a) A car traveled the Budapest-Debrecen route at a speed of $56 \mathrm{~km} /$ hour, and on the return trip along the same route, it traveled at a speed of $72 \mathrm{~km} /$ hour. What was its average speed?

b) Three workers were each given the same number of workpieces to process. The first completed 10, the second 14, and the third 12 workpieces per hour. What is the average hourly productivity in workpieces?
a) We understand average speed as the speed at which, if we travel at a constant rate, we cover a certain distance in the same amount of time as we would if we traveled at a variable speed.

Let the distance between Budapest and Debrecen be $s \mathrm{~km}$, and the time for the round trip be $t$ hours. According to the problem,

$$
t=\frac{s}{56}+\frac{s}{72}
$$

$c$ is the constant speed for the round trip, which takes

$$
\frac{2 s}{c}=t^{\prime}
$$

hours.

By definition, $c$ is the average speed if $t^{\prime}=t$, that is,

$$
\frac{2 s}{c}=\frac{s}{56}+\frac{s}{72}
$$

from which

$$
c=\frac{2}{\frac{1}{56}+\frac{1}{72}}=\frac{1008}{9+7}=63 \mathrm{~km} / \text { hour. }
$$

b) We understand average performance as the performance of a worker who completes the same work as multiple workers together, in the same amount of time as the total time of the multiple workers, assuming that each of the latter workers performs equal work.

If we denote the certain equal work (in this problem, the number of units) by $m$, then the work times of the individual workers are

$$
\frac{m}{10}, \frac{m}{14}, \frac{m}{12} \text { hours. }
$$

Let the sought average performance be $x$ units per hour. According to the above definition,

$$
\frac{3 m}{x}=\frac{m}{10}+\frac{m}{14}+\frac{m}{12}
$$

that is,

$$
x=\frac{3}{\frac{1}{10}+\frac{1}{14}+\frac{1}{12}}=\frac{3 \cdot 420}{42+30+35}=\frac{1260}{107}=11 \frac{83}{107} \sim 11.87 \text { units/hour. }
$$

László Várallyai (Mosonmagyaróvár, Kossuth L. g. II. o. t.)

Note: Many incorrectly solved the problem using the arithmetic mean $\left(\frac{56+72}{2}=64 \mathrm{~km} /\right.$ hour, and $\frac{10+14+12}{3}=12$ units/hour $)$, whereas, as seen above, the harmonic mean (see Erik Dux's article "On Weighted Arithmetic Means" in the April 1956 issue on pages 99-100) was the appropriate method to use.