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

Revision as of 07:40, 6 May 2018

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

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

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

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 (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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All AMC 10 Problems and Solutions

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 == 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?
+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
+<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
+\textbf{(D)}\ 10\qquad
-\mathrm{(E)}\ 11</math>
+\textbf{(E)}\ 11</math>
 == Solution 1 ==