Question

If the numerator of a fraction is
increased by 2 and the denominator is
decreased by 1 it becomes 2/3. If the numerator is increased by 1 and
the denominator is increased by 2, it
becomes 1/3. Find the fraction.

Answer 1

★ Solution :-

Let the numerator be x.

Let the denominator be y.

So, the fraction is,

$\sf \leadsto \dfrac{x}{y}$

We are given that, the numerator is increased by 2 and denominator is decreased by 1. So,

According to the first case,

$\sf \leadsto \dfrac{x + 2}{y - 1} = \dfrac{2}{3}$

$\sf \leadsto 3(x + 2) = 2(y - 1)$

$\sf \leadsto 3x + 6 = 2y - 2$

$\sf \leadsto 3x - 2y = -2 - 6$

$\sf \leadsto 3x - 2y = -8 \: \: --- (i)$

We are also given that the numerator is increased by 1 and the denominator is increased by 2. So,

According to the second case,

$\sf \leadsto \dfrac{x + 2}{y + 2} = \dfrac{1}{3}$

$\sf \leadsto 3(x + 1) = 1(y + 2)$

$\sf \leadsto 3x + 3 = y + 2$

$\sf \leadsto 3x - y = 2 - 3$

$\sf \leadsto 3x - y = -1 \: \: --- (ii)$

By first equation,

$\sf \leadsto 3x - 2y = -8$

$\sf \leadsto 3x = -8 + 2y$

$\sf \leadsto x = \dfrac{-8 + 2y}{3}$

Now, we can find the original value of y.

$\sf \leadsto 3x - y = -1$

$\sf \leadsto 3 \bigg( \dfrac{-8 + 2y}{3} \bigg) - y = -1$

$\sf \leadsto \dfrac{-24 + 6y}{3} - y = -1$

$\sf \leadsto \dfrac{-24 + 6y - 3y}{3} = -1$

$\sf \leadsto \dfrac{-24 + 3y}{3} = -1$

$\sf \leadsto -24 + 3y = -3$

$\sf \leadsto 3y = -3 + 24$

$\sf \leadsto 3y = 21$

$\sf \leadsto y = \dfrac{21}{3}$

$\sf \leadsto y = 7$

Now, we can find the original value of y.

$\sf \leadsto 3x - 2y = -8$

$\sf \leadsto 3x - 2(7) = -8$

$\sf \leadsto 3x - 14 = -8$

$\sf \leadsto 3x = -8 + 14$

$\sf \leadsto 3x = 6$

$\sf \leadsto x = \dfrac{6}{3}$

$\sf \leadsto x = 2$

We know that, the values of x and y are 2 and 7 respectively. So, the fraction is 2/7.

Therefore, the original rational number is 2/7.