Question 54

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

Solution

It is given thatĀ 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 isĀ $$100000\left(\ \frac{\ 100-y}{100}\right)$$.

The population in 2022 isĀ $$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 100+x}{100}\right)$$

It is also given that theĀ population in 2022 was greater than the population in 2020 and the difference between x and y is 10.

Hence,Ā 

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

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

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

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

Hence,Ā $$\left(\ \ 100-y\right)\left\{\left(\ \ 100+y\right)+10\right\}>\ 10000$$

=>Ā $$10000-y^2+1000-10y>\ 10000$$

=>Ā $$y^2+10y<1000$$

=>Ā $$y^2+10y+25<1025$$

=>Ā $$\left(y+5\right)^2=1024\ <\ 1025$$

=>Ā $$\left(y+5\right)^2=32^2$$

=>Ā $$y=27$$

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

The correct option is C

Video Solution

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