The population of a city was 120000 in the year 2009. During next year it increased by 6% but due to an epidemic it decreased by 5% in the following year. what is its population in the year 2011?

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Solution

$Population of the city in 2009, P = 120000 Rate of increase, R = 6 % Time, n = 3 years Then the population of the city in the year 2010 is g i v e n b y Population = P \times {(1 + \frac{R}{100})}^{n} = 120000 \times {(1 + \frac{6}{100})}^{1} = 120000 \times (\frac{100 + 6}{100}) = 120000 \times (\frac{106}{100}) = 120000 \times (\frac{53}{50}) = 2400 \times 53 = 127200 Therefore, the population of the city in 2010 is 127200 . A g a i n, p opulation of the city in 2010, P = 127200 Rate of decrease, R = 5 % Then the population of the city in the year 2011 is g i v e n b y Population = P \times {(1 - \frac{R}{100})}^{n} = 127200 \times {(1 - \frac{5}{100})}^{1} = 127200 \times (\frac{100 - 5}{100}) = 127200 \times (\frac{95}{100}) = 127200 \times (\frac{19}{20}) = 6360 \times 19 = 120840 Therefore, the population of the city in 2011 is 120840 .$

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