Suppose a population of insects quadruples every 6 months an
Solution
(a) Let c be the one months growth factor, then and if initial population of insects be a, then
Population after one month = a.c, Population after four months = a.c4
Population after two months = a.c2 Population after two months = a.c5
Population after three months = a.c3 Population after two months = a.c6
(b) The population, using Six months growth factor for the population of bacteria = 4 a
Thus, a c6 = 4 a or, c = 41/6
(c) Let y be the one year growth factor, then population after 1 year will be
= a.y, where a is the initial population of the insects
Population of insects at the end of first six month = (4 a), and
Population of insects at the end of second six month (i.e. after i year) = 4 (4 a) = 16 a
Thus, ay = 16 a => y = 16
(d) Taking 1 month = 30 days, if d be the one day growth factor, and if initial population of insects be a, then
the population of insects after 30 days = a d30 = a c = a c => d = 41/180 = 0.0056
