Difference between revisions of "2007 AMC 12A Problems/Problem 4"
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== Solution == | == Solution == | ||
* <math>16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14</math> | * <math>16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14</math> | ||
| − | * <math> | + | * <math>\frac{14}2=7\Rightarrow\boxed{A}</math> |
== See also == | == See also == | ||
{{AMC12 box|year=2007|ab=A|num-b=3|num-a=5}} | {{AMC12 box|year=2007|ab=A|num-b=3|num-a=5}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 16:15, 7 February 2011
Problem
Kate rode her bicycle for 30 minutes at a speed of 16 mph, then walked for 90 minutes at a speed of 4 mph. What was her overall average speed in miles per hour?
Solution
See also
| 2007 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 | |
