Question

In 2021, the population was 100000. In 2022 it is decreased by x%. In 2023, it will increase by y%. The population in 2023 is more than that of 2021 and difference between x and y is 10. Minimum population in 2021 was?

Answer 1

Given :
Population of the town in 2020 was 100000.
Population decreased by y% from the year 2020 to 2021 and increased by x% from the year 2021 to 2022 , where both x and y are natural numbers.
Therefore, the population in 2021 is 100000 $(\frac{100-y}{100})$

The population of 2022 is 100000 $(\frac{100-y}{100})(\frac{100+x}{100})$
Now given :
Population in 2022 was greater than the population in 2020 and the difference between x and y is 10.
Therefore,
$100000(\frac{100-y}{100})(\frac{100+x}{100})>100000$, and (x - y) = 10

⇒ $100000(\frac{100-y}{100}(\frac{110+y}{100})>100000$

⇒ $\frac{100-y}{100}(\frac{110+y}{100})>1$
To get the maximum possible value of 2021, we are required to increase the value of y as much as possible.
Therefore, (100 - y) {(100 + y) + 10} > 10000
⇒ 100000 - y2 + 1000 - 10y > 10000
⇒ y2 + 10y < 10000
⇒ (y + 5)2 + 25 < 1025
⇒ (y + 5)2= 322
⇒ y = 27
Therefore, the population of 2021 is :
10000 × (100 - 27) = 73000

So, the correct answer is 73000.