When 3 is added to the denominator and 2 is subtracted from the numerator a fraction becomes $\frac{1}{4}$. And, when 6 is added to numerator and the denominator is multiplied by 3, it becomes $\frac{2}{3}$. Find the fraction.

Given:

When 3 is added to the denominator and 2 is subtracted from the numerator a fraction becomes $\frac{1}{4}$. And, when 6 is added to numerator and the denominator is multiplied by 3, it becomes $\frac{2}{3}$. 

 To do:

We have to find the original fraction.

Solution:

Let the numerator and denominator of the original fraction be $x$ and $y$ respectively.

The original fraction$=\frac{x}{y}$

The fraction becomes $\frac{1}{4}$ when 3 is added to the denominator and 2 is subtracted from the numerator. 

This implies,

New fraction$=\frac{x-2}{y+3}$

According to the question,

$\frac{x-2}{y+3}=\frac{1}{4}$

$4(x-2)=1(y+3)$    (On cross multiplication)

$4x-8=y+3$

$y=4x-8-3$

$y=4x-11$.....(i)

When 6 is added to numerator and the denominator is multiplied by 3, it becomes $\frac{2}{3}$.

This implies,

$\frac{x+6}{3\times y}=\frac{2}{3}$

$3(x+6)=2(3y)$    (On cross multiplication)

$3x+18=6y$

$3x-6y+18=0$

$3x-6(4x-11)+18=0$     (From (i))

$3x-24x+66+18=0$

$-21x+84=0$

$21x=84$

$x=\frac{84}{21}$

$x=4$

$\Rightarrow y=4(4)-11$

$y=16-11$

$y=5$

Therefore, the original fraction is $\frac{4}{5}$.  

Tutorialspoint

Updated on: 10-Oct-2022

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