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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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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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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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New post 01 Oct 2015, 16:28
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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]

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New post 21 Feb 2017, 02:30
Hi can you help me with this problem? The last part of the problem It is not clear for me. I don't know if I have to use 50 fish or 48 fish still in the pond

- 2 fish tagged / 50 in the second catch
- 48 fish are still in the pond

So the equation is:
48 fish tagged in the pond/Tot in the pond = 2 fish tagged out/50 fish catched
Tot in the pond = 48*2/50 = 1200 (aprox. 1250) --> answer is C
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Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]

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New post 30 Apr 2017, 11:41
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?


Hi,

Although solution has already been provided for this. here is my 2 cent.

I think when they say that the percent of tagged fish caught the second time represents the percent of tagged fish in the pond, they are talking about the first 50 that were tagged and left in the pond and not the 50 caught the second time.

Thus, the first 50 tagged fishes are 4% of the entire fish in the pond.

Hope this is clear.
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Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]

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New post 06 May 2017, 17:28
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



We are given that 50 fish were caught, tagged, and returned to the pond, and that a few days later, 50 fish were caught again, of which 2 were tagged. Thus, the percentage of tagged fish is 2/50 = 1/25 = 4%.

Since the the percentage of tagged fish in the second catch approximates the percentage of tagged fish in the pond, the approximate number of fish in the pond is:

0.04(total fish) = 50

total fish = 50/0.04 = 5000/4 = 1250

Answer: C
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Re: In a certain pond, 50 fish were caught, tagged, and returned   [#permalink] 06 May 2017, 17:28
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