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

(Solution 2)
(Solution 2)
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We want to find <math>\left\lceil\frac{128}{12}\right\rceil</math> because there are a whole number of cans.
 
We want to find <math>\left\lceil\frac{128}{12}\right\rceil</math> because there are a whole number of cans.
  
<math>\frac{128}{12} = 10R8\longrightarrow 11\longrightarrow \fbox{E}.</math>
+
<math>\frac{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 12:44, 19 August 2018

Problem

One can, can hold $12$ ounces of soda, what is the minimum number of cans needed to provide a gallon ($128$ ounces) of soda?

$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$

Solution 1

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

Solution 2

We want to find $\left\lceil\frac{128}{12}\right\rceil$ because there are a whole number of cans.

$\frac{128}{12} = 10R8\longrightarrow 11\longrightarrow \fbox{(E)}.$

2009 AMC 10A (ProblemsAnswer KeyResources)
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