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

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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?
  
<math>\mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 9\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14</math>
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<math>\displaystyle \mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 9\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14</math>
  
 
== Solution ==
 
== Solution ==
 
* <math>16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14</math>
 
* <math>16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14</math>
* <math>\frac{14}2=7</math>
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* <math>\displaystyle \frac{14}2=7</math>
 
== 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}}
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[[Category:Introductory Algebra Problems]]

Revision as of 10:47, 9 September 2007

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?

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

Solution

  • $16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14$
  • $\displaystyle \frac{14}2=7$

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