Difference between revisions of "1987 AJHSME Problems/Problem 18"

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<math>\boxed{\text{C}}</math>
 
<math>\boxed{\text{C}}</math>
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==Solution #2==
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First note that of the <math>\frac{1}{2}</math> people remaining in the room, <math>\frac{2}{3}</math> are not dancing. Therefore <math>\frac{1}{2}\cdot\frac{2}{3}= \frac{1}{3}</math> of the original amount of people in the room is <math>12</math>. The answer is <math>\boxed{C}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 19:11, 26 August 2016

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 $\boxed{C}$.

See Also

1987 AJHSME (ProblemsAnswer KeyResources)
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

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