Question

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 $5,06,000$.

Answer 1

Her, Principal $\left(P\right)=5,06,000$
Rate of interest $\left(R\right)=2.5\%$
Time $\left(n\right)=2$ hours
After $2$ hours, number of bacteria,
$\text{Amount,}A=P{(1+\frac{R}{100})}^n$
$=506000{(1+\frac{2.5}{100})}^2$
$=506000{(1+\frac{25}{1000})}^2$
$=506000{(1+\frac{1}{40})}^2$
$=506000{\left(\frac{41}{40}\right)}^2$
$=506000\times \frac{41}{40}\times \frac{41}{40}$
$=5,31,616.25$
Hence, the number of bacteria after two hours is $531616$ (approx.)