The Cannibals of an Island have the habit of eating each other.
One evening, the cannibals threw a dinner party. Six cannibals turned up and they decided to eat each other in turn. So someone was selected for everyone to eat (except the victim!), and when he had been eaten, someone else was selected, and so on.
If it took one cannibal two hours on his own to devour one person, how long was it before just one consumer remained?
Assuming that once a cannibal has eaten another cannibal he now contains the mass of two cannibals and thus takes twice as long to eat (or a proportion of that, depending on how much of another cannibal he has so far eaten), the answer is different.
Assume the mass of a cannibal is C, and any given cannibal can consume a meal at a rate of C every 2 hours.
At the first sitting each of 5 cannibals eats C/5 taking 2/5 hours (=0.4 hr).
At the second sitting each of 4 cannibals eats 1.2C/4 taking 1.2*2/4 hours (=0.6 hr).
At the third sitting each of 3 cannibals eats 1.5C/3 taking 1.5*2/3 hours (=1 hr).
At the fourth sitting each of 2 cannibals eats 2C/2 taking 2*2/2 hours (=2 hr).
At the fifth sitting the last cannibal eats 3C taking 3*2 hours (=6 hr).
Thus the meal finishes in:
0.4 + 0.6 + 1 + 2 + 6 = 10 hours.
This is the same time that would be taken if the if the final cannibal simply sat down and ate each of the other 5 cannibals in turn (2*5 hours).
Solution (add on)
