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

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2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AMC 10 Problems and Solutions

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 ==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?
+{{problem}}
-<math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 5\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 7</math>
 ==Solution==
-Let <math>d</math> be the length of one segment of the race.
+{{solution}}
-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{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==
 {{AMC10 box|year=2008|ab=A|num-b=4|num-a=6}}