Difference between revisions of "2009 AMC 10A Problems/Problem 1"

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== Solution 2 ==
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We can divide <math>128/12</math> and round up because there are a whole number of cans.
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<math>128/12 = 10R8\longrightarrow 11\longrightarrow \fbox{E}.</math>

Revision as of 18:03, 25 December 2017

Problem

One can hold $12$ ounces of soda. What is the minimum number of cans needed to provide a gallon ($128$ ounces) of soda?

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

Solution 1

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

2009 AMC 10A (ProblemsAnswer KeyResources)
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Solution 2

We can divide $128/12$ and round up because there are a whole number of cans.

$128/12 = 10R8\longrightarrow 11\longrightarrow \fbox{E}.$