The denominator of a fraction exceeds its numerator by 1. Two thirds of the fraction exceeds one third of the fraction by $\frac{1}{6}$ . The original fraction is

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Solution

The correct option is C

$\frac{2}{3} \frac{x}{x + 1} = \frac{1 x}{3 (x + 1)} + \frac{1}{6}$

$\frac{x}{3 (x + 1)} = \frac{1}{6}$

2x = x +1

x =1

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$\frac{2}{3} \frac{x}{x + 1} = \frac{1 x}{3 (x + 1)} + \frac{1}{6}$

$\frac{x}{3 (x + 1)} = \frac{1}{6}$

2x = x +1

x =1