Question

In a laboratory the count of bacteria ina certain sxperiment was increasing at the rate of 2.5 per hour.find the bacteria at the end of 2hours if the count was intially 5,06,000.
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Answer 1

Answer:

The count of bacteria at the end of 2 yrs is 5,31,616.

Step-by-step explanation:

Given:

• Initial count of the bacteria = 5,06,000
• The rate at which it is increasing = 2.5%
• Given time = 2 hrs.

Need to find:

• The count of bacteria at the end of 2hours.

Explanation:

As we know that,

$\large \implies{ \blue{V \: = V(1 + \frac{R}{100} ) ^{n}}}$

$\small \implies{V \: = 50600( 1 + \frac{25}{1000} ) ^{2} }$

$\small \implies{V \: = 1265 \times \frac{41}{40} \times \frac{41}{40}}$

$\small \implies{V \: = 531616.25}$

$\small{ \bold{ \implies{V \: = 531616}}}$

Hence,the count of bacteria at the end of 2 yrs is 5,31,616.

Answer 2

$\huge \underline \mathbb {SOLUTION:-}$

AnsWer:

• The count of is 5,31,616 at the end of 2 hours.

Given:

• The count of bacteria in a certain experiment was increasing at the rate of 2.5 per hour.
• Time Given is 2 hours
• The Count was Intially 5,06,000

Need To Find:

• The count of bacteria at the end of 2hours = ?

Explanation:

Formula used here:

• V = V(1 + R/100)^n

Putting the values according to the given formula:

➠ V = 50600(1 + 25/1000)^2

➠ V = 1265 × 41/40 × 41/40

➠ V = 531616.25

➠ V = 531616

Therefore:

• The count of is 5,31,616 at the end of 2 hours.

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