Q. 264.3( 17 Votes )

The bacteria in a

Answer :

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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