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 population, how many fish were in the lake on May 1?
[Solution] The 60 fish marked on May 1, by September 1, possibly remain in the lake as
$$60(1-25 \%)=45 \text { (fish). }$$

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 these 70 fish were in the lake on May 1, which means the number of fish in the lake on May 1 is
$$70(1-40 \%)=42 \text { (fish). }$$

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

Since 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.