13.245. A tank was filled with water over 5 hours. During this time, the amount of water entering the tank decreased by the same factor each hour compared to the previous hour. It turned out that in the first four hours, twice as much water was poured into the tank as in the last four hours. What is the volume of the tank, given that in the first two hours, 48 m$^3$ of water was poured into it?
Solution.

Let $b$ m$^3$ of water enter the tank in the first hour. Then in the second hour, $b q$ m$^3$ enters, where $q>0, q<1$, in the third hour $-b q^{2}$ m$^3$, in the fourth hour $-b q^{3}$ m$^3$, and in the fifth hour $-b q^{4}$ m$^3$.

According to the conditions,

\[
\left\{\begin{array}{l}
b+b q+b q^{2}+b q^{3}=2\left(b q+b q^{2}+b q^{3}+b q^{4}\right) \\
b+b q=48
\end{array}\right.
\]

from which $q=\frac{1}{2}, b=32$ m$^3$.

Then the volume of the tank is

\[
V=\frac{b\left(1-q^{5}\right)}{1-q}=62 \text{ m}^3
\]

Answer: $62 \text{ m}^3$.