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by kishokbabu » Tue Dec 27, 2011 9:22 am
In a certain pond, 50 fish were caught, tagged, and returned to the pond. A few days later, 50 fish were caught again, of which 2 were found to have been tagged. If the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond, what is the approximate no of fish in the pond?

a) 400
b) 625
c) 1250
d) 2500
e) 10000
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by pemdas » Tue Dec 27, 2011 10:05 am
2/50 makes 4% tagged
as 50/X=4% or X=50*25=1250
c
kishokbabu wrote:In a certain pond, 50 fish were caught, tagged, and returned to the pond. A few days later, 50 fish were caught again, of which 2 were found to have been tagged. If the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond, what is the approximate no of fish in the pond?

a) 400
b) 625
c) 1250
d) 2500
e) 10000
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by LalaB » Tue Dec 27, 2011 10:19 am
50/x=2/50

50/x=10/50*5

x=1250
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by Anurag@Gurome » Tue Dec 27, 2011 9:29 pm
kishokbabu wrote:In a certain pond, 50 fish were caught, tagged, and returned to the pond. A few days later, 50 fish were caught again, of which 2 were found to have been tagged. If the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond, what is the approximate no of fish in the pond?

a) 400
b) 625
c) 1250
d) 2500
e) 10000
In the 1st catch, 50 fishes were caught, tagged, and returned to the pond
In the 2nd catch, 2 of 50 fishes caught were already tagged implies (2/50) * 100 = 4% were already tagged.
Total tagged fishes = 50
Therefore, total fishes in pond = (50/4) * 100 = 1250

The correct answer is C.
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by rajivranprasad » Sat Feb 04, 2012 5:56 am
IF x = total fish in the pond
2 fish is tagged ,2 fish = 4 % of 50 caught fish in second trial.
so as per question,4=percentage of remaining tagged fish in the pond
remaining tagged fish in pond= 50-2=48
so, 4=( 48/(x-50))*100
x=1250
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by GMATGuruNY » Sat Feb 04, 2012 6:00 am
I posted a solution here:

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