Difference between revisions of "2008 AMC 10A Problems/Problem 5"

(New page: ==Problem== A triathlete competes in a triathlon in which the swimming, biking, and running segments are all of the same length. The triathlete swims at a rate of 3 kilometers per hour, bi...)
 
(Added Solution)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
{{solution}}
+
Let <math>d</math> be the length of one segment of the race.
 +
 
 +
Average speed is total distance divided by total time. The total distance is <math>3d</math>, and the total time is <math>\frac{d}{3}+\frac{d}{20}+\frac{d}{10}=\frac{20d+3d+6d}{60}=\frac{29d}{60}</math>.
 +
 
 +
Thus, the average speed is <math>3d\div\left(\frac{29d}{60}\right)=\frac{180}{29}</math>. This is closest to <math>6</math>, so the answer is <math>\mathrm{(D)}</math>.
 +
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=A|num-b=4|num-a=6}}
 
{{AMC10 box|year=2008|ab=A|num-b=4|num-a=6}}

Revision as of 22:03, 25 April 2008

Problem

A triathlete competes in a triathlon in which the swimming, biking, and running segments are all of the same length. The triathlete swims at a rate of 3 kilometers per hour, bikes at a rate of 20 kilometers per hour, and runs at a rate of 10 kilometers per hour. Which of the following is closest to the triathlete's average speed, in kilometers per hour, for the entire race?

$\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 5\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 7$

Solution

Let $d$ be the length of one segment of the race.

Average speed is total distance divided by total time. The total distance is $3d$, and the total time is $\frac{d}{3}+\frac{d}{20}+\frac{d}{10}=\frac{20d+3d+6d}{60}=\frac{29d}{60}$.

Thus, the average speed is $3d\div\left(\frac{29d}{60}\right)=\frac{180}{29}$. This is closest to $6$, so the answer is $\mathrm{(D)}$.


See also

2008 AMC 10A (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 10 Problems and Solutions