Question

If 1 is added to both of the numerator and denominator of a fraction, it becomes $\frac{4}{5}$. If however, 5 is subtracted from both numerator and denominator, the fraction $\frac{1}{2}$. Find the fraction.

Answer 1

Let the required fraction be $\frac{x}{y}$.
Then, we have:
$\frac{x+1}{y+1}=\frac{4}{5}$
⇒ 5(x + 1) = 4(y + 1)
⇒ 5x + 5 = 4y + 4
⇒ 5x − 4y = −1 ....(i)

Again, we have:
$\frac{x-5}{y-5}=\frac{1}{2}$
⇒ 2(x − 5) = 1(y − 5)
⇒ 2x − 10 = y − 5
⇒ 2x − y = 5 ....(ii)
On multiplying (ii) by 4, we get:
8x − 4y = 20 ....(iii)
On subtracting (i) from (iii), we get:
3x = (20 − (−1)) = 20 + 1 = 21
⇒ 3x = 21
⇒ x = 7
On substituting x = 7 in (i), we get:
5 × 7 − 4y = −1
⇒ 35 − 4y = −1
⇒ 4y = 36
⇒ y = 9
∴ ​x = 7 and y = 9
Hence, the required fraction is $\frac{7}{9}$.