493. Four pumps of the same capacity, working together, filled the first tanker and a third of the second tanker (of a different volume) in 11 hours. If three pumps had filled the first tanker and then one of them filled a quarter of the second tanker, the work would have taken 18 hours. How many hours would it take for three pumps to fill the second tanker?
$\triangle$ Let one pump fill the first tanker in $x$ hours, and the second tanker in $y$ hours. Then four pumps, working together, will fill the first tanker in $\frac{x}{4}$ hours, and the second in $\frac{y}{4}$ hours.

We have the system of equations:

$$
\left\{\begin{array}{l}
\frac{x}{4}+\frac{y}{4 \cdot 3}=11 \\
\frac{x}{3}+\frac{y}{4}=18
\end{array}\right.
$$

From this, we find that $y=24$ hours. Therefore, three pumps can fill the second tanker in $\frac{24}{3}=8$ hours.

Answer: in 8 hours.