This is a puzzle which I told to my classmates during a talk, a few days before. I did not represent it as a puzzle, but a talk suggesting the importance of Math in general life. This is partially solved for me and I hope you will run your brain-horse to help me solve it completely. If you didn’t notice, this puzzle is not a part of A Trip To Mathematics series. Puzzle which I discussed in the talk was something like this:
“Let I have seven fishes in a huge tank of water —four male and three females. Those were allowed to sex independently but under some conditions. One male is allowed to have intercourse with female, unless other has done so. A male can have intercourse with any number of female fishes possible. If we assume that first female fish could give 100 eggs, second female fish could give 110, third 90. We are also known that a female fish might lay eggs in 21 days since the date of sex with male. The children fish are reproductive only after those are 30 days old. In each bunch of children fish of individual female fish, 60% die. In remaining 40% child fishes, the ratio of male and female is 3:2. If two fishes (male and female) can not do intercourse with each other if those are born to same mother fish, one male is allowed to have intercourse with female, unless other has done so, a male can have intercourse with any number of fishes possible, every three female fishes lay eggs in order of 100,110 and 90 eggs out of which only 40% remain alive having a ratio of male and females of 1:1 and the same rule applies to third, fourth and consecutive generations of fishes; then find the number of total fishes in my tank after one year (365 days).”
I have done many proof-reads of this puzzle and found it valid. Your comments, your ideas and suggestions might help me working more rigorously on this puzzle. This puzzle is neither too hard nor too easy. I will be updating this post frequently as my work on this puzzle is directed towards a correct way.