[loveekansal]

Originally Posted by arun_juee

Hi,

Posting after long time.

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 the 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?

1.400
2.625
3.1250
4.2500
5.10000

Please solve and explain.

P.S.-It may be a sitter but I didn't able to solve.

Later.
Arun

Let total fish be 'X' in pond and let 'P' be the percentage...

now (P/100)X = 50
and (P/100)50 = 2

Solving both, X = 1250

But i guess that the thread related to Quant. would hv been a better place for posting this,...

ATB

[thandamunda]

Hey been a dormant puy for a long time ... here goes my profile

B.Tech (CSE) from JIITU, Noida (Jaypee ), 2007 passout.

Work ex 6 months at Infy

CAT History

CAT 2007 - Disaster .. awful prep totally mismanaged ... which showed in the results

CAT 2008 - Quant is the name of the game ...

[loveekansal]

Originally Posted by adityacooool

Arun,

1. The number of already tagged fish in the 2nd catch was 2. which comes to 4% of the total fishes caught in the 2nd try.
2. The total number of tagged fish in the pond after the secong catch was 98 (50+50-2) [Approximated to 100]
3. so as per the question, the number of the total %age tagged fish in the pond after the second catch= %age of fish found already tagged(in the 2nd catch)
=> 98 == 4% approx.
so total nmbr of fishes = 98/4%approx = 2500 roughly

Hope it was helpful to u and i did not confuse you even further

Its nowhere mentioned in the Q that the fish were tagged after the 2nd catch also...
So the tagged fish count remains same, i.e. 50

[monsterkartik]

Originally Posted by arun_juee

Hi,

Posting after long time.

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 the 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?

1.400
2.625
3.1250
4.2500
5.10000

Please solve and explain.

P.S.-It may be a sitter but I didn't able to solve.

Later.
Arun

I am unclear from the question that whether the fishes from 2nd catch were tagged or not, and because the answer depends upon the number of tagged fish in the pond and hence would like u to clarify the same.

Originally Posted by adityacooool

Arun,

1. The number of already tagged fish in the 2nd catch was 2. which comes to 4% of the total fishes caught in the 2nd try.
2. The total number of tagged fish in the pond after the secong catch was 98 (50+50-2) [Approximated to 100]
3. so as per the question, the number of the total %age tagged fish in the pond after the second catch= %age of fish found already tagged(in the 2nd catch)
=> 98 == 4% approx.
so total nmbr of fishes = 98/4%approx = 2500 roughly

Hope it was helpful to u and i did not confuse you even further

If all the fishes of second catch were tagged then answer as given by u is correct, but if fishes of the second catch were not tagged (i cant make it out from the question) then the answer would be 50/4% = 1250

Cheers

MK

[avinav2712]

Originally Posted by arun_juee

Hi,

Posting after long time.

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 the 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?

1.400
2.625
3.1250
4.2500
5.10000

Please solve and explain.

P.S.-It may be a sitter but I didn't able to solve.

Later.
Arun

Please post QA questions on the relevant thread. A couple of them are mentioned below:

http://www.pagalguy.com/forum/quanti...-marathon.html (Quant Marathon)

http://www.pagalguy.com/forum/quanti...for-cat08.html (official quant thread for cat08)

[dhanu_25]

hey all...

this post may contribute to 'Introduce urself 'thread as I remained silent for whole last yr..
But this yr I have geared up for the IIM race..With 2 yrs work ex at a software company, I terribly want to change my profession...After the last yr weak attempt I am trying to be consistent and regular on PG..this will help me for sure..

PS: Please suggest me some ways to avoid silly mistakes (if any)!

[msr.712]

dear junta ,
count me in the list too. iam also a CAT veteran ,appeared for 3times . and raring to go for CAT 2008.
iam working in oil and gas industry with chemical engg background. having exp of 4 years .
messed up in quants in CAT 2007 ( overall :94 , with quants 40 ,verbal 99.7, and DI :90). hence i started my preperation with quant basics .
lets nail the CAT together.
cheers

[samarth87]

Hi.....Um starting to prepare for CAT'08..can anybody tell me the best institute for getting the course material only????

[arun_juee]

Originally Posted by loveekansal

Its nowhere mentioned in the Q that the fish were tagged after the 2nd catch also...
So the tagged fish count remains same, i.e. 50

This was really my confusion that 4% was 50 or 98?

Later I will post my other queries here.

[arun_juee]

Originally Posted by arun_juee

Hi,

Posting after long time.

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 the 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?

1.400
2.625
3.1250
4.2500
5.10000

Please solve and explain.

P.S.-It may be a sitter but I didn't able to solve.

Later.
Arun

The answer is 3.1250