The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and again increases by 10% in the third hour. Find the bacteria at the end of 3 hours if the count was initially 20000.
Initial count of bacteria, P = 20000
Time, n = 3 hours
Increasing rate, R = 10% 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 at the end of 1st hour,
∴ Count of bacteria = P (1 + R/100)n
= 20000 (1 + 10/100)1
= 20000 (1 + 1/10)
= 20000 (11/10)
= 20000 × 11/10
= 2000 × 11
= 22000
∴ Count of bacteria at the end of 1st hour is 22000.
Now,
Decreasing rate = 10%
∴Count of bacteria at the end of 2nd hour,
∴ Count of bacteria = P (1 + R/100)n
= 22000 (1 - 10/100)1
= 22000 (1 - 1/10)
= 22000 × 9/10
= 2200 × 9
= 19800
∴ Count of bacteria at the end of 2nd hours is 19800.
Now,
Increasing rate = 10%
∴Count of bacteria at the end of 3rd hour,
∴ Count of bacteria = P (1 + R/100)n
= 19800 (1 + 10/100)1
= 19800 (1 + 1/10)
= 19800 (11/10)
= 19800 × 11/10
= 1980 × 11
= 21780
∴ Count of bacteria at the end of 3rd hours is 21780.