In a fraction,the numerator is increased by 2 | vedclass

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Question

(C) Let the original fraction be $\frac{x}{y}$.According to the problem,the numerator is increased by $25 \%$ and the denominator is decreased by $10

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(C) Let the original fraction be $\frac{x}{y}$.According to the problem,the numerator is increased by $25 \%$ and the denominator is decreased by $10 \%$.New numerator = $x + 0.25x = 1.25x$.New denominator = $y - 0.10y = 0.90y$.The new fraction is given as $\frac{5}{9}$,so:$\frac{1.25x}{0.90y} = \frac{5}{9}$To solve for $\frac{x}{y}$,multiply both sides by $\frac{0.90}{1.25}$:$\frac{x}{y} = \frac{5}{9} 	imes \frac{0.90}{1.25}$$\frac{x}{y} = \frac{5}{9} 	imes \frac{90}{125}$$\frac{x}{y} = \frac{5}{9} 	imes \frac{18}{25} = \frac{1}{1} 	imes \frac{2}{5} = \frac{2}{5}$.Thus,the original fraction is $\frac{2}{5}$.

In a fraction,the numerator is increased by $25 \%$ and the denominator is diminished by $10 \%$. The new fraction obtained is $\frac{5}{9}$. The original fraction is:

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