6. A biologist wants to estimate the number of fish in a lake. On May 1, he randomly caught 60 fish and marked them, then released them back into the lake. On September 1, he found that some of the fish were no longer in the lake (due to death or migration), and on September 1, 40% of the fish in the lake were not there on May 1 (due to new births or recent migration into the lake). How many fish were in the lake on September 1?
6. Let the number of fish in the lake on May 1 be $x$, and the number of fish in the lake on September 1 be $y$. According to the problem, we have
$$
y=0.75 x+0.40 y \text {. }
$$

On September 1, the number of tagged fish in the lake is $0.75 \times 60 = 45$. Assuming that the tagged fish on September 1 can represent the whole, we have
$$
\begin{aligned}
3 & =\frac{45}{y} . \\
\therefore y & =1050, x=\frac{0.60 y}{0.75}=840 .
\end{aligned}
$$