Question

It is observed that the population of city X, increases at the rate of 2% per annum. If the population in 2009 is 212,000. Then the population in 2006, was closet to
1) 1,50,000
2) 95,000
3) 2,00,000
4) 2,50,000

Answer 1

answer is option no. 3

Answer 2

3) 2,00,000

The population in 2006, was closet to 2,00,000.

Explanation:

The exponential growth equation is given by :-

$y=A(1+r)^x$ , where  A = initial value , r = rate of growth ( in decimal) and x is time.

As per given , we have

r= 2% = 0.02

Take 2006 as initial year , then for 2009 x= 3.

If the population in 2009 is 212,000.  ,it means

$212000=A(1+0.02)^3\\\\\Rightarrow\ 212000=A(1.061208)\\\\\Rightarrow\ A=\dfrac{2120000}{1.061208}=1997723.34924\approx2,00,000$

Hence, the population in 2006, was closet to 2,00,000.

Thus , the correct answer is 3) 2,00,000.

# Learn more :

The population of a city increases at compounding rate of 8% per year. Find the population in the year 2012 if population in the year 2010 was 2,50,000.

Report by Industry1063 18.07.2018

https://brainly.in/question/4739735