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2009 AMC 10A Problems/Problem 1: Difference between revisions

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== Problem ==
== Problem ==
One can holds <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 can holds <math>12</math> ounces of soda, what is the minimum number of cans needed to provide a gallon (<math>128</math> ounces) of soda?


<math>\mathrm{(A)}\ 7\qquad
<math>\textbf{(A)}\ 7\qquad
\mathrm{(B)}\ 8\qquad
\textbf{(B)}\ 8\qquad
\mathrm{(C)}\ 9\qquad
\textbf{(C)}\ 9\qquad
\mathrm{(D)}\ 10\qquad
\textbf{(D)}\ 10\qquad
\mathrm{(E)}\ 11</math>
\textbf{(E)}\ 11</math>


== Solution ==
== Solution 1 ==
<math>10</math> cans would hold <math>120</math> ounces, but <math>128>120</math>, so <math>11</math> cans are required. Thus, the answer is <math>\mathrm{(E)}</math>.
<math>10</math> cans would hold <math>120</math> ounces, but <math>128>120</math>, so <math>11</math> cans are required. Thus, the answer is <math>\mathrm{\boxed{{(E)}11}}</math>.
 
== Solution 2 ==
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>
 
 
==See Also==


{{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}}

Latest revision as of 16:43, 30 July 2025

Problem

One can holds $12$ 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 $\mathrm{\boxed{{(E)}11}}$.

Solution 2

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

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


See Also

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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