2009 AMC 8 Problems/Problem 15

Revision as of 22:10, 3 September 2019 by Prestonh (talk | contribs) (Solution)

Problem

A recipe that makes $5$ servings of hot chocolate requires $2$ squares of chocolate, $\frac{1}{4}$ cup sugar, $1$ cup water and $4$ cups milk. Jordan has $5$ squares of chocolate, $2$ cups of sugar, lots of water, and $7$ cups of milk. If she maintains the same ratio of ingredients, what is the greatest number of servings of hot chocolate she can make?


$\textbf{(A)}\ 5 \frac18   \qquad \textbf{(B)}\    6\frac14 \qquad \textbf{(C)}\  7\frac12   \qquad \textbf{(D)}\  8 \frac34   \qquad \textbf{(E)}\   9\frac78$

Solution

Assuming excesses of the other ingredients, the chocolate can make $\frac52 \cdot 5=12.5$ servings, the sugar can make $\frac{2}{1/4} \cdot 5 = 40$ servings, the water can make unlimited servings, and the milk can make $\frac74 \cdot 5 = 8.75$ servings. Limited by the amount of milk, Jordan can make at most $\boxed{\textbf{(D)}\ 8 \frac34}$ servings.

See Also

2009 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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