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.

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Solution

$Initial count of bacteria, P = 20000 Rate of increase, R = 10 % Time, n = 3 hours Then the count of bacteria at the end of the first hour is given by Count of bacteria = P \times {(1 + \frac{10}{100})}^{n} = 20000 \times {(1 + \frac{10}{100})}^{1} = 20000 \times (\frac{100 + 10}{100}) = 20000 \times (\frac{110}{100}) = 20000 \times (\frac{11}{10}) = 2000 \times 11 = 22000 Therefore, the count of bacteria at the end of the first hour is 22000 . The count of bacteria at the end of the second hour is given by Count of bacteria = P \times {(1 - \frac{10}{100})}^{n} = 22000 \times {(1 - \frac{10}{100})}^{1} = 22000 \times (\frac{100 - 10}{100}) = 22000 \times (\frac{90}{100}) = 22000 \times (\frac{9}{10}) = 2200 \times 9 = 19800 Therefore, the count of bacteria at the end of the second hour is 19800 . Then the count of bacteria at the end of t h e third hour is i s g i v e n b y Count of bacteria = P \times {(1 + \frac{10}{100})}^{n} = 19800 \times {(1 + \frac{10}{100})}^{1} = 19800 \times (\frac{100 + 10}{100}) = 19800 \times (\frac{110}{100}) = 19800 \times (\frac{11}{10}) = 1980 \times 11 = 21780 Therefore, the count of bacteria at the end of t h e first 3 hours is 21780 .$

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