# Difference between revisions of "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 $$\frac{x}{3}=12\Rightarrow x=36$$

$\boxed{\text{C}}$

## Solution #2

First note that of the $\frac{1}{2}$ people remaining in the room, $\frac{2}{3}$ are not dancing. Therefore $\frac{1}{2}\cdot\frac{2}{3}= \frac{1}{3}$ of the original amount of people in the room is $12$. The answer is $\boxed{C}$.