The population of a town in 2020 was 100000. The population decreased by y% from the year 2020 to 2021, and increased by x% from the year 2021 to 2022, where x and y are two natural numbers. If population in 2022 was greater than the population in 2020 and the difference between x and y is 10, then the lowest possible population of the town in 2021 was

It is given that the population of the town in 2020 was 100000. The population decreased by y% from the year 2020 to 2021 and increased by x% from the year 2021 to 2022, where x and y are two natural numbers.

Hence, the population in 2021 is $$100000\left(\ \frac{\ 100-y}{100}\right)$$.

The population in 2022 is $$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 100+x}{100}\right)$$

It is also given that the population in 2022 was greater than the population in 2020 and the difference between x and y is 10.

Hence, 

$$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 100+x}{100}\right)>\ 100000$$, and (x-y) = 10

=> $$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 110+y}{100}\right)>\ 100000$$

=> $$\ \frac{\ 100-y}{100}\left(\ \frac{\ 110+y}{100}\right)>\ 1$$

To get the minimum possible value of 2021, we need to increase the value of y as much as possible.

Hence, $$\left(\ \ 100-y\right)\left\{\left(\ \ 100+y\right)+10\right\}>\ 10000$$

=> $$10000-y^2+1000-10y>\ 10000$$

=> $$y^2+10y<1000$$

=> $$y^2+10y+25<1025$$

=> $$\left(y+5\right)^2=1024\ <\ 1025$$

=> $$\left(y+5\right)^2=32^2$$

=> $$y=27$$

Hence, the population in 2021 is 100000*(100-27) = 73000

The correct option is C