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by kannans3 » Fri May 20, 2011 7:21 am
Source: OG

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 two 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) 1250
(D) 2500
(E) 10000
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by GMATGuruNY » Fri May 20, 2011 7:55 am
kannans3 wrote:Source: OG

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 two 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) 1250
(D) 2500
(E) 10000
In the second catch, 2/50 = 4/100 = 4% of the caught fish were found to have been tagged.
Since the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond, approximately 4% of all the fish have been tagged.
Total tagged = 50.
Thus, 50 = 4% of the fish in the pond.
Now look at the answer choices. 4% of the correct answer choice must be 50.

Answer choice C: 1250
(.04)*1250 = 50.

The correct answer is C.

The other option is to solve algebraically:
50 = .04x
x = 50/(.04) = 5000/4 = 1250.
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by gmattesttaker2 » Sun Mar 31, 2013 11:05 am
GMATGuruNY wrote:
kannans3 wrote:Source: OG

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 two 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) 1250
(D) 2500
(E) 10000
In the second catch, 2/50 = 4/100 = 4% of the caught fish were found to have been tagged.
Since the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond, approximately 4% of all the fish have been tagged.
Total tagged = 50.
Thus, 50 = 4% of the fish in the pond.
Now look at the answer choices. 4% of the correct answer choice must be 50.

Answer choice C: 1250
(.04)*1250 = 50.

The correct answer is C.

The other option is to solve algebraically:
50 = .04x
x = 50/(.04) = 5000/4 = 1250.

Hello,

I was able to follow till where it says:

Thus, 50 = 4% of the fish in the pond.

I was getting confused here since the question says "the percent of fish in the second catch (i.e. 4%) approximates the percent of tagged fish in the pond". I was just thinking how 50 is the percent of tagged fish in the pond? Maybe should the question be saying "the percent of fish in the second catch (i.e. 4%) approximates the number of tagged fish in the pond (i.e. 50)?". Again, this was just a thought. Thanks for your help.

Best Regards,
Sri
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by Anju@Gurome » Sun Mar 31, 2013 5:20 pm
kannans3 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 two 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?
Let us assume there are N fishes in the pond.
Hence, percentage of tagged fish in the pond = (50/N)*100

Percentage of tagged fish in the second catch = (2/50)*100
Now, percentage of tagged fish in the second catch approximates the percentage of tagged fish in the pond.
So, (2/50)*100 ≈ (50/N)*100
--> N = 50*50/2 = 1250

The correct answer is C.
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by GMATGuruNY » Mon Apr 01, 2013 4:03 am
gmattesttaker2 wrote:
GMATGuruNY wrote:
kannans3 wrote:Source: OG

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 two 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) 1250
(D) 2500
(E) 10000
In the second catch, 2/50 = 4/100 = 4% of the caught fish were found to have been tagged.
Since the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond, approximately 4% of all the fish have been tagged.
Total tagged = 50.
Thus, 50 = 4% of the fish in the pond.
Now look at the answer choices. 4% of the correct answer choice must be 50.

Answer choice C: 1250
(.04)*1250 = 50.

The correct answer is C.

The other option is to solve algebraically:
50 = .04x
x = 50/(.04) = 5000/4 = 1250.

Hello,

I was able to follow till where it says:

Thus, 50 = 4% of the fish in the pond.

I was getting confused here since the question says "the percent of fish in the second catch (i.e. 4%) approximates the percent of tagged fish in the pond". I was just thinking how 50 is the percent of tagged fish in the pond? Maybe should the question be saying "the percent of fish in the second catch (i.e. 4%) approximates the number of tagged fish in the pond (i.e. 50)?". Again, this was just a thought. Thanks for your help.

Best Regards,
Sri
50 fish were caught again, of which two were found to have been tagged.
This statement indicates the PROPORTION of tagged fish in the SECOND CATCH:
(tagged fish in the second catch)/(total number of fish in the second catch) = 2/50 = 1/25.

The percent of tagged fish in the second catch approximates the percent of tagged fish in the pond.
In other words:
The PROPORTION of tagged fish in the ENTIRE POND is about equal to the PROPORTION of tagged fish in the SECOND CATCH.
Thus, the proportion of tagged fish in the ENTIRE pond ≈ 1/25.

In a certain pond, 50 fish were caught, TAGGED, and returned to the pond.
Since these 50 tagged fish are about equal to 1/25 of the total number of fish in the entire pond, we get:
50 ≈ (1/25)x
x ≈ 1250.

The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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