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Difference between revisions of "1990 AIME Problems/Problem 6"

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== Solution ==
 
== Solution ==
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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>.
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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 ==

Revision as of 14:38, 15 March 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 $70$ fish caught in September, $40\%$ were not there in May, so $42$ fish were there in May. Of the $60$ fish tagged in May, $25\%$ 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 (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
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