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In a certain pond, 50 fish were caught, tagged, and returned [#permalink]
 18 Aug 2011, 11:50
Question Stats:
78% (01:46) correct
21% (03:10) wrong based on 3 sessions
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 number of fish in the pond? (A) 400 (B) 625 (C) 1,250 (D) 2,500 (E) 10,000
Last edited by Bunuel on 24 Nov 2012, 05:04, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Fish - World Problem!! [#permalink]
24 Nov 2012, 03:28
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Sachin9 wrote: so, x * 4% = 50
This is how I solved too.. This works but I dont think this is right...
4% is actually equal to percent of tagged fish in the pond..
Could somebody please confirm if this is right? This is correct. You are assuming that the total number of fish in the pond is x 4% of x = 50 (Number of tagged fish is 4% of the total fish) You get x = 1250 So total fish in the pond = 1250
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Re: Fish - World Problem!! [#permalink]
18 Aug 2011, 11:56
DeeptiM 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 number of fish in the pond?
(A) 400
(B) 625
(C) 1,250
(D) 2,500
(E) 10,000 2/50 re-caught means that 50 fish represents 1/25th of all fish in the pond. 50*25 = 1250, C
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Re: Fish - World Problem!! [#permalink]
18 Aug 2011, 12:57
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Re: Fish - World Problem!! [#permalink]
23 Nov 2012, 23:38
so, x * 4% = 50This is how I solved too.. This works but I dont think this is right... 4% is actually equal to percent of tagged fish in the pond.. Could somebody please confirm if this is right?
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Re: Fish - World Problem!!
[#permalink]
23 Nov 2012, 23:38
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