Practice Test Discussion

If the numerator of a fraction is increased by $\frac{1}{4}$ and the denominator is decreased by $\frac{1}{3}$ , the new fraction obtained is $\frac{33}{64}$ . what was the original fraction?

[IOB, 2010]

108

107

110

Cannot be determined

None of these

Explanation

Let the original fraction be $\frac{x}{y}$

Then,

$= \frac{x + \frac{x}{4}}{y - \frac{y}{3}} = \frac{33}{64} = \frac{\frac{4 x + x}{4}}{\frac{2 y}{3}} = \frac{33}{64} = \frac{x}{y} = \frac{33 \times 8}{320 \times 3} = \frac{x}{y} = \frac{11}{40}$

Hence, Option E is correct.

Arithmetics

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