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

Latest revision as of 11:48, 27 November 2024

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 $n = \boxed{840}$

Solution 3(easy to comprehend)

Let $k$ be the number of fish in the lake on May 1, and $n$ be the number of fish in the lake on September 1. Because the biologist believes that 40% of $n$ were not there on May 1, we have the equality $\frac{6}{10}\cdot n = \frac{3}{4} \cdot k$ which reduces to $n=\frac{5}{4} \cdot k$ Then, since we have 3 out of 70 fishies tagged on September 1st, we can write the equality $\frac{3}{70}=\frac{60\cdot\frac{3}{4}}{\frac{5\cdot k}{4}}$ because 25% of the 60 fishes tagged on May 1st are now gone, for the numerator, and for the deonominator we have $n$ , which we found is $\frac{5}{4}\cdot k$ . Solving, we get $k= \boxed{840}$

~MathCosine

Video Solution

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

@@ Line 11: / Line 11: @@
 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>
+==Solution 3(easy to comprehend)==
+Let <math>k</math> be the number of fish in the lake on May 1, and <math>n</math> be the number of fish in the lake on September 1. Because the biologist believes that 40% of <math>n</math> were not there on May 1, we have the equality <cmath>\frac{6}{10}\cdot n = \frac{3}{4} \cdot k</cmath> which reduces to <cmath>n=\frac{5}{4} \cdot k</cmath> Then, since we have 3 out of 70 fishies tagged on September 1st, we can write the equality <cmath>\frac{3}{70}=\frac{60\cdot\frac{3}{4}}{\frac{5\cdot k}{4}}</cmath> because 25% of the 60 fishes tagged on May 1st are now gone, for the numerator, and for the deonominator we have <math>n</math>, which we found is <math>\frac{5}{4}\cdot k</math>. Solving, we get <math>k= \boxed{840}</math>
+~MathCosine
 ==Video Solution==

1990 AIME (Problems • Answer Key • Resources)
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