Difference between revisions of "2004 AMC 8 Problems/Problem 3"

 
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==See Also==
 
==See Also==
 
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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 $18$ people. If they shared, how many meals should they have ordered to have just enough food for the $12$ of them?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 18$

Solution

Set up the proportion $\frac{12\ \text{meals}}{18\ \text{people}}=\frac{x\ \text{meals}}{12\ \text{people}}$. Solving for $x$ gives us $x= \boxed{\textbf{(A)}\ 8}$.

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
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

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