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.

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Answer 1

5,31,616 bacteria

Explanation:

Solution:

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

After 2 hours, number of bacteria,

Amount (A) = P\left(1+\frac{R}{100}\right)^nP(1+100R)n

= 506000\left(1+\frac{2.5}{100}\right)^2506000(1+1002.5)2

= 506000\left(1+\frac{25}{1000}\right)^2506000(1+100025)2

= 506000\left(1+\frac{1}{40}\right)^2506000(1+401)2

= 506000\left(\frac{41}{40}\right)^2506000(4041)2

= 506000\times\frac{41}{40}\times\frac{41}{40}506000×4041×4041

= 5,31,616.25

Hence, number of bacteria after two hours are 531616 (approx.).