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

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== Problem ==
 
== Problem ==
One can hold <math>12</math> ounces of soda. What is the minimum number of cans needed to provide a gallon (<math>128</math> ounces) of soda?
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One can, can hold <math>12</math> ounces of soda, what is the minimum number of cans needed to provide a gallon (<math>128</math> ounces) of soda?
  
<math>\mathrm{(A)}\ 7\qquad
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<math>\textbf{(A)}\ 7\qquad
\mathrm{(B)}\ 8\qquad
+
\textbf{(B)}\ 8\qquad
\mathrm{(C)}\ 9\qquad
+
\textbf{(C)}\ 9\qquad
\mathrm{(D)}\ 10\qquad
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\textbf{(D)}\ 10\qquad
\mathrm{(E)}\ 11</math>
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\textbf{(E)}\ 11</math>
  
 
== Solution 1 ==
 
== Solution 1 ==

Revision as of 07:40, 6 May 2018

Problem

One can, can hold $12$ 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 $\mathrm{(E)}$.

Solution 2

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

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

2009 AMC 10A (ProblemsAnswer KeyResources)
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