A water tank with a capacity of $120 \mathrm{hl}$ has three pipes: the first two are inlet pipes, and the third is an outlet pipe. The first pipe fills the tank one hour faster than the second. If the first and third pipes are open, the tank fills in $7 \frac{1}{2}$ hours, and if the second and third pipes are open, it takes 20 hours to fill the tank. How many hours will it take to fill the tank if all three pipes are open at the same time?
To find the sought time, we do not need the data on the amount of water. If the first pipe fills the tank in $x$ hours, the second in $x+1$ hours, and the third empties it in $y$ hours, then according to the problem:

$$
\frac{1}{x}-\frac{1}{y}=\frac{1}{7.5}
$$

$$
\frac{1}{x+1}-\frac{1}{y}=\frac{1}{20}
$$

Subtract (2) from (1):

$$
\frac{1}{x+1}-\frac{1}{x}=-\frac{1}{12}
$$

or

$$
x^{2}+x-12=0
$$

from which the positive value of $x$ is 3; thus $x=3, x+1=4, y=5$.

If all three pipes are open at the same time, then:

$$
\frac{1}{3}+\frac{1}{4}-\frac{1}{5}=\frac{1}{t}
$$

from which

$$
\frac{1}{t}=\frac{23}{60}
$$

or the sought time $t=\frac{60}{23}$ hours.

(Weisz Ármin, Pécs.)

Number of solutions: 43.