Difference between revisions of "2009 AMC 10A Problems/Problem 1"
I like pie (talk | contribs) (Added solution) |
I like pie (talk | contribs) m (Fixed year in AMC box) |
||
Line 11: | Line 11: | ||
<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{(E)}</math>. | ||
− | {{AMC10 box|year= | + | {{AMC10 box|year=2009|ab=A|before=First Question|num-a=2}} |
Revision as of 13:47, 15 February 2009
Problem
One can holds ounces of soda. What is the minimum number of cans needed to provide a gallon ( ounces) of soda?
Solution
cans would hold ounces, but , so cans are required. Thus, the answer is .
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 |