Question

A fraction becomes $\dfrac{9}{{11}}$ if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes $\dfrac{5}{6}$, find the fraction.

Answer 1

Hint: In the above question, first convert the given word statement into equation, for this we need to assume the fraction that is $\dfrac{x}{y}$. Then we will form an equation from the given information in the question. According to the question we will get two equations. When we solve these equations, we will easily get the value of x and y which we have assumed and then we will get the fraction.

Complete step-by-step answer: Let the numerator be x and denominator be y, therefore the fraction be $\dfrac{x}{y}$
According to the question if 2 is added to both numerator and denominator then the fraction becomes$\dfrac{9}{{11}}$.
Thus, $\dfrac{{x + 2}}{{y + 2}} = \dfrac{9}{{11}}$
$ \Rightarrow 11(x + 2) = 9(y + 2)$
$ \Rightarrow 11x + 22 = 9y + 18$
$ \Rightarrow 11x - 9y = - 4$ …(i)
Now, if 3 is added to both the numerator and denominator then the fraction becomes$\dfrac{5}{6}$.
Thus, $\dfrac{{x + 3}}{{y + 3}} = \dfrac{5}{6}$
$ \Rightarrow 6(x + 3) = 5(y + 3)$
$ \Rightarrow 6x + 18 = 5y + 15$
$ \Rightarrow 6x - 5y = - 3$ …(ii)
Now, solving equations (i) and (ii), using elimination method
Multiplying equation (i) by 5, we get;
$\left( {11x - 9y = - 4} \right) \times 5$
$ \Rightarrow 55x - 45y = - 20$
$55x - 45y + 20 = 0$ … (iii)
Multiplying equation (ii) by 9, we get;
$\left( {6x - 5y = - 3} \right) \times 9$
$ \Rightarrow 54x - 45y = - 27$
$54x - 45y + 27 = 0$ …(iv)
Subtracting equation (iv) from equation (iii), we get;
$55x - 45y + 20 - 54x + 45y - 27 = 0$(Both 45y will be cancelled because they have plus and minus sign)
$\therefore x = 7$
The value of the numerator is 7.
Now, putting the value of x in equation (i), we get;
$11x - 9y = - 4$
$ \Rightarrow 11 \times 7 - 9y = - 4$
$
   \Rightarrow - 9y = - 4 - 77 \\
  \therefore y = 9 \\
 $
The value of the denominator is 9.
The values obtained also satisfy the given statements.
Hence, the required fraction is$\dfrac{7}{9}$.

Note: According to the given question we are supposed to find the fraction. So we will always assume it in the form of $\dfrac{x}{y}$not only ‘x’ because fraction is always in the form of $\dfrac{x}{y}$where x represents numerator and y represents denominator.