2004 AMC 8 Problems/Problem 3

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 $x= \boxed{\textbf{(A)}\ 8}$ .

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2004 AMC 8 Problems/Problem 3

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