Difference between revisions of "2009 AMC 10A Problems/Problem 1"
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\mathrm{(E)}\ 11</math> | \mathrm{(E)}\ 11</math> | ||
| − | == Solution == | + | == Solution 1 == |
<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>. | <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>. | ||
{{AMC10 box|year=2009|ab=A|before=First Question|num-a=2}} | {{AMC10 box|year=2009|ab=A|before=First Question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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| + | == Solution 2 == | ||
| + | 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 17:02, 25 December 2017
Problem
One can hold 
ounces of soda. What is the minimum number of cans needed to provide a gallon (
ounces) of soda?
Solution 1
cans would hold 
ounces, but 
, so 
cans are required. Thus, the answer is 
.
| 2009 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
Solution 2
We can divide 
and round up because there are a whole number of cans.
