Question

the numerator of a fraction is 3 less than its denominator . if 2 is added to both the numerator and denominator , then the sum of new fraction and original fraction is 29/20 . find the original fraction

Answer 1

Let the numerator be x and denominator be y.
Then, fraction will be $\frac{x}{y}$.
According to the question,
$y=x+3$ ...(i)
$$\begin{gathered} \frac{x+2}{y+2}+\frac{x}{y}=\frac{29}{20} \\ \Rightarrow \frac{xy+2y+xy+2x}{y^2+2y}=\frac{29}{20} \\ \Rightarrow \frac{2x\left(x+3\right)+2\left(x+3\right)+2x}{{\left(x+3\right)}^2+2\left(x+3\right)}=\frac{29}{20}\left[\mathrm{From}\mathrm{equation}\left(\mathrm{i}\right)\right] \\ \Rightarrow \frac{2x^2+6x+2x+6+2x}{x^2+6x+9+2x+6}=\frac{29}{20} \\ \Rightarrow 20\left(2x^2+10x+6\right)=29\left(x^2+8x+15\right) \\ \Rightarrow 40x^2+200x+120=29x^2+232x+435 \\ \Rightarrow 11x^2-32x-315=0 \\ \Rightarrow \left(11x+45\right)\left(x-7\right)=0 \\ \Rightarrow x=-\frac{45}{11},7 \\ \mathrm{If}x=-\frac{45}{11}\mathrm{then}y=-\frac{45}{11}+3=-\frac{12}{11}\mathrm{and}\mathrm{fraction}=\frac{45}{12}\left(\mathrm{Which}\mathrm{dose}\mathrm{not}\mathrm{satisfiy}\mathrm{the}\mathrm{condition}\mathrm{given}\mathrm{in}\mathrm{the}\mathrm{question}\right) \\ \mathrm{If}x=7\mathrm{then}y=7+3=10\mathrm{and}\mathrm{fraction}=\frac{7}{10}. \\ \mathrm{Hence},\mathrm{the}\mathrm{original}\mathrm{fraction}=\frac{7}{10} \end{gathered}$$