AIME 1990 · 第 6 题

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 $n = \boxed{840}$

Solution 3(easy to comprehend)

Let $k$ be the number of fish in the lake on May 1, and $n$ be the number of fish in the lake on September 1. Because the biologist believes that 40% of $n$ were not there on May 1, we have the equality

$$\frac{6}{10}\cdot n = \frac{3}{4} \cdot k$$ which reduces to

$$n=\frac{5}{4} \cdot k$$ Then, since we have 3 out of 70 fishies tagged on September 1st, we can write the equality

$$\frac{3}{70}=\frac{60\cdot\frac{3}{4}}{\frac{5\cdot k}{4}}$$ because 25% of the 60 fishes tagged on May 1st are now gone, for the numerator, and for the deonominator we have $n$, which we found is $\frac{5}{4}\cdot k$. Solving, we get $k= \boxed{840}$

~MathCosine

Video Solution

https://www.youtube.com/watch?v=oyMP8NdGtB8