Ex 8.3, 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.Given,
Initial count of bacteria = 506000
It is increasing at the rate of 2.5% per hour
Here, 2.5% is the compounded Rate.
So we use the formula
A = P (1+𝑅/100)^𝑛
Here,
A = Count of bacteria at the end of 2 hours
P = Initial Count of bacteria = 5,06,000
R = 2.5%
N = Number of hours = 2
Putting Values in formula,
A = 5,06,000 (1+2.5/100)^2
= 506000 × (1+25/(10 × 100))^2
= 506000 × (1+25/1000)^2
= 506000 × ((1000 + 25)/1000)^2
= 506000 × (1025/1000)^2
= 506000 × 1050625/(1000 × 1000 )
= (506 × 1050625)/(1000 )
= 531616250/1000
= 531616.25
Since number of bacteria cannot be in decimals
∴ Count of bacteria at the end of 2 hours is 5,31,616 (approx)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.