Difference between revisions of "2009 AMC 10A Problems/Problem 1"
(→Solution 1) |
(→Solution 1) |
||
Line 10: | Line 10: | ||
== 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{(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>. | ||
− | |||
== Solution 2 == | == Solution 2 == |
Revision as of 18:04, 25 December 2017
Problem
One 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 can divide and round up because there are a whole number of cans.
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.