2008 AMC 10A Problems/Problem 6

Problem

Tony swims for the first 2/11 of a triathlon.Then bikes for 2/3 of the remaining time. He then runs for the last 33 minutes. How long is the triathlon?

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 $\frac{3}{\frac{1}{3}+\frac{1}{20}+\frac{1}{10}}=\frac{3}{\frac{29}{60}}=\frac{180}{29}\approx6\Rightarrow\boxed{\mathrm{(D)}\ 6}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2008 AMC 10A Problems/Problem 6

Contents

Problem

Solution

Solution 2

See also