Difference between revisions of "2004 AMC 8 Problems/Problem 20"

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<math>\textbf{(A)}\ 12\qquad \textbf{(B)}\ 18\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36</math>
 
<math>\textbf{(A)}\ 12\qquad \textbf{(B)}\ 18\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36</math>
  
==Solution==
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==Solution 1==
Working backwards, if <math>3/4</math> of the chairs are taken and <math>6</math> are empty, then there are three times as many taken chairs as empty chairs, or <math>3 \cdot 6 = 18</math>. If <math>x</math> is the number of people in the room and <math>2/3</math> are seated, then <math>\frac23 x = 18</math> and <math>x = \boxed{(D) 27}</math>.
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Working backwards, if <math>3/4</math> of the chairs are taken and <math>6</math> are empty, then there are three times as many taken chairs as empty chairs, or <math>3 \cdot 6 = 18</math>. If <math>x</math> is the number of people in the room and <math>2/3</math> are seated, then <math>\frac23 x = 18</math> and <math>x = \boxed{(\text{D}) 27}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2004|num-b=19|num-a=21}}
 
{{AMC8 box|year=2004|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:00, 5 October 2016

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 1

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 $x = \boxed{(\text{D}) 27}$.

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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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