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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Solution

Her, Principal (P) = 5,06,000 ,
Rate of interest (R) = 2.5%
Time (n) = 2 hours

After 2 hours, number of bacteria,

$A m o u n t (A)$
$A = P {(1 + \frac{R}{100})}^{n}$
$= 506000 {(1 + \frac{2.5}{100})}^{2}$
$= 506000 {(1 + \frac{25}{1000})}^{2}$
$= 506000 {(1 + \frac{1}{40})}^{2}$
$= 506000 {(\frac{41}{40})}^{2}$
$= 506000 \times \frac{41}{40} \times \frac{41}{40}$
$= 5, 31, 616.25$

Hence, number of bacteria after two hours are 531616 (approx.)

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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.

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.