Tried this population question but getting confused with the % values. Can someone help?

Question

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

A. 74000
B. 75000
C. 72000
D. 73000

A. 74000
B. 75000
C. 72000
D. 73000

V Gupta · Accepted Answer

This is a percentage-based population change question, but it gets easier once you fix x = y + 10.

Use the formula: 100000 &times; (1 &ndash; y/100) &times; (1 + x/100) > 100000
Cancel out 100000 and plug in x = y + 10
It becomes: (1 &ndash; y/100)(1 + (y + 10)/100) > 1
Solve that inequality and you&rsquo;ll get (y + 5)^2 < 1025
That gives y &le; 27, so max drop = 27%
2021 population = 100000 &times; (1 &ndash; 27/100) = 73000

Answer is Option D. Just needed to handle the inequality slowly and plug in x properly.

T Ronak · Answer

The key to this question is using the condition that the population in 2022 is more than in 2020. Start by writing the expression for 2022 population: Population in 2022 = 100000 × (1 – y/100) × (1 + x/100) We’re told this is > 100000, and also that x = y + 10. Cancel 100000 and plug x = y + 10 into the inequality: (1 – y/100)(1 + (y+10)/100) > 1 Multiply and simplify: (100 – y)(100 + y + 10) > 10000 This gives a quadratic that solves to (y + 5)^2 < 1025 So y + 5 < 32 → y = 27 max To find minimum 2021 population, use the highest possible y. Population = 100000 × (1 – 27/100) = 73000 , Answer is Option D.