Question

Twice of the numerator of a fraction is $1$ more than the denominator. If $5$ is added to both numerator and denominator, the fraction is $\frac{7}{60}$​ more than the original one. Find the original fraction.

Answer 1

Step 1: Finding the quadratic equation in $x$

Let the numerator of a fraction be $x$.

Given that, twice the numerator of the fraction is $1$ more than the denominator.

So, the denominator of the fraction is $2x-1$.

Also, given if $5$ is added to both numerator and denominator, the fraction is $\frac{7}{60}$​ more than the original one.

So, we have a new equation as

$$\begin{gathered} \frac{x+5}{2x-1+5}=\frac{7}{60}+\frac{x}{2x-1} \\ \Rightarrow \frac{x+5}{2x+4}=\frac{14x-7+60x}{60\left(2x-1\right)} \\ \Rightarrow 60\left(x+5\right)\left(2x-1\right)=\left(74x-7\right)\left(2x+4\right) \\ \Rightarrow 60\left(2x^2-x+10x-5\right)=148x^2+296x-14x-28 \\ \Rightarrow 120x^2+540x-300=148x^2+282-28 \\ \Rightarrow 28x^2-258x+272=0 \\ \Rightarrow 14x^2-129x+136=0 \end{gathered}$$

Step 2: Determining the original fraction

By factorization method,

$$\begin{gathered} 14x^2-129x+136=0 \\ \Rightarrow 14x^2-112x-17x+136=0 \\ \Rightarrow 14x\left(x-8\right)-17\left(x-8\right)=0 \\ \Rightarrow \left(x-8\right)\left(14x-17\right)=0 \\ \Rightarrow x=8,\frac{17}{14} \end{gathered}$$

The numerator of a fraction cannot be $\frac{17}{14}$.

So, the numerator of fraction $x=8$.

and denominator of the fraction $\left(2x-1\right)=15$.

Hence, the fraction is $\frac{8}{15}$.