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>60</math> fish tagged in May, <math>25%</math> are no longer there in September, so <math>45</math> remain. Since the percentage of tagged fish in September is proportional to the percentage of tagged fish in May, <math>\frac{3}{42} = \frac{45}{x} \Longrightarrow \boxed{x = 630}</math>. |
== See also == | == See also == | ||
{{AIME box|year=1990|num-b=5|num-a=7}} | {{AIME box|year=1990|num-b=5|num-a=7}} | ||
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+ | [[Category:Intermediate Combinatorics Problems]] |
Revision as of 20:20, 26 October 2007
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, $40%$ (Error compiling LaTeX. Unknown error_msg) were not there in May, so fish were there in May. Of the fish tagged in May, $25%$ (Error compiling LaTeX. Unknown error_msg) are no longer there in September, so remain. Since the percentage of tagged fish in September is proportional to the percentage of tagged fish in May, .
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 |