Question 1159185
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120 minutes comprise 8 times 15-minute periods.


So, 120 minutes comprise 120/15 = 8 doubling periods.


When we go one doubling period back, from 120 minutes to 105 minutes, we should divide the current population size by the factor of 2.


When we go 8 doubling periods back, from 120 minutes to the very beginning, we should divide the population of 60000 by {{{2^8}}}, 

which gives the <U>initial population</U>  = {{{60000/2^8}}} = 234.375 = 234 (rounded).


From 120 minutes to 5 full hours, we have 3 hours = 3*4 = 12 doubling periods.


Therefore, the size of the bacterial population after 5 hours is  {{{60000*2^12}}} = 245760000.
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Solved.