Difference between revisions of "2009 AMC 10A Problems/Problem 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?
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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>\textbf{(A)}\ 7\qquad
 
<math>\textbf{(A)}\ 7\qquad
Line 15: Line 15:
  
 
<math>\frac{128}{12} = 10R8\longrightarrow 11\longrightarrow \fbox{(E)}.</math>
 
<math>\frac{128}{12} = 10R8\longrightarrow 11\longrightarrow \fbox{(E)}.</math>
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==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}}
 
{{MAA Notice}}

Latest revision as of 09:00, 8 June 2021

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 $\mathrm{\boxed{(E)}}$.

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 (ProblemsAnswer KeyResources)
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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