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 $2$ hour if the count was initially $5, 06, 000$ .

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Solution

Initial count of bacteria $= 5, 06, 000$
Increasing rate $= 2.5$ %/hr
Count of bacteria after $1$ hr.
$\begin{matrix} \Rightarrow 5, 06, 000 + \frac{2.5}{100} \times 5, 06, 000 = 5, 18, 650 \end{matrix}$
Count of bacteria after $2$ hr
$\begin{matrix} \Rightarrow 5, 18, 650 + \frac{2.5}{100} \times 518650 = 531616.25 \end{matrix}$

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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 $5, 06, 000$ .

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 hour if the count was initially 5,06,000.

Initial count of bacteria =5,06,000Increasing rate =2.5%/hrCount of bacteria after 1 hr.⇒5,06,000+2.5100×5,06,000=5,18,650 Count of bacteria after 2hr⇒5,18,650+2.5100×518650=531616.25

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 $2$ hour if the count was initially $5, 06, 000$ .