okay, this is a weird problem...thanks to anyone who can solve it
A certain strain of bacteria is living in a can. They are good at reproducing and are capable of doubling their population every 60 seconds. Unfortunately, they only live for 1 hour. The first bacterium invaded the can at 10:00 PM and at 10:57 PM one bacterium shouts, "We've been alive for 57 minutes and the can is only one-eighth full. Party."
--A--How much longer will the party go on?
As 11:00 approaches, an expedition is formed to locate a new can to live. A returning bacterium announces, "Hooray, we have located 3 more empty cans. Thats three times more space than we've ever had in our entire existence.
--B--How much more time has the expedition provided?
If the time is measured in minutes the amount of bacteria can be calculated by:
Originally Posted by doctorgk
After 60 minutes the first bacterium will die. There are missing 3 minutes to this moment. Since 2³ = 8 the population of bacteria will increase 8 times and will fill the can completely!
... and the party will become a funeral (Crying)
Ok so the can is able to hold bacteria since at minute n there are bacteria and after 57 minutes the can is 1/8 full. Provided that a dead bacteria disintigrates into the ether upon death and doesn't take up any room in the can then at 11:00 there will be bacteria in the can, (the first guy died). In the next minute however the population will double and only 1 more bacterium dies (the 1 created at 10:01). Since the party will end at 11:01 (4 minutes after the first said "Party").
Given 3 times the space, the group is able to support bacteria.
-the least value in this inequality is the population at 11:01, the greatest value is the population at 11:02, so it bought them 1 minute.
What do you think? I guess I am assuming the doubling happens instantly at the end of the minute and immediately after the bacteria 1 hour old die. (more of a mind game interpretation than a population growth).