Difference between revisions of "2011 AMC 12A Problems/Problem 3"
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== Problem == | == Problem == | ||
== Solution == | == Solution == | ||
− | To find how many small bottles we need, we can simply divide 500 by 35. This simplifies to 100/7=14+2/7. Since the answer must be an integer greater than 14, we have to round up to 15 bottles= | + | To find how many small bottles we need, we can simply divide 500 by 35. This simplifies to 100/7=14+2/7. Since the answer must be an integer greater than 14, we have to round up to 15 bottles=<math>\boxed{E}</math> |
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=2|num-a=4|ab=A}} | {{AMC12 box|year=2011|num-b=2|num-a=4|ab=A}} |
Revision as of 20:22, 9 February 2011
Problem
Solution
To find how many small bottles we need, we can simply divide 500 by 35. This simplifies to 100/7=14+2/7. Since the answer must be an integer greater than 14, we have to round up to 15 bottles=
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
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 12 Problems and Solutions |