The count of a certain breed of bacteria was found to increase at the rate of 4% per hour and then decrease by 2% per hour. Find the bacteria at the end of 2 hours if the count was 2,00,000 initially.

Correct option is C

Given:
Initial count = 2,00,000
Growth rate R1 = 4%
Decay rate R2 = 2%
Formula Used:
Final Count = $$P(1 + \frac{R1}{100})(1 - \frac{R2}{100})$$​​
Solution:
Bacteria after 1st hour = 200000 $$\times$$​ (1 + 0.04) = 208000
Bacteria after 2nd hour = 208000 $$\times$$​ (1 - 0.02) = 208000 $$\times$$​ 0.98
Final Count = 203840
So the correct answer is (c)