Question

The count of bacteria in a certain experiment was increasing at the rate of 2%per hour. Find the bacteria at the end of 2 hours if the count was initially 500000.

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Answer 1

Question :

Q. The count of bacteria in a certain experiment was increasing at the rate of 2%per hour. Find the bacteria at the end of 2 hours if the count was initially 5,00,000.

Given :

• The rate in which the bacteria is increasing is 2%per hr.
• Present count of the bacteria is 5,00,000 with respect to the time of 2 hrs.

To find :

• The initial (present) count of bacteria after 2 hrs (by following the given situations) .

Solution :

Here,

The given values are taken as :

➠ Initial count of the bacteria

• P (principal) i. e., 5,00,000

➠ Increasing rate :

• r (rate) i.e., 2% per hr.

➠Time taken :

• n (time) i.e., 2 hrs.

We know that,

➠ A(amount) = P(1 +$\sf \frac{r}{100} {)}^{n}$

Procedure :

Substituting the given values as follows :

➠ A = 5,00,000(1 + $\sf \frac{2}{100} {)}^{2}$

➠ A = $\rm{\cancel{5,00,000}}(\sf \frac{102}{\cancel{100}} \times\frac{102}{\cancel{100}} {)}^{}$

➠ A = 5 × 102 × 102

➠ A = 52,202. ans.

Hence,

The bacteria at the end of 2 hours will be :

An unit of = 52,202 ans.