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 06:41, 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. Unknown error_msg)\frac{128}{12} = 10R8\longrightarrow 11\longrightarrow \fbox{E}.$
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