Difference between revisions of "2006 AMC 12B Problems/Problem 5"
| Line 8: | Line 8: | ||
== Solution == | == Solution == | ||
The speed that Bob is catching up to John is <math>5-3=2</math> miles per hour. Since Bob is one mile behind John, it will take <math>\frac{1}{2} \Rightarrow \text{(A)}</math> of an hour to catch up to John. | The speed that Bob is catching up to John is <math>5-3=2</math> miles per hour. Since Bob is one mile behind John, it will take <math>\frac{1}{2} \Rightarrow \text{(A)}</math> of an hour to catch up to John. | ||
| − | |||
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=4|num-a=6}} | {{AMC12 box|year=2006|ab=B|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 20:47, 1 August 2020
Problem
John is walking east at a speed of 3 miles per hour, while Bob is also walking east, but at a speed of 5 miles per hour. If Bob is now 1 mile west of John, how many minutes will it take for Bob to catch up to John?
Solution
The speed that Bob is catching up to John is 
miles per hour. Since Bob is one mile behind John, it will take 
of an hour to catch up to John.
See also
| 2006 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
