Difference between revisions of "2006 AMC 12B Problems/Problem 5"

 
 
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== Problem ==
 
== Problem ==
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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?
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<math>
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\text {(A) } 30 \qquad \text {(B) } 50 \qquad \text {(C) } 60 \qquad \text {(D) } 90 \qquad \text {(E) } 120
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</math>
  
 
== Solution ==
 
== Solution ==
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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 ==
* [[2006 AMC 12B Problems]]
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{{AMC12 box|year=2006|ab=B|num-b=4|num-a=6}}
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{{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?

$\text {(A) } 30 \qquad \text {(B) } 50 \qquad \text {(C) } 60 \qquad \text {(D) } 90 \qquad \text {(E) } 120$

Solution

The speed that Bob is catching up to John is $5-3=2$ miles per hour. Since Bob is one mile behind John, it will take $\frac{1}{2} \Rightarrow \text{(A)}$ of an hour to catch up to John.

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
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

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