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.
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== See also ==
 
== See also ==
 
* [[2006 AMC 12B Problems]]
 
* [[2006 AMC 12B Problems]]

Revision as of 08:30, 14 November 2007

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