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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Answer 1

Answer:

$\huge\tt{\underline{Solution :-}}$

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.

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Answer 2

Answer:

$\huge{\underline{\mathtt{\red{A}\pink{N}\green{S} \blue{W} \purple{E} \orange{R}: }}}$

$Here, \\ Principal (P) = 5,06,000, \\ Rate \: of \: Interest (R) \: = 2.5 \%, \\ Time (n) = 2 hours \\ \red{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}} \\ After \: 2 \: hours, \: number \: of \: bacteria, \\ Amount (A) = P{(1 + \frac{R}{100} )}^{n} \\ = 506000{(1 + \frac{2.5}{100} )}^{2} \\ = 506000{(1 + \frac{25}{1000} )}^{2} \\ = 506000{(1 + \frac{1}{40} )}^{2} \\ = 506000{( \frac{41}{40} )}^{2} \\= 506000 \times \frac{41}{40} \times \frac{41}{40} \\ = 5,31,616.25 \\ \green{\rule{45pt}{7pt}}\green{\rule{45pt}{7pt}}\green{\rule{45pt}{7pt}}\green{\rule{45pt}{7pt}} \\ Hence, \: number \: of \: bacteria \: two \: hours \: are \: 531616 \: (approx.).$