In the picture, you can see a pool with a long step on one of its walls. We started filling the empty pool with a constant flow and observed the water level. After

![](https://cdn.mathpix.com/cropped/2024_04_17_ac63ecbffe7e6b0bc0e7g-2.jpg?height=309&width=614&top_left_y=451&top_left_x=755)

8 min, the water level had risen to a height of $20 \mathrm{~cm}$ and had not yet reached the level of the step. After $23 \mathrm{~min}$ of filling, the water level was at $55 \mathrm{~cm}$ and the step had already been submerged for some time. After 35.5 min of filling, the pool was filled to a height of $80 \mathrm{~cm}$. At what height $h$ from the bottom of the pool is the step located?

(L. Šimünek)
Rise rate during the time $0-8 \mathrm{~min}: 20: 8=2.5(\mathrm{~cm} / \mathrm{min})$.

Rise rate during the time $23-35.5 \mathrm{~min}:(80-55):(35.5-23)=25: 12.5=2(\mathrm{~cm} / \mathrm{min})$.

The time it takes for the water level to rise to height $h$ is denoted as $t$ and is calculated using the equation:

$$
t \cdot 2.5+(23-t) \cdot 2=55
$$

from which $t=18$. It therefore took 18 minutes for the water level to rise to height $h$. For the entire duration, it rose at a rate of $2.5 \mathrm{~cm} / \mathrm{min}$. Therefore,

$$
h=18 \cdot 2.5=45(\mathrm{~cm}) .
$$

(rise rate to height $h \ldots 1$.., rise rate above height $h$... 1 b., setting up the equation + solving it ... $1+1$ b., correct height $h$ ... 2 b.)