# 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.

Given,

Initial count of bacteria, P = 5, 06,000 (Principal Amount)

Rate of increase, R = 2.5% per hour.

Time period, n = 2 hours

(This question is done in a similar manner as compound interest)

Number of bacteria after 2 hours = $P(1+\frac{R}{100})^n$

$= 506000(1+\frac{2.5}{100})^2 \approx 531616$

Therefore, the number of bacteria at the end of 2 hours will be 531616 (approx)

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