By using the formula, find the amount and compound interest on:

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

$Growth of bacteria in a culture (R_{1}) = 10 % in first hour Decrease in next hour (R_{2}) = 10 % Increase in the third hour (R_{3}) = 10 % Bacteria in the beginning = 20000 ∴ Bacteria after 3 hours = P (1 + \frac{R_{1}}{100}) (1 - \frac{R_{2}}{100}) (1 + \frac{R_{3}}{100}) = 20000 (1 + \frac{10}{100}) (1 - \frac{10}{100}) (1 + \frac{10}{100}) = 20000 \times \frac{11}{10} \times \frac{9}{10} \times \frac{11}{10} = 21780$

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By using the formula, find the amount and compound interest on: 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.

Growth of bacteria in a culture(R1)=10% in first hourDecrease in next hour(R2)=10%Increase in the third hour(R3)=10%Bacteria in the beginning = 20000∴ Bacteria after 3 hours= P(1+R1100)(1−R2100)(1+R3100)=20000(1+10100)(1−10100)(1+10100)=20000×1110×910×1110=21780

By using the formula, find the amount and compound interest on:

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.