2009 AMC 10A Problems/Problem 1

Problem

One can holds $12$ ounces of soda, what is the minimum number of cans needed to provide a gallon ($128$ ounces) of soda?

$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$

Solution 1

$10$ cans would hold $120$ ounces, but $128>120$, so $11$ cans are required. Thus, the answer is $\mathrm{\boxed{{(E)}11}}$.

Solution 2

We want to find $\left\lceil\frac{128}{12}\right\rceil$ because there are a whole number of cans.

$\frac{128}{12} = 10R8\longrightarrow 11\longrightarrow \fbox{(E)}.$

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
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First Question
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Problem 2
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All AMC 10 Problems and Solutions

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