Difference between revisions of "1990 AIME Problems/Problem 6"
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== Solution == | == Solution == | ||
| − | Of the <math>70</math> fish caught in September, <math>40\%</math> were not there in May, so <math>42</math> fish were there in May | + | Of the <math>70</math> fish caught in September, <math>40\%</math> were not there in May, so <math>42</math> fish were there in May. Since the percentage of tagged fish in September is proportional to the percentage of tagged fish in May, <math>\frac{3}{42} = \frac{60}{x} \Longrightarrow \boxed{x = 840}</math>. |
| + | |||
| + | (Note the 25% death rate does not affect the answer.) | ||
== See also == | == See also == | ||
Revision as of 16:46, 5 August 2008
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 
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.)
See also
| 1990 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
