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In a certain pond, 50 fish were caught, tagged, and returned

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In a certain pond, 50 fish were caught, tagged, and returned [#permalink]

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New post 18 Aug 2011, 11:50
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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
[Reveal] Spoiler: OA

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]

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New post 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]

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New post 18 Aug 2011, 12:57
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total fish = x
percentage of second catch = (2/50)*100 = 4%
so, x * 4% = 50
x = 1250 ans.
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Re: Fish - World Problem!! [#permalink]

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New post 23 Nov 2012, 23:38
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?
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Re: Fish - World Problem!! [#permalink]

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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: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]

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New post 09 Dec 2013, 17:48
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Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]

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In a certain pond, 50 fish were caught, tagged, and returned [#permalink]

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New post 01 Oct 2015, 12:35
VeritasPrepKarishma wrote:
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


Why is the total number of tagged fish 50 as opposed to 98. I got 98 by adding the number of tagged fish in the first catch and the number of tagged fish in the second catch. -> 50+48 = 98
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Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]

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HI tinnyshenoy,

This prompt is about ratios (and in the broader sense, it's an example of 'representative sampling').

To start, we're told that 50 fish are caught, tagged and returned to the pond. There are now an UNKNOWN number of total fish in the pond, but 50 of them are 'tagged.'

Later on, 50 fish are again caught, but 2 of them are ALREADY TAGGED. We're told that the percent of fish IN THIS GROUP that are tagged is approximately = the TOTAL percent of ALL fish that are tagged....With this information, we can set up a ratio...

2/50 = ratio of tagged fish in this sample
50/X = ratio of tagged fish in the pond

2/50 = 50/X

Now we can cross-multiply and solve for X...

2X = 2500
X = 1250

Final Answer:
[Reveal] Spoiler:
C


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Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]

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New post 01 Oct 2015, 19:49
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tinnyshenoy wrote:
VeritasPrepKarishma wrote:
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


Why is the total number of tagged fish 50 as opposed to 98. I got 98 by adding the number of tagged fish in the first catch and the number of tagged fish in the second catch. -> 50+48 = 98


Very simply - the 50 fish caught in the second catch were not tagged. They were just caught and it was observed that 2 of them are tagged. The leftover 48 were not tagged. The second catch was only to find the approximate percentage of tagged fish in the pond (a technique called sampling).

For example: In a large population, it is difficult to find the number of people with a certain trait, say red hair. So you pick up 100 people at random (unbiased selection) and see the number of people who have red hair. Say, 12 have red hair. So you can generalise that approximately 12% of the whole population has red hair.

Here, since counting the number of total fish in the pond is hard, they tagged 50 and let them disperse evenly in the population. Then they caught 50 and found 2 to be tagged. So approximately 4% of the fish were tagged. So 50 is 4% of the entire fish population of the pond. Note that the method uses huge approximation because of the small sample number. If 1 more tagged fish were caught among the 50, it would change the approximated fish population number by a huge amount. But they have given us that "the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond" so we can make this approximation.
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Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]

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New post 27 Mar 2016, 12:49
Is it a 600 level question or more?
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Re: In a certain pond, 50 fish were caught, tagged, and returned   [#permalink] 27 Mar 2016, 12:49
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