Difference between revisions of "2009 AMC 10A Problems/Problem 1"
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== Solution 1 == | == 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>. | ||
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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> | ||
{{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}} |
Revision as of 18:03, 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 .
Solution 2
We can divide and round up because there are a whole number of cans.
2009 AMC 10A (Problems • Answer Key • Resources) | ||
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