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.

Answer 1

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

$$\begin{gathered} \frac{(x-1)}{y}=\frac{1}{3} \\ \Rightarrow y=\left(3x-3\right)\to \left(1\right) \end{gathered}$$

According to the second given condition,

$$\begin{gathered} \frac{x}{(y+8)}=\frac{1}{4} \\ \Rightarrow y+8=4x \\ \Rightarrow y=4x-8\to \left(2\right) \end{gathered}$$

Combining (1) and (2) we get,

$\begin{array}{rcl}3x-3& =& 4x-8\\ & \Rightarrow & 3x-4x=-8+3\\ -x& =& -5\\ x& =& 5\end{array}$

from (1)

$$\begin{gathered} y=15-3 \\ y=12 \end{gathered}$$

So, the fraction is $\frac{5}{12}$

Hence, the fraction is $\frac{5}{12}$.