The population of a city increases at the rate of 4% per annum. There is an additional increase of 1% per annum due to influx of job seekers. The percent increase in population after 2 years is:

The population of a city increases at the rate of 4% per annum. There is an additional increase of 1% per annum due to influx of job seekers. The percent increase in population after 2 years is:
1). 10.25%
2). 12.5%
3). 15.755
4). 8.25%

If the original population of a locality be P and the annual growth rate be r%, then the population of the locality after n years-

$(= P{\left( {1 + \frac{r}{{100}}} \right)^n})$

The population of a city increases at the rate of 4% per annum. There is an additional increase of 1% per annum due to influx of job seekers.

According to the question, total percentage increase = (4 + 1)% = 5%

Let initial population = 100,

∴ Population after 2 years $(= 100{\left( {1 + \frac{5}{{100}}} \right)^2} = 100{\left( {\frac{{21}}{{20}}} \right)^2} = \frac{{441}}{4} = 110.25)$

Increase in population = 110.25 – 100 = 10.25

∴ Percent increase = (10.25/100) × 100% = 10.25%

Hence, the required percentage increase in the population of the city is 10.25%.

Let the population of the city be 100

By question, increase in population% annually is 5%

Therefore , after 1st year the population becomes 105

After 2nd year the population become:105+5/100×105=105+21/4=105+5.25=110.25

Hence, population increase after to year is 110.25-100=10.25.which is 10.25%(10.25/100×100)%

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