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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==See Also== | ==See Also== | ||
Latest revision as of 17:43, 30 July 2025
Contents
Problem
One can holds 
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
