CAT 2023 Slot 3 Solved Paper

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, \text{ and } (x-y)=10$$\Rightarrow 100000 \left( \frac{100-y}{100} \right) \left( \frac{110+y}{100} \right) > 100000$$\Rightarrow \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.$\text{ Hence, } (100-y)\{(100+y)+10\} > 10000$$\Rightarrow 10000 - y^2 + 1000 - 10y > 10000$$\Rightarrow y^2 + 10y < 1000$$\Rightarrow y^2 + 10y + 25 < 1025$$\Rightarrow (y+5)^2 = 1024 < 1025$$\Rightarrow (y+5)^2 = 32^2$$\Rightarrow y = 27$Hence, the population in 2021 is $100000 \cdot (100-27) = 73000$The correct option is $C$