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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