Difference between revisions of "2009 AMC 10A Problems/Problem 1"
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== Solution 2 == | == Solution 2 == | ||
− | We | + | We want to find <math>\left\lceiling\frac{128}{12}\right\rceiling because there are a whole number of cans. |
− | <math>128 | + | </math>\frac{128}{12} = 10R8\longrightarrow 11\longrightarrow \fbox{E}.$ |
{{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 07:41, 6 May 2018
Problem
One can, 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 want to find $\left\lceiling\frac{128}{12}\right\rceiling because there are a whole number of cans.$ (Error compiling LaTeX. ! Undefined control sequence.)\frac{128}{12} = 10R8\longrightarrow 11\longrightarrow \fbox{E}.$
2009 AMC 10A (Problems • Answer Key • Resources) | ||
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