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

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In a certain pond, 50 fish were caught, tagged, and returned to the po  [#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


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Originally posted by Bunuel on 18 Aug 2011, 10:50.
Last edited by Bunuel on 06 Nov 2018, 01:47, edited 2 times in total.
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In a certain pond, 50 fish were caught, tagged, and returned to the po  [#permalink]

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New post 10 Jan 2014, 06:10
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SOLUTION

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

The percent of tagged fish in the second catch is 2/50*100 = 4%.

We are told that 4% approximates the percent of tagged fish in the pond. Since there are 50 tagged fish, then we have 0.04x = 50 --> x = 1,250.

Answer: C.
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Re: In a certain pond, 50 fish were caught, tagged, and returned to the po  [#permalink]

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New post 18 Aug 2011, 11:57
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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: In a certain pond, 50 fish were caught, tagged, and returned to the po  [#permalink]

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New post 23 Nov 2012, 22:38
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: In a certain pond, 50 fish were caught, tagged, and returned to the po  [#permalink]

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New post 24 Nov 2012, 02:28
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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: In a certain pond, 50 fish were caught, tagged, and returned to the po  [#permalink]

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New post 01 Oct 2015, 11:35
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 to the po  [#permalink]

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New post 01 Oct 2015, 15:28
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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

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

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New post 01 Oct 2015, 18:49
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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 to the po  [#permalink]

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New post 29 Jul 2016, 10:02
Hey Bunuel i have a query in this question
Here why arent we taking the number of tagged fishes as 48 as 2 fishes that were caught arent returned to the pond.
we have no evidence that the two fishes were returned to the pond
so (4/100)*number of fishes = 48
although no option is provided i think its answer should be 1200

Where am i wrong ?

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

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New post 29 Jul 2016, 10:23
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stonecold wrote:
Hey Bunuel i have a query in this question
Here why arent we taking the number of tagged fishes as 48 as 2 fishes that were caught arent returned to the pond.
we have no evidence that the two fishes were returned to the pond
so (4/100)*number of fishes = 48
although no option is provided i think its answer should be 1200

Where am i wrong ?

cc-EMPOWERgmatRichC


True. If one of the options would have been 1200, then this could be a concern.

But I believe GMAC people are intelligent enough to avoid such options that cause confusions.

So, since we donot have any option that matches per your assumption, we need to consider 50 tagged fishes only.

Hope it clears your query!!
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Re: In a certain pond, 50 fish were caught, tagged, and returned to the po  [#permalink]

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New post 29 Jul 2016, 10:28
abhimahna wrote:
stonecold wrote:
Hey Bunuel i have a query in this question
Here why arent we taking the number of tagged fishes as 48 as 2 fishes that were caught arent returned to the pond.
we have no evidence that the two fishes were returned to the pond
so (4/100)*number of fishes = 48
although no option is provided i think its answer should be 1200

Where am i wrong ?

cc-EMPOWERgmatRichC


True. If one of the options would have been 1200, then this could be a concern.

But I believe GMAC people are intelligent enough to avoid such options that cause confusions.

So, since we donot have any option that matches per your assumption, we need to consider 50 tagged fishes only.

Hope it clears your query!!



Nopes bruh..!
I disagree . I think we are really missing something here .
Abhishek009 has stopped doing quant i guess else he would have been the first one to respond on this
Lets wait for Bunuel to respond
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Re: In a certain pond, 50 fish were caught, tagged, and returned to the po  [#permalink]

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New post 29 Jul 2016, 11:08
Hi stonecold,

There are a few different ways to address the interpretational 'issue' that you're talking about:

To start, regardless of how you interpret the prompt, the question ultimately asks for the APPROXIMATE number of fish in the pond - your way of interpreting the question leads to the SAME correct answer. Thus, it doesn't really matter that you didn't get to 1250 exactly.

Second, IF the question expected you to calculate a result by 'ignoring' the 50 fish that were caught the second time, then it would have asked for the number of REMAINING fish in the pond (which the prompt did not do). By extension, you would have to question your interpretation of the wording (and possibly consider an alternative interpretation - which you seem to refuse to do). While minor interpretational 'biases' can sometimes happen in GMAT questions, the answer choices will be written so that there can only be one correct answer. If "your answer" isn't there, then you have to consider how you interpreted the given information and adjust your work accordingly.

Third, from a ratio standpoint, the 50 fish 'subgroup' that was caught the second time IS, mathematically speaking, part of the total number of fish in the pond (so you can't ignore that group when calculating the total number of fish in the pond).

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

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New post 06 May 2017, 16:28
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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 to the po  [#permalink]

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New post 07 Oct 2017, 19:04
ScottTargetTestPrep wrote:

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



This equation assumes that there are 50 fish tagged in the total population. We do not know that. The only thing we know is that that the percentage of tagged fish in the second catch is 4%. The question says that 4% approximates the number of tagged fish in the pond. So this is the true equation we have:

Quote:
.04 (total) = tagged fish


We are missing two variables. We don't know the total fish and we don't know the tagged fish.

If there are 16 tagged fish, then choice A is correct.
If there are 26 tagged fish, then choice B is correct, and etc.

If we assume that the number of fish in the second catch (50) is the number of fish tagged, then yes the total fish would be 1250. However, that's not what the question provides. I think this question is written poorly.
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Re: In a certain pond, 50 fish were caught, tagged, and returned to the po  [#permalink]

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New post 12 Oct 2017, 16:54
joondez wrote:
ScottTargetTestPrep wrote:

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



This equation assumes that there are 50 fish tagged in the total population. We do not know that. The only thing we know is that that the percentage of tagged fish in the second catch is 4%. The question says that 4% approximates the number of tagged fish in the pond. So this is the true equation we have:

Quote:
.04 (total) = tagged fish


We are missing two variables. We don't know the total fish and we don't know the tagged fish.

If there are 16 tagged fish, then choice A is correct.
If there are 26 tagged fish, then choice B is correct, and etc.

If we assume that the number of fish in the second catch (50) is the number of fish tagged, then yes the total fish would be 1250. However, that's not what the question provides. I think this question is written poorly.


So the question states that "In a certain pond, 50 fish were caught, tagged, and returned to the pond."

From this sentence, we can deduce that there are indeed a total of 50 tagged fish in the pond. The only way to have some other number of tagged fish in the pond is if there were already some number of tagged fish in the pond (in which case, the question would have told us so) or if either more fish were tagged afterward or some of the tagged fish were removed from the pond (again, we would have been told). Since we have no such information, we cannot assume that there might be some other number of tagged fish in the pond.

Perhaps you are missing the fact that 50 fish are caught TWICE: first all of them are tagged, and the second time, the tagged fish are counted.
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Re: In a certain pond, 50 fish were caught, tagged, and returned to the po  [#permalink]

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New post 14 Dec 2017, 08:57
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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


The concept here is that the 50 fish that were caught the second time are REPRESENTATIVE of the entire fish population in the pond.
In other words, the RATIO of the # of tagged fish to total fish in second sample = the RATIO of the # of tagged fish in pond to total fish in pond

That is: (# of tagged fish caught the second time)/(total # of fish caught the second time) = (# of tagged fish in pond)/(total # of fish in pond)
Let x = total # of fish in pond
We get: 2/50 = 50/x
Cross multiply to get: 2x = (50)(50)
Solve: x = 1250

Answer: C

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