In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the number of bacteria at the end of 2 hours if the count was initially 5, 06,000.

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Solution

A = P × ${[1 + \frac{R}{100}]}^{n}$

From given problem substitute

P = 506000

R = 2.5

n = 2

When these values are substituted,

We get A = 531616.25

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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 number of bacteria at the end of 2 hours if the count was initially 5, 06,000.

A = P × [1+R100]n From given problem substitute P = 506000 R = 2.5 n = 2 When these values are substituted, We get A = 531616.25

A = P × ${[1 + \frac{R}{100}]}^{n}$

From given problem substitute

P = 506000

R = 2.5

n = 2

When these values are substituted,

We get A = 531616.25