The population of a town in $2020$ was 100000. The population decreased by $y \%$ from the year $2020$ to 2021, and increased by $\mathrm{x} \%$ 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

Given Information:

Population in 2020 = 100000

Population decreases by y% from 2020 to 2021

Population increases by x% from 2021 to 2022

Population in 2022 > Population in 2020

x - y = 10 (where x, y are natural numbers)

Goal: Find the minimum possible population in 2021

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Let us express the population changes mathematically:

Population in 2021 = $100000 \times \left(1 - \tfrac{y}{100}\right) = 100000 \times \tfrac{100-y}{100}$

Population in 2022 = Population in 2021 × $\left(1 + \tfrac{x}{100}\right) = 100000 \times \tfrac{100-y}{100} \times \tfrac{100+x}{100}$

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Since population in 2022 > population in 2020:

$100000 \times \dfrac{100-y}{100} \times \dfrac{100+x}{100} > 100000$

$\dfrac{(100-y)(100+x)}{10000} > 1$

$(100-y)(100+x) > 10000$

Since $x = y + 10$:

$(100-y)(100+y+10) > 10000$

$(100-y)(110+y) > 10000$

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Expanding the left side:

$(100-y)(110+y) = 11000 + 100y - 110y - y^2 = 11000 - 10y - y^2$

The inequality becomes:

$11000 - 10y - y^2 > 10000$

$1000 - 10y - y^2 > 0$

$y^2 + 10y - 1000 < 0$

Using the quadratic formula:

$y = \dfrac{-10 \pm \sqrt{100 + 4000}}{2} = \dfrac{-10 \pm \sqrt{4100}}{2}$

Since $\sqrt{4100} \approx 64.03$:

$y = \dfrac{-10 \pm 64.03}{2}$

This gives us: $y < \dfrac{-10 + 64.03}{2} = 27.015$

Since y must be a natural number: $y \leq 27$

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To minimize the population in 2021, we need to maximize the decrease (maximize y).

Maximum value of $y = 27$

Therefore, $x = y + 10 = 37$

Minimum population in 2021 = $100000 \times \dfrac{100-27}{100} = 100000 \times \dfrac{73}{100} = 73000$

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Therefore, the minimum possible population in 2021 is 73000.