9. (1990 American Mathematics Invitational Competition) 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 on 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?
9. The 60 fish marked on May 1, by September 1, the possible number still in the lake is $60(1-25\%)=45$ (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 $70(1-40\%)=42$ (fish).
Thus, the number of fish from May 1 in the lake on September 1 is $42 \div \frac{1}{15}=630$ (fish).

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

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