Difference between revisions of "2004 AMC 8 Problems/Problem 20"
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==Solution== | ==Solution== | ||
| − | 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 | + | 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>. |
==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 11:25, 18 April 2015
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 
empty chairs, how many people are in the room?
Solution
Working backwards, if 
of the chairs are taken and are empty, then there are three times as many taken chairs as empty chairs, or 
. If 
is the number of people in the room and 
are seated, then 
and 
.
See Also
| 2004 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
