1994 AJHSME Problems/Problem 13

Problem

The number halfway between $\dfrac{1}{6}$ and $\dfrac{1}{4}$ is

$\text{(A)}\ \dfrac{1}{10} \qquad \text{(B)}\ \dfrac{1}{5} \qquad \text{(C)}\ \dfrac{5}{24} \qquad \text{(D)}\ \dfrac{7}{24} \qquad \text{(E)}\ \dfrac{5}{12}$

Solution

The number halfway between is the average.

$\frac{\frac16 + \frac14}{2} = \frac{\frac{2}{12} + \frac{3}{12}}{2} = \boxed{\text{(C)}\ \frac{5}{24}}$

1994 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AJHSME/AMC 8 Problems and Solutions

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1994 AJHSME Problems/Problem 13

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