Difference between revisions of "2007 AMC 12A Problems/Problem 4"

 
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== Problem ==
 
== 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?
 
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?
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<math>\mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 9\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14</math>
  
 
== Solution ==
 
== Solution ==
* 16*(30/60)+4*(90/60)=14
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<cmath>16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14</cmath>
* 14/2=7
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<cmath>\frac{14}2=7\Rightarrow\boxed{A}</cmath>
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== See also ==
 
== See also ==
* [[2007 AMC 12A Problems/Problem 1 | Previous problem]]
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{{AMC12 box|year=2007|ab=A|num-b=3|num-a=5}}
* [[2007 AMC 12A Problems/Problem 3 | Next problem]]
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* [[2007 AMC 12A Problems]]
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 19:36, 27 October 2013

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?

$\mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 9\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$

Solution

\[16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14\] \[\frac{14}2=7\Rightarrow\boxed{A}\]

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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All AMC 12 Problems and Solutions

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