CAT 2023 Slot 3 Solved Paper

It is given that the population of the town in 2020 was 100000 . The population decreased by $\mathit{y}\%$ from the year 2020 to 2021 and increased by $\mathit{x}\%$ from the year 2021 to 2022, where $\mathit{x}$ and $\mathit{y}$ are two natural numbers.
Hence, the population in 2021 is $100000\left(\frac{100-\mathit{y}}{100}\right)$.
The population in 2022 is $100000\left(\frac{100-\mathit{y}}{100}\right)\left(\frac{100+\mathit{x}}{100}\right)$
It is also given that the population in 2022 was greater than the population in 2020 and the difference between $\mathrm{x}$ and $\mathrm{y}$ is 10 .
Hence,

$\Rightarrow 100000\left(\frac{100-\mathit{y}}{100}\right)\left(\frac{110+\mathit{y}}{100}\right)>100000$
$\Rightarrow \frac{100-\mathit{y}}{100}\left(\frac{110+\mathit{y}}{100}\right)>1$
To get the minimum possible value of 2021 , we need to increase the value of $\mathit{y}$ as much as possible.

$\Rightarrow 10000-{\mathit{y}}^2+1000-10\mathit{y}>10000$



$\Rightarrow {\left(\mathit{y}+5\right)}^2={32}^2$
$\Rightarrow \mathit{y}=27$
Hence, the population in 2021 is $100000*\left(100-27\right)=73000$
The correct option is $\mathrm{C}$