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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 7.3, 2 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.