Two cylindrical tanks filled with water to the brim started to be pumped out by identical pumps at a constant rate precisely at noon. At 2 PM, the water level in both tanks was the same. At 5 PM, the first tank was empty, and by 8 PM, the second tank was also empty. If the second tank is 10 meters tall, how tall is the first one?
Solution. From the text of the problem, it is clear that the first tank emptied in 5 hours, while the second tank emptied in 8 hours. Since the pumping is uniform, the time to pump out is proportional to the height of the water pumped out (we don't need to talk about volume because the cross-section is the same everywhere in a cylinder). Therefore, at 14:00 (2 hours later), $\frac{2}{5}$ of the water in the first tank was pumped out, while $\frac{2}{8}$, or $\frac{1}{4}$, of the water in the second tank was pumped out. This means that $\frac{3}{5}$ of the height $x$ of the water in the first tank is equal to $\frac{3}{4}$ of the height of the water in the second tank, which is $7.5 \mathrm{~m}$. Thus,

$$
\frac{3 x}{5}=7.5
$$

from which $x=12.5 \mathrm{~m}$.

Therefore, the first tank is 12.5 meters high.