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2008 AMC 10A Problem 6

2008 AMC 10A Problem 6AMC 10 Step by Step

Name: Date: October 28, 2026

Ratios, Percents & Averages · difficulty 2 of 5 · about 2 min · position P1-10

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?

A) $3$
B) $4$
C) $5$
D) $6$
E) $7$

Answer

Key insight

Average speed is total distance over total time. With 60 km legs, the times are 20, 3 and 6 hours, so the average is 180 over 29, about 6.

Solution

The legs have the same length, but we are not told what it is. The answer cannot depend on it, so pick a length that divides nicely by $3$ , $20$ and $10$ . Let each leg be $60$ km.
Find the time for each leg, using time = distance ÷ speed:

Leg	Distance	Speed	Time
Swim	$60$ km	$3$ km/h	$20$ h
Bike	$60$ km	$20$ km/h	$3$ h
Run	$60$ km	$10$ km/h	$6$ h

3. Add up. Total distance is $3 \times 60 = 180$ km. Total time is $20 + 3 + 6 = 29$ hours.
4. Average speed is total distance ÷ total time:
$\frac{180}{29} \approx 6.2\ \text{km/h}.$

The closest choice is $\boxed{\textbf{(D)}\ 6}$ .

Why this works

Average speed is not the average of the speeds. It is total distance over total time, and the slow part of a trip eats up most of the time. Here the swim takes $20$ of the $29$ hours, so the average lands close to the swimming speed. Choosing a distance that every speed divides into keeps all the arithmetic in whole numbers.

The trap

Averaging the three speeds to get 11. The slow swim takes far more time than the other legs, so it pulls the average down.

Concise solution

The answer does not depend on the length of a leg, so choose a length divisible by $3$ , $20$ and $10$ : let each segment be $60$ km.

Time spent on each leg:
$\text{swim: } \frac{60}{3} = 20\text{ h},\qquad \text{bike: } \frac{60}{20} = 3\text{ h},\qquad \text{run: } \frac{60}{10} = 6\text{ h}.$

Total distance $180$ km, total time $29$ h, so the average speed is
$\frac{180}{29} \approx 6.2\ \text{km/h}.$

The closest choice is $\boxed{\textbf{(D)}\ 6}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)
Prealgebra, Chapter 13: Data and Statistics (§13.1)

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer