Question

The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1 , the numerator becomes half the denominator. Determine the fraction.

Answer 1

Let the numerator $$ =x $$ and the denominator $$ =y $$
So, the fraction $$ =\dfrac{x}{y} $$
According to the question,
Condition I:
$$ x+y=2 y-3 $$
$$ \Rightarrow x+y-2 y=-3 $$
$$ \Rightarrow x-y=-3 \ldots$$(i)
Condition II:
$$ \dfrac{x-1}{y-1}=\dfrac{1}{2} $$
$$ \Rightarrow 2(x-1)=y-1 $$
$$ \Rightarrow 2 x-2=y-1 $$
$$ \Rightarrow 2 x-y=1 \ldots(\text { ii }) $$
On subtracting Eq.(i) from Eq. (ii), we get
$$ 2 x-y-x+y=1+3 $$
$$ \Rightarrow x=4 $$
On putting the value of x in Eq. (i), we get
$$ 4-y=-3 $$
$$ \Rightarrow y=7 $$
So, the numerator is 4 and the denominator is 7 Hence, the fraction is
$$ \dfrac{4}{7} $$