Math, asked by vaibhavmulticraftpla, 13 hours ago

A certain number of fish are swimming in a pond. A fisherman casts his net and 1/3 rd of the fish are caught in the net. After removing the catch from the net, he throws a second time. This time 1\4 th of the fish remaining in the pond are caught in the net. When he throws the net a third time, 1\5 th of the fish remaining in the pond are caught in the net. If 300 fish are still swimming in the pond, what is the original number of fish in the pond ?

Answers

Answered by p09ss
1

Answer:

18000

Step-by-step explanation:

300÷1/5=1500

1500÷1/4=6000

6000÷1/3=18000

Answered by caffeinated
0

Answer: The original number of fish in the pond is 450.

To calculate:

The original number of fish in the pond

Given:

  • 1/3rd caught first time.
  • 1/4th of the remaining caught the second time.
  • 1\5th of the remaining caught the third time.
  • 300 fishes are still in the pond.

Calculations:

Let the total number of fish in the pond be 'x'.

First time

  • Fish caught first time = 1/3 * x = 1x/3.
  • Remaining = x - 1/3x = 2x/3.

Second time

  • Fish caught second time = 1/4 * 2x/3 =1x/6.
  • Remaining = 2x/3 - 1x/6 = x/2.                

Third time

  • Fish caught third time = 1/5 * x/2 =x/10.
  • Remaining = x/2 - x/10 = 2x/3.  

According to the question: 2x/3 = 300

                                              2x = 900.

                                                x =450.

Thus, the original number of fish in the pond is 450.  

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