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. [2 Marks]

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Solution

The initial count of bacteria is given as 5,06,000
Bacteria at the end of 2 hours:
$= 506000 {(1 + \frac{2.5}{100})}^{2} = 506000 \times {(1 + \frac{1}{40})}^{2} = 506000 \times {(\frac{41}{40})}^{2} = 531616.25$
Therefore, the count of bacteria at the end of 2 hours will be 5,31,616 (approx.)
[2 Mark]

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Q. Question 11
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.

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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. [2 Marks]

The initial count of bacteria is given as 5,06,000 Bacteria at the end of 2 hours: = 506000(1 + 2.5100)2 = 506000×(1 + 140)2 = 506000×(4140)2 = 531616.25 Therefore, the count of bacteria at the end of 2 hours will be 5,31,616 (approx.) [2 Mark]