Difference between revisions of "2009 AMC 10A Problems/Problem 1"
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<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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{{AMC10 box|year=2009|ab=A|before=First Question|num-a=2}} | {{AMC10 box|year=2009|ab=A|before=First Question|num-a=2}} | ||
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Revision as of 14:32, 1 May 2021
Contents
Problem
One can, can hold ounces of soda, what is the minimum number of cans needed to provide a gallon ( ounces) of soda?
Solution 1
cans would hold ounces, but , so cans are required. Thus, the answer is .
Solution 2
We want to find because there are a whole number of cans.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.