Difference between revisions of "2004 AMC 8 Problems/Problem 3"
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| − | Set up the proportion <math>\frac{12\text{meals}}{18\text{people}}=\frac{x\text{meals}}{12\text{people}}</math>. Solving for <math>x</math> gives us <math>x= | + | Set up the proportion <math>\frac{12\ \text{meals}}{18\ \text{people}}=\frac{x\ \text{meals}}{12\ \text{people}}</math>. Solving for <math>x</math> gives us <math>x= \boxed{\textbf{(A)}\ 8}</math>. |
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| + | ==See Also== | ||
| + | {{AMC8 box|year=2004|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:54, 4 July 2013
Problem
Twelve friends met for dinner at Oscar's Overstuffed Oyster House, and each ordered one meal. The portions were so large, there was enough food for 
people. If they shared, how many meals should they have ordered to have just enough food for the 
of them?
Solution
Set up the proportion 
. Solving for 
gives us 
.
See Also
| 2004 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
