Question

A fraction becomes equal to $\dfrac{4}{5}$ if $1$ is added to both the numerator and the denominator. If, however, $5$ is subtracted from both the numerator and denominator, the fraction becomes equal to $\dfrac{1}{2}$. What is the fraction?

Answer 1

Hint: Approach the solution by considering a fraction and proceed with further simplification by applying the given conditions.

Let us consider the fraction as $\dfrac{x}{y}$
By applying the ${1^{st}}$ condition we will get
$ \Rightarrow \dfrac{{x + 1}}{{y + 1}} = \dfrac{4}{5}$
On further simplification we will get
$\
   \Rightarrow 5\left( {x + 1} \right) = 4\left( {y + 1} \right) \\
   \Rightarrow 5x + 5 = 4y + 4 \\
\ $
$ \Rightarrow 5x - 4y = - 1$$ \to (1)$
And now by applying ${2^{nd}}$condition we will get
$\
   \Rightarrow \dfrac{{x - 5}}{{y - 5}} = \dfrac{1}{2} \\
    \\
\ $
On further simplification we will get
$\
   \Rightarrow 2(x - 5) = 1(y - 5) \\
   \Rightarrow 2x - 10 = y - 5 \\
   \Rightarrow 2x - y = - 5 + 10 \\
\ $
$2x - y = 5 \to (2)$
For further calculation multiply equation $(2) \times 4$
$\
   \Rightarrow 4 \times (2x - y = 5) \\
   \Rightarrow 8x - 4y = 20 \to (3) \\
\ $
Now subtract equation $(3)$ from equation $(1)$ we get
$x = 7$
Putting $x = 7$ in equation $(2)$ we get
$\
   \Rightarrow 2x - y = 5 \\
   \Rightarrow 2(7) - y = 5 \\
   \Rightarrow y = 9 \\
\ $
Here we got both $x\& y$ values
Therefore required fraction is $\dfrac{x}{y} = \dfrac{7}{9}$
Note: Apply the conditions in a step-by-step process with the proper approach to get the answer as the given problem is full of simplification.