question_answer 27)

                In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours, if the count was initially 5, 06, 000.

Answer:

                P = 5, 06, 000 R = 2.5% per hour \[\therefore \]  \[A=P{{\left( 1+\frac{R}{100} \right)}^{n}}\] \[=5,06,000{{\left( 1+\frac{2.5}{100} \right)}^{2}}\] \[=5,06,000{{\left( 1+\frac{1}{40} \right)}^{2}}\] \[=5,06,000{{\left( \frac{41}{40} \right)}^{2}}\] \[=5,06,000\times \frac{41}{40}\times \frac{41}{40}\] = 5,31, 616 (approx.) Hence, the count of bacteria at the end of 2 hours is 5, 31, 616 (approx.).