# The bacteria in a

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.

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