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

Let the fraction $=\frac{x}{y}$

Then, According to the given question,

$2x=y+1$

$x=\frac{y+1}{2}........\left(1\right)$

Now, again according to the given question,

$\frac{x+5}{y+5}=\frac{7}{60}+\frac{x}{y}$

$\Rightarrow \frac{x+5}{y+5}=\frac{7y+60x}{60y}$

$\Rightarrow 60xy+300y=7y^2+60xy+35y+300x$

$\Rightarrow 7y^2+35y+300x=300y$

$\Rightarrow 7y^2+35y+300x-300y=0$

Put the value of $x$ by equation (1) to (2) and we get,

$7y^2+35y+300\left(\frac{y+1}{2}\right)-300y=0$

$\Rightarrow 7y^2+35y+150y+150-300y=0$

$\Rightarrow 7y^2-115y+150=0$

$\Rightarrow 7y^2-(105+10)y+150=0$

$\Rightarrow 7y^2-105y-10y+150=0$

$\Rightarrow 7y(y-15)-10(y-15)=0$

$\Rightarrow (y-15)(7y-10)=0$

$\Rightarrow y-15=07y-10=0$

$\Rightarrow y=15,y=\frac{10}{7}$

Put the value of $y=15$ in equation (1) and we get,

$x=\frac{y+1}{2}$

$x=\frac{15+1}{2}$

$x=8$

Hence, the fraction is $=\frac{x}{y}=\frac{8}{15}$ ans.