Aptitude Discussion

Q. |
In a certain water body, 50 fish were caught, tagged and released to the same water body. A few days later, 50 fish were caught again, of which 2 were found to have been tagged on previous occasion. If the percent of tagged fish in the second catch approximates the percent of tagged fish in water body, what is the approximate number of fish in the water body: |

✖ A. |
10,000 |

✖ B. |
625 |

✔ C. |
1250 |

✖ D. |
2500 |

**Solution:**

Option(**C**) is correct

Let the number of fish be $x$ then:

\(\left(\dfrac{50\times 100}{x}\right)=\left(\dfrac{48\times 100}{x-50}\right)\)

\(\dfrac{50}{x}=\dfrac{48}{x-50}\)

\(x=1250\)

**Edit:** For a detailed alternative solution, check comment by **PRATYUSH ANAND.**

**Edit 2:** For yet another alternative solution, check comment by **Sravan Reddy.**

**ABHIJEET**

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**Sravan Reddy**

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In one catch of 50 fishes, you got 2 tagged fishes -> In 25 catches, you can find all 50 tagged fish. That means in 25 times, you will get all the fish in the water body.

Hence total fish $= 25*50 = 1250$

**PRATYUSH ANAND**

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Hi 2 all,

Given in Question:-

1. 1st catch -> 50 fish and were tagged .

2. 2nd catch -> 50 fish and 48 were tagged because 2 fish were already tagged.

As per Question--

%age of tagged fish in the 2nd catch =%age of tagged fish in water body.

Let the total fish in water body be x.

Now

i)%age of tagged fish in the 2nd catch

means -->

(Tagged fish on 2nd catch= 48)/(No. of fish not tagged after 1st catch of fish are done=x-50))*100

=(48/(x-50))*100

ii)%age of tagged fish in water body

means -->

(Tagged fish on 1st catch= 50)/(Total no. of fish in water body)*100

=(50/x)*100

So if you read the main line in question, on that basis equation is made-

If the percent of tagged fish in the second catch approximates the percent of tagged fish in water body.

Therefore combining both equations-

i.e (48/(x-50))*100=(50/x)*100

=>x=1250.

Hence no. of fish in water body are 1250.

48 fishes were tagged later