The count of bacteria in a certain experiment was increasing at the rate of 2% per hour. Find the bacteria at the end of 2 hours if the count was initially 500000.
Count of bacteria, P = 500000
Time, n = 2 hours
Increasing rate, R = 2% per hour
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time]
∴ Count of bacteria = P (1 + R/100)n
= 500000 (1 + 2/100)2
= 500000 (102/100)2
= 500000 × 102/100 × 102/100
= 50 × 102 × 102
= 520200
∴ Count of bacteria at the end of 2 hours is 520200.