1990 AIME Problems/Problem 6
Contents
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 fish caught in September, were not there in May, so fish were there in May. Since the percentage of tagged fish in September is proportional to the percentage of tagged fish in May, .
(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 percent of fish have been added such that , or . Solving for , we get that , so the total number of fish in September is , or times the total number of fish in May.
Since of the fish in September were tagged, , where is the number of fish in May. Solving for , we see that
See also
1990 AIME (Problems • Answer Key • Resources) | ||
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Followed by Problem 7 | |
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