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

(Problem 1)
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== Problem 1 ==
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== Problem ==
One can holds <math>12</math> ounces of soda. What is the minimum number of cans needed to provide a gallon (128 ounces) of soda?
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One can holds <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>
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<math>\mathrm{(A)}\ 7\qquad
\mathrm{(A)}\ 7
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\mathrm{(B)}\ 8\qquad
\qquad
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\mathrm{(C)}\ 9\qquad
\mathrm{(B)}\ 8
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\mathrm{(D)}\ 10\qquad
\qquad
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\mathrm{(E)}\ 11</math>
\mathrm{(C)}\ 9
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\qquad
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== Solution ==
\mathrm{(D)}\ 10
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<math>10</math> cans would hold <math>120</math> ounces, but <math>128>120</math>, so <math>11</math> cans are required. Thus, the answer is <math>\mathrm{(E)}</math>.
\qquad
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\mathrm{(E)}\ 11
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{{AMC10 box|year=2008|ab=A|before=First Question|num-a=2}}
</math>
 

Revision as of 14:46, 15 February 2009

Problem

One can holds $12$ 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

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

2008 AMC 10A (ProblemsAnswer KeyResources)
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First Question
Followed by
Problem 2
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All AMC 10 Problems and Solutions
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