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

Revision as of 09:43, 6 May 2018

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

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

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

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 (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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 == Solution 2 ==
-We want to find <math>\left\lfloor\frac{128}{12}\right\rfloor+1</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>