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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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Solution
Let the number of bacteria initially = P
And number of bacteria at the end of 2 hours = A
Given initial number of bacteria (P) = 5,06,000
Time (n) = 2 years
Rate (R) = 2.5% p.a.
Number of bacteria at the end of 2 hours
(A) = P `(1+"R"/100)^"n"`
= 5,06,000 `(1+2.5/100)^2`
= 5,06,000 `(1+1/40)^2`
= 5,06,000 `(41/40)^2`
= 5,31,616.25
Hence, number of bacteria at the end of 2 hours (A) = 5,31,616.25
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