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Hint:- For solving a given problem we use the compound interest formula. Compound interest is the interest calculated on the initial principal including the interest from the previous periods on the deposit.

Given: Initial count of bacteria(P) = 5,06,000, Increasing rate of bacteria(R)=2.5% ,Time = 2 hours

After two hours , the amount of bacteria can be given by using the compound interest formula i.e. \[{\text{Amount = P}}{\left( {1 + \dfrac{{\text{R}}}{{100}}} \right)^{\text{n}}}\] , where n is the time. Putting the values in the equation , we get

\[{\text{Amount = 506000}}{\left( {1 + \dfrac{{2.5}}{{100}}} \right)^2}\]

\[{\text{Amount = 506000}}{\left( {1 + 0.025} \right)^2}\]

\[{\text{Amount = 506000}} \times {\text{1}}{\text{.050625}}\]

\[{\text{Amount = }}\]531616.25

Hence, the number of bacteria present at the end of 2 hours will be approximately 5,31,616.

Note:- In these types of questions , the key point is to understand whether a simple interest or compound interest formula is required. In case of simple interest the principle remains constant for all periods.

__Complete step-by-step solution -__Given: Initial count of bacteria(P) = 5,06,000, Increasing rate of bacteria(R)=2.5% ,Time = 2 hours

After two hours , the amount of bacteria can be given by using the compound interest formula i.e. \[{\text{Amount = P}}{\left( {1 + \dfrac{{\text{R}}}{{100}}} \right)^{\text{n}}}\] , where n is the time. Putting the values in the equation , we get

\[{\text{Amount = 506000}}{\left( {1 + \dfrac{{2.5}}{{100}}} \right)^2}\]

\[{\text{Amount = 506000}}{\left( {1 + 0.025} \right)^2}\]

\[{\text{Amount = 506000}} \times {\text{1}}{\text{.050625}}\]

\[{\text{Amount = }}\]531616.25

Hence, the number of bacteria present at the end of 2 hours will be approximately 5,31,616.

Note:- In these types of questions , the key point is to understand whether a simple interest or compound interest formula is required. In case of simple interest the principle remains constant for all periods.

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