Question

# A fraction becomes $\frac{1}{3}$ when $1$ is subtracted from the numerator and it becomes $\frac{1}{4}$ when $8$ is added to its denominator. Find the fraction.

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Solution

## Let's find the fraction with given details.Let us assume the numerator to be $x$ and the denominator to be $y$.Hence the fraction becomes = $\frac{x}{y}$According to the given condition$\frac{\left(x-1\right)}{y}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}⇒y=\left(3x-3\right)\to \left(1\right)$According to the second given condition,$\frac{x}{\left(y+8\right)}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒y+8=4x\phantom{\rule{0ex}{0ex}}⇒y=4x-8\to \left(2\right)$Combining (1) and (2) we get,$\begin{array}{rcl}3x-3& =& 4x-8\\ & ⇒& 3x-4x=-8+3\\ -x& =& -5\\ x& =& 5\end{array}$from (1)$y=15-3\phantom{\rule{0ex}{0ex}}y=12$So, the fraction is $\frac{5}{12}$Hence, the fraction is $\frac{5}{12}$.

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