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 $\frac{4}{5}$.
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