# 1990 AIME Problems/Problem 6

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## Problem

A biologist wants to calculate the number of fish in a lake. On May 1 she catches a random sample of 60 fish, tags them, and releases them. On September 1 she catches a random sample of 70 fish and finds that 3 of them are tagged. To calculate the number of fish in the lake on May 1, she assumes that 25% of these fish are no longer in the lake on September 1 (because of death and emigrations), that 40% of the fish were not in the lake May 1 (because of births and immigrations), and that the number of untagged fish and tagged fish in the September 1 sample are representative of the total population. What does the biologist calculate for the number of fish in the lake on May 1?

## Solution 1

Of the $70$ fish caught in September, $40\%$ were not there in May, so $42$ fish were there in May. Since the percentage of tagged fish in September is proportional to the percentage of tagged fish in May, $\frac{3}{42} = \frac{60}{x} \Longrightarrow \boxed{x = 840}$.

(Note the 25% death rate does not affect the answer because both tagged and nontagged fish die.)

## Solution 2

First, we notice that there are 45 tags left, after 25% of the original fish have went away/died. Then, some $x$ percent of fish have been added such that $\frac{x}{x+75} = 40 \%$, or $\frac{2}{5}$. Solving for $x$, we get that $x = 50$, so the total number of fish in September is $125 \%$, or $\frac{5}{4}$ times the total number of fish in May.

Since $\frac{3}{70}$ of the fish in September were tagged, $\frac{45}{5n/4} = \frac{3}{70}$, where $n$ is the number of fish in May. Solving for $n$, we see that $n = \boxed{840}$