1994 AJHSME Problems/Problem 13

Revision as of 22:40, 27 October 2016 by Alxz (talk | contribs) (Fixed a mistake)

Problem

Uh Oh Unluky 13!

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.

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

See Also

1994 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS