Practice Test Discussion

What fraction must be subtracted from the sum of $\frac{1}{4}$ and $\frac{1}{6}$ to have an average of $\frac{1}{12}$ of all the three fractions?

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{4}$

$\frac{1}{6}$

None of these

Explanation

According to the question,

Required fraction

$= \frac{1}{4} + \frac{1}{6} - 3 \times \frac{1}{12} = \frac{1}{4} + \frac{1}{6} - \frac{1}{4} = \frac{1}{12}$

Hence, Option D is correct.

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