2008 AMC 10A Problems/Problem 6

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 $\frac{d}{3}+\frac{d}{20}+\frac{d}{10}=\frac{29d}{60}$ .

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

Solution 2

Since the three segments are all the same length, the triathlete's average speed is the harmonic mean of the three given rates. Therefore, the average speed is $\frac{3}{\frac{1}{3}+\frac{1}{120}+\frac{1}{10}}=\frac{3}{\frac{29}{60}}=\frac{180}{29}\approx6\Rightarrow\boxed{\mathrm{(D)}\ 6}$ .

2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2008 AMC 10A Problems/Problem 6

Contents

Problem

Solution

Solution 2

See also