All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 The Cannibals (Posted on 2004-02-04)
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?

 See The Solution Submitted by Ravi Raja Rating: 1.2857 (7 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Potential Solution | Comment 12 of 14 |
I worked through the problem and got the same answers as Dave as well as many others. After re-reading the problem I agree with tim that the semantics are a bit tricky, and we need to drop the final iteration of the cannibal being eaten.

The first cannibal was eaten in 120/5 minutes, or 24 minutes. The second cannibal weighing 1.2C...or 1.2 times the original mass of the cannibals, was eaten in 120*1.2/4, or 36 minutes. The 4 remaining cannibals had 1.2C/4 (.3) added to their current weight of 1.2C and now weighed 1.5.
This goes on until two, not one cannibal is left. The reason again being an issue of semantics, as the question is looking for one remaining consumer.
When all the figures are added up I got an even 4 hours.
24 mins for the first at 1.0C
36 mins for the second at 1.2C
60 mins for the third at 1.5 C
and 120 mins for the last at 2.0C.

This leaves 2 cannibals that both way 3x their starting body weight, and if one was to eat the other it'd take 3.0C*120/1 minutes - 6 hours in other words. So in the end, if the one was to eat all 5 others it would as others have said, take ten hours, but since he's only eating 4 others, it will take him only 4 hours.

 Posted by Seth on 2004-02-08 16:10:23

 Search: Search body:
Forums (0)