1990 AIME Problems/Problem 6

Revision as of 21:20, 26 October 2007 by Azjps (talk | contribs) (solution)

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

Of the $70$ fish caught in September, $40%$ (Error compiling LaTeX. Unknown error_msg) were not there in May, so $42$ fish were there in May. Of the $60$ fish tagged in May, $25%$ (Error compiling LaTeX. Unknown error_msg) are no longer there in September, so $45$ remain. Since the percentage of tagged fish in September is proportional to the percentage of tagged fish in May, $\frac{3}{42} = \frac{45}{x} \Longrightarrow \boxed{x = 630}$.

See also

1990 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AIME Problems and Solutions