ldots \ldots$                           ...(ii)

On multiplying both sides of Eq. (ii) by 100 , we get

$10000 x=32.3232 \ldots \ldots . .$                ...........(iii)

On subtracting Eq. (ii) from Eq. (iii), we get

$10000 x-100 x=32.32 \ldots \ldots-0.3232 \ldots$

$\Rightarrow$        $9900 x=32$

$\therefore$           $x=\frac{32}{9900}=\frac{8}{2475}$            [dividing numerator and denominator by 4 ]

(viii) Let                  $x=0.404040$              $\ldots$ (i)

On multiplying both sides of Eq. (i) by 100 , we get

$100 x=40.4040$            ...(ii)

On subtracting Eq. (i) from Eq. (ii), we get

$100 x-x=40.4040 \ldots \ldots-0.404040 \ldots .$

$\Rightarrow$        $99 x=40$

$\therefore$              $x=\frac{40}{99}$