Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2008 AMC 10A Problems/Problem 6"

m (Typo: 1/120 should be 1/20)
(→‎Problem)
Line 1: Line 1:
 
==Problem==
 
==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?
+
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?
−
 
 
−
<math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 5\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 7</math>
 
  
 
==Solution==
 
==Solution==

Revision as of 17:48, 16 June 2020

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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_6&oldid=125668"