Arnaldo, Bráulio, and Carlos will participate in a street race. After several weeks, they were discussing their strategies. Arnaldo runs the first half of the total distance of the race at $9 \mathrm{~km} / \mathrm{h}$ and the second half at $11 \mathrm{~km} / \mathrm{h}$. Bráulio runs one-third of the distance at $9 \mathrm{~km} / \mathrm{h}$, the second third at $10 \mathrm{~km} / \mathrm{h}$, and finally, the last third at $11 \mathrm{~km} / \mathrm{h}$. Carlos uses a different strategy from the first two; he runs half of his total race time at $9 \mathrm{~km} / \mathrm{h}$ and the final half of the time at $11 \mathrm{~km} / \mathrm{h}$. Determine the order of the total times of Arnaldo, Bráulio, and Carlos at the finish line of the race.
Solution

We will call the times, measured in hours, of Arnaldo, Bráulio, and Carlos $t_{A}, t_{B}$, and $t_{C}$, respectively. Let $6 d$ be the total distance of the race, measured in kilometers. Since Carlos runs half the time at $9 \mathrm{~km} / \mathrm{h}$ and half at $11 \mathrm{~km} / \mathrm{h}$, then

$$
6 d=9 \cdot \frac{t_{C}}{2}+11 \cdot \frac{t_{C}}{2}=10 \cdot t_{C}
$$

With this, we can set up equations and compare the three times.

$$
\begin{aligned}
t_{A} & =\frac{3 d}{9}+\frac{3 d}{11} \\
& =\frac{600 d}{990} \\
t_{B} & =\frac{2 d}{9}+\frac{2 d}{10}+\frac{2 d}{11} \\
& =\frac{598 d}{990} \\
t_{C} & =\frac{6 d}{10} \\
& =\frac{594 d}{990}
\end{aligned}
$$

We conclude that $t_{C}<t_{B}<t_{A}$, that is, Carlos's time is less than Bráulio's, which in turn is less than Arnaldo's.