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

The correct option is 73000.

Let us write down the population of the town step-by-step.

Let the population in the year 2020 be $P_{2020}=100000$.

The population decreased by $y$% from 2020 to 2021. Therefore, the population in 2021, $P_{2021}$, is:
$P_{2021}=100000\times \left(1-\frac{y}{100}\right)=1000(100-y)$

The population increased by $x$% from 2021 to 2022. Therefore, the population in 2022, $P_{2022}$, is:
$P_{2022}=P_{2021}\times \left(1+\frac{x}{100}\right)=1000(100-y)\left(1+\frac{x}{100}\right)=10(100-y)(100+x)$

We are given that the population in 2022 is greater than the population in 2020:
$P_{2022}>P_{2020}$

Substituting the values, we get:
$10(100-y)(100+x)>100000$
$\Rightarrow (100-y)(100+x)>10000$

We want to find the lowest possible population in 2021, which is $1000(100-y)$. To minimize this value, we must maximize the natural number $y$.

The difference between $x$ and $y$ is 10, meaning $\left|x-y\right|=10$. This gives us two cases:

Case 1: $x-y=10\Rightarrow x=y+10$

Substituting $x=y+10$ into our inequality:
$(100-y)(100+y+10)>10000$
$\Rightarrow (100-y)(110+y)>10000$
$\Rightarrow 11000-10y-y^2>10000$
$\Rightarrow y^2+10y-1000<0$

We find the roots of the corresponding quadratic equation $y^2+10y-1000=0$ using the quadratic formula:
$y=\frac{-10\pm \sqrt{{10}^2-4\left(1\right)(-1000)}}{2}=\frac{-10\pm \sqrt{4100}}{2}=-5\pm 5\sqrt{41}$

Since $\sqrt{41}\approx 6.403$, the positive root is:
$y\approx -5+32.015=27.015$

Since the inequality requires $y^2+10y-1000<0$, we must have $y<27.015$. Since $y$ is a natural number, the maximum integer value it can take is:
$y=27$

Case 2: $y-x=10\Rightarrow x=y-10$

Substituting $x=y-10$ into our inequality:
$(100-y)(100+y-10)>10000$
$\Rightarrow (100-y)(90+y)>10000$
$\Rightarrow 9000+10y-y^2>10000$
$\Rightarrow y^2-10y+1000<0$

For the expression $y^2-10y+1000$, the discriminant is:
$D={(-10)}^2-4\left(1\right)\left(1000\right)=100-4000=-3900<0$

Since the discriminant is negative and the leading coefficient is positive, the quadratic $y^2-10y+1000$ is always positive for all real numbers. Thus, there are no real solutions for this case.

Consequently, the maximum value of $y$ is 27. The lowest possible population of the town in 2021 is:
$P_{2021}=1000(100-27)=1000\times 73=73000$