Question

The sum of the numerator and denominator of a certain fraction is $11$. If $1$ is added to both numerator and denominator, the fraction increases by $\frac{3}{56}$. Find the fraction.

Answer 1

Step 1: Assume the fraction

Let the numerator of the fraction be $x$.

The denominator of the fraction is $11-x$.

The fraction will be $\frac{x}{11-x}$.

Step 2: According to the question

$\begin{array}{rcl}\frac{x+1}{(11-x)+1}& =& \frac{3}{56}+\frac{x}{11-x}\\ \frac{x+1}{(11-x)+1}-\frac{x}{11-x}& =& \frac{3}{56}\\ \frac{x+1}{(12-x)}-\frac{x}{11-x}& =& \frac{3}{56}\\ \frac{(x+1)(11-x)-x(12-x)}{(12-x)(11-x)}& =& \frac{3}{56}\\ \frac{11x+11-x^2-x-12x+x^2}{132-11x-12x+x^2}& =& \frac{3}{56}\\ \frac{11-2x}{132-23x+x^2}& =& \frac{3}{56}\\ 616-112x& =& 396-69x+3x^2\\ 3x^2+43x-220& =& 0\end{array}$

Step 3: Use the quadratic formula

$\begin{array}{rcl}3x^2+43x-220& =& 0\\ x& =& \frac{-43\pm \sqrt{{43}^2-4\times 3\times (-220)}}{2\times 3}\\ x& =& \frac{-43\pm \sqrt{1849+2640}}{2\times 3}\\ x& =& \frac{-43\pm \sqrt{4489}}{6}\\ x& =& \frac{-43\pm 67}{6}\\ x& =& 4\text{or}\frac{-55}{3}\end{array}$

Neglect the negative value because the numerator is a positive integer.

The numerator of the fraction is $4$.

The denominator of the fraction is $11-4=7$.

Hence, the required fraction is $\frac{4}{7}$.