SOLUTION: Two crafty bacteria fall into a pot of milk which has recently been sterilized. They reproduce at a rate of 4% per day. Determine how many bacteria will be present after 100 days.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Two crafty bacteria fall into a pot of milk which has recently been sterilized. They reproduce at a rate of 4% per day. Determine how many bacteria will be present after 100 days.       Log On


   



Question 1199062: Two crafty bacteria fall into a pot of milk which has recently been sterilized. They reproduce at a rate of 4% per day. Determine how many bacteria will be present after 100 days.
A. 100
B. 800
C. 0
D. 2
E. 109
F. 345

Found 2 solutions by ikleyn, josgarithmetic:
Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
.

As the problem is posed, its meaning is dark,

because the growth rate of 4% per day gives/keeps the population of 2 bacteria unchangeable in every next day,
since the number of bacteria can only be integer.

From the other side, the problem says that the growth rate is 4%, not zero.

The problem is badly posed in statistical sense, since the announced growth rate
relation / (formula) does not work for so small initial number of bacteria.


            As the problem is posed,  it is self-contradictory and,  therefore,  is  DEFECTIVE.


            Alternatively, it can be a kind of an entertainment puzzles
            that are not  Math problems,  but are a kind of traps,  instead.



Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
p=2%281.04%29%5Ed for p population size and d days

p=2%281.04%29%5E100
highlight%28p=101%29