4. A reservoir of water in a national park is replenished with 1 hectoliter of water every day. A herd of 38 elephants will drink all the water from the reservoir in one day, while a herd of 8 elephants will drink all the water from the reservoir in 5 days. If each elephant drinks the same amount of water on average, how many days, starting from today, can one elephant drink all the water from the reservoir? (The elephants start drinking water from the reservoir after it has been replenished with water.)
Solution. Every day, the amount of water in the reservoir increases by 1 hectoliter, or 100 liters of water. In one day, the amount of water in the reservoir is $V+100$, where $V$ represents the amount of water that was in the reservoir. According to the first condition, this is the amount of water that 38 elephants drink daily. If we denote by $s$ the amount of water (in liters) that one elephant drinks daily, then from the conditions of the problem, we can form the following system of equations:

$$
\left\{\begin{array}{l}
V+100=38 s \\
V+5 \cdot 100=5 \cdot 8 s
\end{array}\right.
$$

The solution to this system is $s=200, V=7500$.

Thus, the reservoir contained $7500 \mathrm{l}$ of water. Since 100 l of water is added to the reservoir every day, and one elephant drinks an average of 200 l of water daily, we conclude that one elephant reduces the amount of water by 100 l per day. Therefore, it will take $7500: 100=75$ days for the reservoir to be emptied.

## IX Section