Question -
Answer -
Let y be the number ofbacteria at any instant t.
Itis given that the rate of growth of the bacteria is proportional to the numberpresent.
Integratingboth sides, we get:
Let y0 be the number ofbacteria at t = 0.
⇒ log y0 = C
Substitutingthe value of C in equation (1), we get:
Also,it is given that the number of bacteria increases by 10% in 2 hours.
Substitutingthis value in equation (2), we get:
Therefore,equation (2) becomes:
Now, let the time when the number ofbacteria increases from 100000 to 200000 be t1.
⇒ y = 2y0 at t = t1
Fromequation (4), we get:
Hence, in 
hoursthe number of bacteria increases from 100000 to 200000.