Ncert solutions

Ncert solutions

Ncert solutions

Grade 8

Q11) 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.

Solution:

Here, Principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time (n) = 2 hours

After 2 hours, number of bacteria,

Amount (A) = P\left(1+\frac{R}{100}\right)^n

= 506000\left(1+\frac{2.5}{100}\right)^2

= 506000\left(1+\frac{25}{1000}\right)^2

= 506000\left(1+\frac{1}{40}\right)^2

= 506000\left(\frac{41}{40}\right)^2

= 506000\times\frac{41}{40}\times\frac{41}{40}

= 5,31,616.25

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

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Book a free class nowSolution:

Here, Principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time (n) = 2 hours

After 2 hours, number of bacteria,

Amount (A) = P\left(1+\frac{R}{100}\right)^n

= 506000\left(1+\frac{2.5}{100}\right)^2

= 506000\left(1+\frac{25}{1000}\right)^2

= 506000\left(1+\frac{1}{40}\right)^2

= 506000\left(\frac{41}{40}\right)^2

= 506000\times\frac{41}{40}\times\frac{41}{40}

= 5,31,616.25

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

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