Question : The population of the city was 12,50,000 in the year 2010. It increased at the rate of 2% per annum during the years 2011 and 2012. In the year 2013, it decreased by 1%. Find the population of the city at the end of 2013.
Option 1: 1,28,54,952
Option 2: 13,50,000
Option 3: 13,87,495
Option 4: 12,87,495
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Correct Answer: 12,87,495
Solution : The population of the city in 2010 is 1250000. Then, Population, $P$ = 1250000 Rate of increment, $R$ = 2% We have to find the population at the end of the year 2012, i.e. time, $n$ = 2 years Population after 2 years is given by New Population $= P(1+\frac{R}{100})^n$ $=1250000(1+\frac{2}{100})^2$ $=1250000\times \frac{51}{50}\times \frac{51}{50}$ $=13,00,500$ Again it decreased by 1% $R = –1$% New Population $= P(1-\frac{R}{100})$ $=1300500(1-\frac{1}{100})$ $=1300500\times \frac{99}{100}$ $=12,87,495$ Hence, the correct answer is 12,87,495.
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