1987 AJHSME Problems/Problem 18

Problem

Half the people in a room left. One third of those remaining started to dance. There were then $12$ people who were not dancing. The original number of people in the room was

$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 42 \qquad \text{(E)}\ 72$

Solution

Let the original number of people in the room be $x$. Half of them left, so $\frac{x}{2}$ of them are left in the room.

After that, one third of this group is dancing, so $\frac{x}{2}-\frac{1}{3}\left( \frac{x}{2}\right) =\frac{x}{3}$ people are not dancing.

This is given to be $12$, so \[\frac{x}{3}=12\Rightarrow x=36\]

$\boxed{\text{C}}$

See Also

1987 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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