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(A) Let the numerator be $x$ and the denominator be $x+3$.The fraction is represented as $\frac{x}{x+3}$.According to the problem,if both the numerato

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$\frac{8}{11}$

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(A) Let the numerator be $x$ and the denominator be $x+3$.The fraction is represented as $\frac{x}{x+3}$.According to the problem,if both the numerator and the denominator are increased by $4$,the new fraction becomes $\frac{4}{5}$.So,$\frac{x+4}{(x+3)+4} = \frac{4}{5}$.$\frac{x+4}{x+7} = \frac{4}{5}$.Cross-multiplying gives: $5(x+4) = 4(x+7)$.$5x + 20 = 4x + 28$.$5x - 4x = 28 - 20$.$x = 8$.Therefore,the numerator is $8$ and the denominator is $8+3 = 11$.The original fraction is $\frac{8}{11}$.

The denominator of a fraction is $3$ more than the numerator. If the numerator as well as the denominator is increased by $4,$ the fraction becomes $\frac{4}{5}$. What was the original fraction?

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