Question

The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by $\frac{1}{15}$. Find the fraction. [CBSE 2012]

Answer 1

Let the denominator of the required fraction be x.

Numerator of the required fraction = x − 3

∴ Original fraction = $\frac{x-3}{x}$
If 1 is added to the denominator, then the new fraction obtained is $\frac{x-3}{x+1}$.

According to the given condition,
$$\begin{gathered} \frac{x-3}{x+1}=\frac{x-3}{x}-\frac{1}{15} \\ \Rightarrow \frac{x-3}{x}-\frac{x-3}{x+1}=\frac{1}{15} \\ \Rightarrow \frac{\left(x-3\right)\left(x+1\right)-x\left(x-3\right)}{x\left(x+1\right)}=\frac{1}{15} \\ \Rightarrow \frac{x^2-2x-3-x^2+3x}{x^2+x}=\frac{1}{15} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{x-3}{x^2+x}=\frac{1}{15} \\ \Rightarrow x^2+x=15x-45 \\ \Rightarrow x^2-14x+45=0 \\ \Rightarrow x^2-9x-5x+45=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x\left(x-9\right)-5\left(x-9\right)=0 \\ \Rightarrow \left(x-5\right)\left(x-9\right)=0 \\ \Rightarrow x-5=0\mathrm{or}x-9=0 \\ \Rightarrow x=5\mathrm{or}x=9 \end{gathered}$$

When x = 5,
$\frac{x-3}{x}=\frac{5-3}{5}=\frac{2}{5}$

When x = 9,
$\frac{x-3}{x}=\frac{9-3}{9}=\frac{6}{9}=\frac{2}{3}$ (This fraction is neglected because this does not satisfies the given condition.)

Hence, the required fraction is $\frac{2}{5}$.