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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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Updated on: 24 Nov 2012, 05:04
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25% (medium)

Question Stats:

78% (01:11) correct 22% (01:44) wrong based on 444 sessions

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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

Originally posted by DeeptiM on 18 Aug 2011, 11:50.
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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24 Nov 2012, 03:28
3
2
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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Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! ##### General Discussion Manager Status: Quant 50+? Joined: 02 Feb 2011 Posts: 104 Concentration: Strategy, Finance Schools: Tuck '16, Darden '16 Re: Fish - World Problem!! [#permalink] ### Show Tags 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 Director Status: No dream is too large, no dreamer is too small Joined: 14 Jul 2010 Posts: 549 Re: Fish - World Problem!! [#permalink] ### Show Tags 18 Aug 2011, 12:57 2 total fish = x percentage of second catch = (2/50)*100 = 4% so, x * 4% = 50 x = 1250 ans. _________________ Collections:- PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html 100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html Senior Manager Status: Gonna rock this time!!! Joined: 22 Jul 2012 Posts: 477 Location: India GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29 WE: Information Technology (Computer Software) Re: Fish - World Problem!! [#permalink] ### Show Tags 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? _________________ hope is a good thing, maybe the best of things. And no good thing ever dies. Who says you need a 700 ?Check this out : http://gmatclub.com/forum/who-says-you-need-a-149706.html#p1201595 My GMAT Journey : http://gmatclub.com/forum/end-of-my-gmat-journey-149328.html#p1197992 Intern Joined: 17 Aug 2015 Posts: 5 In a certain pond, 50 fish were caught, tagged, and returned [#permalink] ### Show Tags 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 EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12185 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink] ### Show Tags 01 Oct 2015, 16:28 1 2 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: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: In a certain pond, 50 fish were caught, tagged, and returned  [#permalink]

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01 Oct 2015, 19:49
4
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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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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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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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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06 May 2017, 17:28
1
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

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

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07 Oct 2017, 20:04
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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In a certain pond, 50 fish were caught, tagged, and returned  [#permalink]

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12 Oct 2017, 17:54
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  [#permalink]

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14 Dec 2017, 09:57
Top Contributor
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

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

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29 Mar 2018, 11:25
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

guys 2 fish out of 50 is 4% or 0.04 ?

i solved like this

0.04 = 50
100 = x

cross multiply, solve for x , so i got 12500 ...why ?
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Re: In a certain pond, 50 fish were caught, tagged, and returned  [#permalink]

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29 Mar 2018, 12:01
1
dave13 wrote:
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

guys 2 fish out of 50 is 4% or 0.04 ?

i solved like this

0.04 = 50
100 = x

cross multiply, solve for x , so i got 12500 ...why ?

Hi dave13

If 4% is 0.04, then 100% is 1

So the right calculation would be something like this
4 ------ 50
100 ---- x

$$4x = 100*50$$

Solving for x, x = $$\frac{100*50}{4} = 25*50 = 1250$$ (Option C)

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

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29 Mar 2018, 14:35
How can we assume that there are no previously tagged fish in the pond (before the 50 fish are tagged)?

The way i understand the question, the pond has 50 or more tagged fish.
Re: In a certain pond, 50 fish were caught, tagged, and returned &nbs [#permalink] 29 Mar 2018, 14:35
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