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OG10 #228 Tagged Fish

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by oquiella » Fri Oct 16, 2015 8:24 am
In a certain pond, 50 fish were caught, tagged, and returned to the pond. A few days later, 50 fish weere caught again,of which 2 were found to have been tagged. If the percent of tagged in the second catch 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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by Brent@GMATPrepNow » Fri Oct 16, 2015 8:52 am
oquiella wrote:In a certain pond, 50 fish were caught, tagged, and returned to the pond. A few days later, 50 fish weere caught again,of which 2 were found to have been tagged. If the percent of tagged in the second catch 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 can use equivalent ratios here.
We're told that the proportion of tagged fish in the 50-fish SAMPLE is equal to the the proportion of tagged fish in the ENTIRE pond.

So, (# tagged fish)/50 = (total # tagged fish)/(total # of fish in pond)
Let x = # of fish in the entire pond.

We get: 2/50 = 50/x
Cross multiply to get 2x = (50)(50)
x = 1250
Answer: C
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by DavidG@VeritasPrep » Mon Oct 19, 2015 6:07 am
oquiella wrote:In a certain pond, 50 fish were caught, tagged, and returned to the pond. A few days later, 50 fish weere caught again,of which 2 were found to have been tagged. If the percent of tagged in the second catch 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


Whats your approach?
And there's always good old-fashioned back-solving. We know 2/50 fish were tagged. This reduces to 1/25. We also know that the total number of fish tagged was 50. So 1/25 of the correct answer should give us 50.

Test D. 1/25 of 2500 is 100. This number is too big, as 50 fish were tagged.

Test B. 1/25 of 625 is 25. This number is too small, as 50 fish were tagged.

If D is too large and B is too small, the answer has to be C. (And, indeed, 1/25 of 1250 is 50.)
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by Aman verma » Mon Oct 19, 2015 7:03 am
oquiella wrote: 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?


Whats your approach?
(2/50)X 100=4%

(50/4)X 100=1250 (C)
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by Jeff@TargetTestPrep » Thu Dec 14, 2017 6:48 am
oquiella wrote:In a certain pond, 50 fish were caught, tagged, and returned to the pond. A few days later, 50 fish weere caught again,of which 2 were found to have been tagged. If the percent of tagged in the second catch 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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by Brent@GMATPrepNow » Thu Dec 14, 2017 8:57 am
oquiella wrote:In a certain pond, 50 fish were caught, tagged, and returned to the pond. A few days later, 50 fish weere caught again,of which 2 were found to have been tagged. If the percent of tagged in the second catch 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

Cheers,
Brent
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