Difference between revisions of "2004 AMC 8 Problems/Problem 16"

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Revision as of 01:00, 5 July 2013

Problem

Two $600$ mL pitchers contain orange juice. One pitcher is $1/3$ full and the other pitcher is $2/5$ full. Water is added to fill each pitcher completely, then both pitchers are poured into one large container. What fraction of the mixture in the large container is orange juice?

$\textbf{(A)}\ \frac18 \qquad \textbf{(B)}\ \frac{3}{16} \qquad \textbf{(C)}\ \frac{11}{30} \qquad \textbf{(D)}\ \frac{11}{19}\qquad \textbf{(E)}\ \frac{11}{15}$

Solution

The first pitcher contains $600 \cdot \frac13 = 200$ mL of orange juice. The second pitcher contains $600 \cdot \frac25 = 240$ mL of orange juice. In the large pitcher, there is a total of $200+240=440$ mL of orange juice and $600+600=1200$ mL of fluids, giving a fraction of $\frac{440}{1200} = \boxed{\textbf{(C)}\ \frac{11}{30}}$.

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AJHSME/AMC 8 Problems and Solutions

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