A hectoliter barrel is full of wine; from it, 1 liter of liquid flows out within 24 hours, which is replaced by 1 liter of water flowing in during the same time. After how much time will the content of the barrel be half wine and half water?
Let's examine the problem in general first; the barrel contains $h$ liters of wine and a certain amount $l$ liters flows out over a period of time.

After the first outflow, $r_{1}=h-l$ liters of wine remain. If the deficiency is made up with water, for every liter of the new liquid, $\frac{h-l}{h}$ liters of wine are present, and for the $l$ liters of liquid that flow out again, $\frac{l(h-l)}{h}$ liters of wine. Thus, the remaining pure wine is

$$
r_{2}=(h-l)-\frac{l}{h}(h-l)=(h-l)\left(1-\frac{l}{h}\right)=\frac{(h-l)^{2}}{h}
$$

In the third cycle, the pure wine in the outflowing liquid is $\frac{l}{h^{2}}(h-l)^{2}$, and the remaining amount is therefore

$$
r_{3}=\frac{(h-l)^{2}}{h}-\frac{l}{h} \frac{(h-l)^{2}}{h}=\frac{(h-l)^{2}}{h}\left(1-\frac{l}{h}\right)=\frac{(h-l)^{3}}{h^{2}}
$$

The $n$-th remaining amount is therefore

$$
r^{n}=\frac{(h-l)^{n}}{h^{n-1}}=h\left(\frac{h-l}{h}\right)^{n}=h\left(1-\frac{l}{h}\right)^{n}
$$

and this must be equal to $\frac{h}{2}$.

Accordingly, substituting the given values,

$$
\begin{gathered}
0.99^{n}=0.5 \\
n \log 0.99=\log 0.5 \\
n=\frac{\log 0.5}{\log 0.99}=\frac{1-0.69897}{1-0.99564} \\
=\frac{0.30103}{0.00436}=\frac{30100}{436}=69 \text{ days }
\end{gathered}
$$

as a first approximation.

(Zsigmond Stark, 8th grade student, Pécs.)

The problem was also solved by: Viktor Hónig, Kaposvár; Sándor Kugel and Mihály Seidner, Losonc; István Sztrapkovits, 8th grade student, Sátoraljaújhely; Gusztáv Szabó, 6th grade student, Győr.