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
Explanation:
Population at the end of 2021 = 1,00,000 × (1 - x/100) Population at the end of 2022 = 1,00,000 × (1 - y/100) × (1 + x/100) ...(1)
∴ Overall % change = -y + x + (-y × x)/100 [using formula = a + b + ab/100]
Since population in 2022 is greater than that in 2020 ⇒ x > y. ∴ x = y + 10
Hence, the population decreases by y% and then increases by (y + 10)% ∴ Overall % change = -y + (y + 10) + (-y × (y + 10))/100 = 10 - y(y + 10)/100 [using formula = a + b + ab/100]
Overall change should be positive ∴ Overall % change = 10 - y(y + 10)/100 > 0 ⇒ y(y + 10) < 1000
Higheset value of y satisfying the above inequality is 27.
∴ The population will decrease by a maximum of 27%. ⇒ Least population in 2021 = 1,00,000 × (1 - 27%) = 73,000.
Hence, option (a).
Concept: Successive Percentage Change Formula
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