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.

531616

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528976

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567190

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543974

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Solution

The correct option is A
531616

A = P × ${[1 + \frac{R}{100}]}^{n}$

Given

P = 506000

R = 2.5

n = 2

A = 506000 × ${[1 + \frac{2.5}{100}]}^{2}$

A = 531616 approx

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

The correct option is A 531616 A = P × [1+R100]n Given P = 506000 R = 2.5 n = 2 A = 506000 × [1+2.5100]2 A = 531616 approx

The correct option is A
531616

A = P × ${[1 + \frac{R}{100}]}^{n}$

Given

P = 506000

R = 2.5

n = 2

A = 506000 × ${[1 + \frac{2.5}{100}]}^{2}$

A = 531616 approx