Difference between revisions of "2004 AMC 8 Problems/Problem 20"
Quantummech (talk | contribs) m (→Solution) |
Bobthemath (talk | contribs) (→Solution) |
||
| Line 4: | Line 4: | ||
<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== | + | ==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{(\text{D}) 27}</math>. | 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>. | ||
Revision as of 21:00, 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 
empty chairs, how many people are in the room?
Solution 1
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. 
