Question 10.

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,

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

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

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

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

Hence, $(100-y)\{(100+y) + 10\} \gt 10000$

=> $10000 - y^2 + 1000 - 10y \gt 10000$

=> $y^2 + 10y \lt 1000$

=> $y^2 + 10y + 25 \lt 1025$

=> $(y + 5)^2 = 1024 \lt 1025$

=> $(y + 5)^2 = 32^2$

=> $y = 27$

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