In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour.

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Find the bacteria at the end of 2 hours if the count was in itially 5,06,000.

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This problem can be solved on the basis of CI problems.

Let principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time, n = 2 hours

After 2 hours, number of bacteria ,= Comparing Quantities

Comparing Quantities

=5,31,616.25

Hence,numberofbacteriaaftertwohoursare531616(approx.)

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