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>
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<cmath>16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14</cmath>
* <math>\frac{14}2=7\Rightarrow\boxed{A}</math>
+
<cmath>\frac{14}2=7\Rightarrow\boxed{A}</cmath>
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== 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}}

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
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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