Difference between revisions of "2008 AMC 10A Problems/Problem 6"
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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>. | 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>. | ||
| − | + | ||
| + | ==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 <cmath>\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}</cmath>. | ||
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==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=A|num-b=5|num-a=7}} | {{AMC10 box|year=2008|ab=A|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 21:01, 2 September 2020
Contents
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?
Solution
Let 
be the length of one segment of the race.
Average speed is total distance divided by total time. The total distance is 
, and the total time is 
.
Thus, the average speed is 
. This is closest to 
, so the answer is 
.
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 ![\[\frac{3}{\frac{1}{3}+\frac{1}{20}+\frac{1}{10}}=\frac{3}{\frac{29}{60}}=\frac{180}{29}\approx6\Rightarrow\boxed{\mathrm{(D)}\ 6}\]](http://latex.artofproblemsolving.com/2/7/3/27301fb8f762ab0e29ede7c8d7e2e9c2538e9eaf.png)
.
See also
| 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. 
