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?
Try it yourself first — the explanation is most useful after you’ve attempted it.
View mode:
Toolkit + CCSS Solution
Understand
Restated: A race has three equal-length parts: swim, bike, run. The athlete moves at 3 km/h swimming, 20 km/h biking, and 10 km/h running. Find which choice is closest to the average speed for the whole race.
Givens: Swim, bike, and run segments are all the same length.; Swim rate is 3 km/h.; Bike rate is 20 km/h.; Run rate is 10 km/h.
Unknowns: The average speed for the entire race, in km/h.
Understand
Restated: A race has three equal-length parts: swim, bike, run. The athlete moves at 3 km/h swimming, 20 km/h biking, and 10 km/h running. Find which choice is closest to the average speed for the whole race.
Givens: Swim, bike, and run segments are all the same length.; Swim rate is 3 km/h.; Bike rate is 20 km/h.; Run rate is 10 km/h.
The segment length is never given, so I name it with a variable. Because each part is the same length, that length appears in both the total distance and the total time, and it cancels. The unit relation distance = rate x time turns each rate into a travel time, and the whole race becomes total distance over total time.
Execute — Answer: D
#4 Introduce a Variable 6.RP.A.2Step 1
Name the segment length
Let $d$ be the length of one segment, in kilometers.
Then swim, bike, and run each cover $d$ km, so the total distance is $3d$ km.
The exact value of $d$ will not matter, because it will cancel.
$$\text{total distance} = d + d + d = 3d$$
💡 Give the unknown length a name so you can carry it through and watch it cancel.
#8 Analyze the Units 6.NS.A.1Step 2
Turn each rate into a time
Speed is distance divided by time, so time is distance divided by speed.
For a segment of length $d$: swimming takes $d/3$ hours, biking takes $d/20$ hours, and running takes $d/10$ hours.
💡 Break total time into three pieces, then a common denominator lets them add cleanly.
#4 Introduce a Variable 7.RP.A.1Step 4
Divide distance by time
Average speed is total distance divided by total time: $3d$ divided by $\frac{29d}{60}$.
Dividing by a fraction means multiplying by its reciprocal, and the $d$ cancels, leaving $\frac{180}{29}$ km/h.
$$\text{avg speed} = \frac{3d}{\frac{29d}{60}} = 3d \cdot \frac{60}{29d} = \frac{180}{29}$$
💡 Because the length is the same everywhere, it divides out and the answer never depended on it.
#3 Eliminate Possibilities 6.NS.A.1Step 5
Round to the closest choice
Evaluate $\frac{180}{29} \approx 6.21$ km/h.
Among the choices $3, 4, 5, 6, 7$, the closest value is $6$, so the answer is (D).
$$\frac{180}{29} \approx 6.21 \;\Rightarrow\; 6$$
💡 Once you have the exact fraction, pick the listed number nearest to it.
[1]
#4 6.RP.A.2Let $d$ be the length of one segment, in kilometers. Then swim, bike, and run ea
[2]
#8 6.NS.A.1Speed is distance divided by time, so time is distance divided by speed. For a s
[3]
#7 5.NF.A.1The total time is the sum of the three segment times. Use a common denominator o
[4]
#4 7.RP.A.1Average speed is total distance divided by total time: $3d$ divided by $\frac{29
[5]
#3 6.NS.A.1Evaluate $\frac{180}{29} \approx 6.21$ km/h. Among the choices $3, 4, 5, 6, 7$,
Review
Reasonableness: The average speed $\frac{180}{29} \approx 6.21$ sits between the slowest rate (3) and the fastest (20), which any average must. It is much closer to the slow rates than to 20, which fits intuition: the swim is so slow that it dominates the time, dragging the average down toward the low end. 6 km/h is the closest choice, confirming (D).
Alternative: Since the three segments have equal length, the average speed equals the harmonic mean of the three rates: $\frac{3}{\frac{1}{3} + \frac{1}{20} + \frac{1}{10}} = \frac{3}{\frac{29}{60}} = \frac{180}{29} \approx 6.21$, giving the same answer without introducing $d$ explicitly.
CCSS standards used (min grade 7)
6.RP.A.2 Understand the concept of a unit rate and use rate language (Reading each speed as kilometers per hour and treating average speed as a rate.)
6.NS.A.1 Interpret and compute quotients of fractions and solve word problems (Turning each rate into a travel time by dividing distance by speed, and evaluating 180/29.)
5.NF.A.1 Add and subtract fractions with unlike denominators (Summing the three segment times over a common denominator of 60.)
7.RP.A.1 Compute unit rates associated with ratios of fractions (Dividing total distance 3d by the complex fraction 29d/60 to get the average speed.)
⭐ Average speed is total distance over total time, not the average of the speeds, so the slow swim pulls the whole race down toward 6 km/h.
⭐ Average speed is total distance over total time, not the average of the speeds, so the slow swim pulls the whole race down toward 6 km/h.