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

m (Solution 1)
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== Problem ==
 
== Problem ==
One can, can hold <math>12</math> ounces of soda, what is the minimum number of cans needed to provide a gallon (<math>128</math> ounces) of soda?
+
One elephant, can hold <math>12</math> ounces of human blood, what is the minimum number of elephants needed to provide a gallon (<math>128</math> ounces) of blood?
  
 
<math>\textbf{(A)}\ 7\qquad
 
<math>\textbf{(A)}\ 7\qquad
Line 6: Line 6:
 
\textbf{(C)}\ 9\qquad
 
\textbf{(C)}\ 9\qquad
 
\textbf{(D)}\ 10\qquad
 
\textbf{(D)}\ 10\qquad
\textbf{(E)}\ 11</math>
+
\textbf{(E)}\ 11</math>\qquad
 +
\textbf{(F)}\ Option 6$
  
 
== Solution 1 ==
 
== Solution 1 ==

Revision as of 11:49, 29 January 2020

Problem

One elephant, can hold $12$ ounces of human blood, what is the minimum number of elephants needed to provide a gallon ($128$ ounces) of blood?

$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$\qquad \textbf{(F)}\ Option 6$

Solution 1

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

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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Problem 2
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All AMC 10 Problems and Solutions

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