What if a single bacterium landed in a nutrient filled test

What if a single bacterium landed in a nutrient filled test tube at 10:00 am? Since bacteria use nutrients and reproduce by doubling, at 10:01am there are 2 bacteria. Now the binary fusion continues and at 10:02am there are 4 bacteria, and 10:03am there are 8 bacteria and so on. In other words, the bacteria consume the nutrients present and double every minute.

By 11:00am the test tube is full. Now since there are no longer nutrients present, the bacteria begin to die.

A. how many bacteria are in the test tube at 10:06am

B. how many bacteria are in the test tube at 10:20am

C. how many bacteria are in the test tube at 10:50am

D. if the test tube is full at 11:00am what fraction of the test tube was full at 10:59am?

E. what fraction of the tube was full at 10:58?

Solution

so given bacteria is growing in binary fusion

10:01am there are 2^1 bacteria

at 10:02am there are 2^2= 4 bacteria

at 10:03am there are 2^3= 8 bacteria

a). at 10:06am there are 2^6 = 64 bacteria

b). at 10:20am there are 2^20 = 1048576 bacteria

c). at 10:50am there are 2^50 = 1.125899907 x 10^15

D).

test tube is full at 11 there are 2^60 =1.152921505*10^18

at 10:59 there are = 2^59 =5.764607523 *10^17

fraction = bacteria present at 10:59 / bacteria present at 11:00

=0.5

so half of vessal are present at 10:59

E).similerly

fraction = bacteria at 10:58 / bacteria at 11:00

=2.88*10^17/11.53*10^17

=2.88/11.53

=0.25