Question

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.

Answer 1

Given:

Intial count of Bacteria (P) = 5,06,000, increasing Rate of Bacteria(R) = 2.5%, Time = 2 hours

After 2 hours, number of bacteria,

Amount (A) = P(1+R/100)^n

A= 506000(1+2.5/100)^n

A= 506000(1+25/1000)²


A= 506000(1+1/40)²

A= 506000(41/40)²

A= 506000(41×41/40×40)

A= 506000×1681/ 1600

A=(5060 × 1681) /16

A= 316.25×1681

A= 5,31,616.25

Hence, the number of bacteria after two hours are 531616 (approx.).

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Hope this will help you....

Answer 2

Solution :-

Bacteria count initially = 506000

Rate of increasing per hour = 2.5 %

Time = 2 hours

This question can be solved through the formula of compound interest.

⇒ A = P*(1 + R/100)ⁿ

⇒ A = 506000*(1 + 2.5/100)²

⇒ 506000*(1/1 + 25/1000)²

⇒ 506000*(1/1 + 1/40)²

Taking LCM of the denominators and then solving it.

⇒ 506000*41/40*41/40

⇒ 531616.25

As the number of bacteria cannot be in decimal, so the number of bacteria at the end of 2 hours is 531616.

Answer.