CAT 2023

Slot 3

QA

Question & Solution

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