Difference between revisions of "1987 AJHSME Problems/Problem 18"
5849206328x (talk | contribs) (New page: ==Problem== Half the people in a room left. One third of those remaining started to dance. There were then <math>12</math> people who were not dancing. The original number of people in...) |
(→Solution) |
||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 14: | Line 14: | ||
<math>\boxed{\text{C}}</math> | <math>\boxed{\text{C}}</math> | ||
| + | |||
| + | |||
| + | ==Solution #2== | ||
| + | |||
| + | 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== | ||
| − | + | {{AJHSME box|year=1987|num-b=17|num-a=19}} | |
| − | |||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:11, 26 August 2016
Contents
Problem
Half the people in a room left. One third of those remaining started to dance. There were then 
people who were not dancing. The original number of people in the room was
Solution
Let the original number of people in the room be 
. Half of them left, so 
of them are left in the room.
After that, one third of this group is dancing, so 
people are not dancing.
This is given to be , so ![\[\frac{x}{3}=12\Rightarrow x=36\]](http://latex.artofproblemsolving.com/7/8/3/783c0b7a3c13d6d8ec3fcef9d13a4b6db510f949.png)
Solution #2
First note that of the 
people remaining in the room, 
are not dancing. Therefore 
of the original amount of people in the room is . The answer is 
.
See Also
| 1987 AJHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
