11. In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.

We have: Initial count of bacteria = 5,06,000

Rate = 2.5% per hour

n = 2 hours

So, we have:

num. of bacteria(end of 2 hours)=Num. of count of bacteria initially\(\times(1+\frac{R}{100})^n\)

\( =5,06,000\times(1+\frac{2.5}{100})^2\)

\( =5,06,000\times(\frac{41}{40})^2\)

\( =5,06,000\times(\frac{1681}{1600})^2\)

\( =5,31,616.25\)

Hence we have, the number of bacteria after two hours = 5,31,616

(approx).