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Difference between revisions of "1990 AIME Problems/Problem 6"
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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?  
 
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?  
  
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== Solution ==
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== Solution 1==
 
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>.
 
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>.
  
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(Note the 25% death rate does not affect the answer b/c both tagged and nontagged fish die.)
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(Note the 25% death rate does not affect the answer because both tagged and nontagged fish die.)
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== Solution 2 ==
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First, we notice that there are 45 tags left, after 25% of the original fish have went away/died. Then, some <math>x</math> percent of fish have been added such that <math>\frac{x}{x+75} = 40 \% </math>, or <math>\frac{2}{5}</math>. Solving for <math>x</math>, we get that <math>x = 50</math>, so the total number of fish in September is <math>125 \%</math>, or <math>\frac{5}{4}</math> times the total number of fish in May.
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Since <math>\frac{3}{70}</math> of the fish in September were tagged, <math>\frac{45}{5n/4} = \frac{3}{70}</math>, where <math>n</math> is the number of fish in May. Solving for <math>n</math>, we see that <math>n = \boxed{840}</math>
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==Video Solution==
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https://www.youtube.com/watch?v=oyMP8NdGtB8
  
 
== 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]]
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[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:16, 26 June 2022

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 $70$ fish caught in September, $40\%$ were not there in May, so $42$ fish were there in May. Since the percentage of tagged fish in September is proportional to the percentage of tagged fish in May, $\frac{3}{42} = \frac{60}{x} \Longrightarrow \boxed{x = 840}$.

(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 $x$ percent of fish have been added such that $\frac{x}{x+75} = 40 \%$, or $\frac{2}{5}$. Solving for $x$, we get that $x = 50$, so the total number of fish in September is $125 \%$, or $\frac{5}{4}$ times the total number of fish in May.

Since $\frac{3}{70}$ of the fish in September were tagged, $\frac{45}{5n/4} = \frac{3}{70}$, where $n$ is the number of fish in May. Solving for $n$, we see that $n = \boxed{840}$

Video Solution

https://www.youtube.com/watch?v=oyMP8NdGtB8

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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