Question

If $2$ is added to the numerator and denominator of a fraction, it becomes $9/10.$ If $3$ is subtracted from the numerator and denominator of the same fraction, it becomes $4/5.$ What is the fraction?

• A. $\frac{5}{6}$
• B. $\frac{6}{7}$
• C. $\frac{7}{8}$
• D. $\frac{8}{9}$

Answer 1

The correct option is C $\frac{7}{8}$

Let the fraction is $\frac{x}{y}$

According to the condition 1:
When $2$ is added to the numerator and denominator of a fraction, then it equals $\frac{9}{10}$
$\frac{x+2}{y+2}=\frac{9}{10}$
$\Rightarrow 10(x+2)=9(y+2)\Rightarrow 10x+20=9y+18\Rightarrow 10x-9y=-2\dots \left(i\right)$

According to the condition 2:
When $3$ is subtracted from the numerator and denominator of the same fraction, then it equals $\frac{4}{5}$
$\frac{x-3}{y-3}=\frac{4}{5}$
$5(x-3)=4(y-3)5x-15=4y-125x-4y=3$
Multilying both the sides by $2$
$2(5x-4y)=2\left(3\right)$
$10x-8y=6\dots \left(ii\right)$

Now, subtracting $\left(ii\right)from\left(i\right)$
$10x-9y=-2$
$10x-8y=6$
(-)_____________
$-y=-8$
$-y=-8$
$y=8$

Substituting the value of $y=8$ in $\left(i\right),$ we get
$10x-9\left(8\right)=-2\Rightarrow 10x=-2+72\Rightarrow 10x=70\Rightarrow x=7$

$x=7,y=8$
Hence, the required fraction is $\frac{7}{8}.$