Question

If $3$ is subtracted from the numerator as well as the denominator of a fraction, the reduced form of the new fraction so obtained is $\frac{4}{9}$. If $3$ is added to the numerator as well as the denominator of the same fraction, the reduced form of the new fraction so obtained is $\frac{2}{3}$. Find the original fraction.

Answer 1

Let the fraction be $\frac{N}{D}$

$3$ is subtracted to both of the numerator and denominator

$⟹\frac{N-3}{D-3}=\frac{4}{9}9N-27=4D-12N=\frac{4D+15}{9}$

$3$ is added to both of the numerator and denominator

$⟹\frac{N+3}{D+3}=\frac{2}{3}3N+9=2D+63\left(\frac{4D+15}{9}\right)+9=2D+64D+15+27=6D+182D=24D=12N=\frac{48+15}{9}N=\frac{63}{9}=7$

So the original fraction is $\frac{7}{12}$