2004 AMC 8 Problems/Problem 20

Problem

Two-thirds of the people in a room are seated in three-fourths of the chairs. The rest of the people are standing. If there are $6$ empty chairs, how many people are in the room?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 18\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$

Solution

Working backwards, if $3/4$ of the chairs are taken and $6$ are empty, then there are three times as many taken chairs as empty chairs, or $3 \cdot 6 = 18$ . If $x$ is the number of people in the room and $2/3$ are seated, then $\frac23 x = 18$ and $x = \boxed{(D) 27}$ .

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AJHSME/AMC 8 Problems and Solutions

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2004 AMC 8 Problems/Problem 20

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