Difference between revisions of "2009 AMC 10A Problems/Problem 1"
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== Solution 1 == | == 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{\boxed{(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{\text{(E)}11}}</math>. |
== Solution 2 == | == Solution 2 == | ||
Revision as of 13:37, 22 November 2024
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
