Question

When 3 is added to the denominator and 2 is subtracted from the numerator a fraction becomes 1/4 when 6 is added to numerator and denominator is multiplied by 3 it becomes 2/3 find the fraction.

Class 10 CBSE
Pair of linear eqautions in two variables.
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Answer 1

Answer:

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Answer 2

GIVEN :-

• When three is added to the denominator and two is subtracted

       from the numerator the fraction becomes  $\rm \dfrac{1}{4}$ .

• When six is added to numerator and denominator is multiplied

       by three the fraction becomes  $\rm\dfrac{2}{3}$ .

TO FIND :-

• The fraction .

SOLUTION :-

    Let ,

     Numerator = x

     Denominator = y

    According to the first condition ,

       $\longrightarrow\sf \dfrac{x-2}{y+3}=\dfrac{1}{4} \\\\\longrightarrow 4(x-2)=y+3 \\\\\longrightarrow 4x-8 =y+3 \\\\\longrightarrow 4x-y = 11 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..................(1)$

    According to the second condition ,

      $\longrightarrow\sf \dfrac{x+6}{3y}=\dfrac{2}{3}\\\\\longrightarrow 3(x+6)=2\times 3y \\\\\longrightarrow 3x+18=6y\\\\\longrightarrow 3x-6y=-18 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..................(2)$

    Eq (1) × 6 ,

      $\longrightarrow\sf 24x-6y=66 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..................(3)$

   Eq (3) - Eq (2) ,

    $\longrightarrow\sf 21x=84\\\\\longrightarrow \bf x = 4$

   

  Substituting $\rm x= 4$ in Eq (1) ,

    $\longrightarrow\sf 4 \times 4 -y = 11\\\\\longrightarrow 16-y=11\\\\\longrightarrow -y = -5\\\\\longrightarrow \bf y = 5$

     $\bf\underline{\underline{ FRACTION = \dfrac{4}{5}}}$