7・13 A biologist wants to calculate 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 randomly caught 70 fish again and found that 3 of them were marked. He assumes that 25% of the fish in the lake on May 1 were no longer in the lake by September 1 (due to death or migration), and 40% of the fish in the lake on September 1 were not in the lake on May 1 (due to new births or recent migration). Assuming that the fish caught on September 1 are representative of the entire lake, how many fish were in the lake on May 1?
[Solution] The 60 fish marked on May 1, by September 1, the possible number still in the lake is
$$
60(1-25 \%)=45 \text { (fish). }
$$

Of these 45 fish, 3 marked fish were caught, so the proportion of marked fish caught out of all marked fish is $\frac{3}{45}=\frac{1}{15}$.

These three marked fish were found among the 70 fish caught, and 60% of the 70 fish were in the lake on May 1, i.e., the number of fish in the lake on May 1 is
$$
70(1-40 \%)=42 \text { (fish). }
$$

Thus, the number of fish from May 1 in the lake on September 1 is
$$
42 \div \frac{1}{15}=630 \text { (fish). }
$$

And only $(1-25 \%)$ of the fish from May 1 are still in the lake on September 1. Let $x$ be the number of fish in the lake on May 1, then
$$
\begin{array}{c}
x(1-25 \%)=630, \\
x=840 \text { (fish). }
\end{array}
$$

Therefore, there were 840 fish in the lake on May 1.