Question

A fraction becomes 1/3 if 1 is subtracted from both its numerator and denominator. It 1 is added to both the numerator and denominator, it becomes 1/2. Find the fraction.

Answer 1

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is ${\displaystyle \frac{x}{y}}$

If 1 is subtracted from both numerator and the denominator, the fraction becomes ${\displaystyle \frac{1}{3}}$. Thus, we have

${\displaystyle \frac{x-1}{y-1}=\frac{1}{3}}$

${\displaystyle \Rightarrow 3(x-1)=y-1}$

${\displaystyle \Rightarrow 3x-3=y-1}$

${\displaystyle \Rightarrow 3x-y-2=0}$

If 1 is added to both numerator and the denominator, the fraction becomes ${\displaystyle \frac{1}{2}}$. Thus, we have

${\displaystyle \frac{x+1}{y+1}=\frac{1}{2}}$

${\displaystyle \Rightarrow 2(x+1)=y+1}$

${\displaystyle \Rightarrow 2x+2=y+1}$

${\displaystyle \Rightarrow 2x-y+1=0}$

So, we have two equations

${\displaystyle 3x-y-2=0}$

${\displaystyle 2x-y+1=0}$

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

${\displaystyle \frac{x}{(-1)\times 1-(-1)\times (-2)=}}$ ${\displaystyle \frac{-y}{3\times 1-2\times 2(-2)}=}$ ${\displaystyle \frac{1}{3\times (-1)-2\times (-1)}}$

${\displaystyle \Rightarrow \frac{x}{-1-2}=\frac{-y}{3+4}=\frac{1}{-3+2}}$

${\displaystyle \Rightarrow \frac{x}{-3}=\frac{-y}{7}=\frac{1}{-1}}$

${\displaystyle \Rightarrow \frac{x}{3}=\frac{y}{7}=1}$

${\displaystyle \Rightarrow x=3,y=7}$

Hence, the fraction is ${\displaystyle \frac{3}{7}}$