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 $\boxed{C}$ .

1987 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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.

Page

Toolbox

Search

1987 AJHSME Problems/Problem 18

Contents

Problem

Solution

Solution #2

See Also