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 the new fraction and the original fraction is ’$\frac{29}{20}$'. Find the original fraction.

Answer 1

Let the denominator be '$\mathrm{x}$'.

Here, it is given that, the numerator is $3$ less than the denominator, So, the numerator$=\mathrm{x}-3$

$\therefore$ The fraction$=\frac{\mathrm{x}-3}{\mathrm{x}}$

It is given that if '$2$' is added to both the numerator and denominator, then the sum of the new fraction and the original fraction is $\frac{29}{20}$

$\therefore$Adding '$2$' to the numerator and denominator, the new fraction becomes, $\frac{\mathrm{x}-3+2}{\mathrm{x}+2}=\frac{\mathrm{x}-1}{\mathrm{x}+2}$

$\therefore \frac{\mathrm{x}-3}{\mathrm{x}}+\frac{\mathrm{x}-1}{\mathrm{x}+2}=\frac{29}{20}$

$\begin{array}{rcl}\frac{\left(\mathrm{x}-3\right)\left(\mathrm{x}+2\right)+\left(\mathrm{x}-1\right)\mathrm{x}}{\mathrm{x}\left(\mathrm{x}+2\right)}& =& \frac{29}{20}\\ 20\left[{\mathrm{x}}^2+2\mathrm{x}-3\mathrm{x}-6+{\mathrm{x}}^2-\mathrm{x}\right]& =& 29\left[{\mathrm{x}}^2+2\mathrm{x}\right]\\ 20\left[2{\mathrm{x}}^2-2\mathrm{x}-6\right]& =& 29{\mathrm{x}}^2+58\mathrm{x}\\ 40{\mathrm{x}}^2-40\mathrm{x}-120& =& 29{\mathrm{x}}^2+58\mathrm{x}\\ 11{\mathrm{x}}^2-98\mathrm{x}-120& =& 0\\ 11{\mathrm{x}}^2-110\mathrm{x}+12\mathrm{x}-120& =& 0\\ \left(11\mathrm{x}+12\right)\left(\mathrm{x}-10\right)& =& 0\end{array}$

$\therefore \mathrm{x}=\frac{-12}{11}$ or $\mathrm{x}=10$

Here, we will reject '$\mathrm{x}=\frac{-12}{11}$'and will accept $\mathrm{x}=10$

So, $\frac{\mathrm{x}-3}{\mathrm{x}}=\frac{7}{10}$

Therefore, the original fraction is $\frac{7}{10}$.