Question

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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.

Answer 1

Answer:

A

531616

We have, P

0

= Original count of bacteria =506000,

Rate of increase =R=2.5% per hour,

Time =2 hours

∴ Bacteria count after 2 hours = P

P=P

0

(1+

100

R

)

T

=506000×(1+

100

2.5

)

2

=506000×

100

102.5

×

100

102.5

=531616.25=531616 (approx)

Answer 2

Given :

• Initial count of bacteria = 5,06,000
• Rate of increase = 2.5% per hours
• Time = 2hours

To Find :

The count of bacteria after 2 hours .

Formula to be applied :

Here in this situation, The formula of CI will be used.

Which says,

Amount = Principle(1 + Rate/100)^Time

Here, The amount will change to final count of bacteria and Principle is the initial count of bacteria.

Solution :

Given, initial count of bacteria = 5,06,000

Rate = 2.5% per hours

Time = 2hours

Hence, the final count

$\tt{ \implies fc = 506000(1 + \frac{2.5}{100} {)}^{2} }$

$\tt{ \implies fc = 506000(1 + \frac{1}{40} {)}^{2} }$

$\tt{ \implies fc = 506000 \times \frac{1681}{1600} }$

$\implies \tt{fc = 316.25 \times 1681}$

$\red{ \underline{ \boxed{ \tt{ \therefore \: final \: count ≈ 531616.25}}}}$

Hence, Final count of Bacteria ≈ 5,31,616 bacteria.

__________________________

Reason :

Why I used compound interest formula ?

Because, when the question is related to population of any living thing, we use the formula of compound interest.