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. 73000
• D. 72000

Answer 1

The Correct Option is C

Problem resolution proceeds incrementally. Data points are as follows:

• Initial population (2020): 100000.
• Population decline in 2021: y%.
• Population increase in 2022: x%.
• 2022 population exceeded 2020 population (> 100000).
• The relationship between x and y is defined by x = y + 10.

Population dynamics are formulated as equations:

• 2021 Population: \(P_{2021} = 100000 \times \left(1 - \frac{y}{100}\right)\)
• 2022 Population: \(P_{2022} = P_{2021} \times \left(1 + \frac{x}{100}\right)\)

Given the condition \(P_{2022} > 100000\), substitute \(P_{2021}\):

\(100000 \times \left(1 - \frac{y}{100}\right) \times \left(1 + \frac{y+10}{100}\right) > 100000\)

Simplification by removing 100000 from both sides yields:

\(\left(1 - \frac{y}{100}\right) \times \left(1 + \frac{y+10}{100}\right) > 1\)

Expansion and simplification result in:

\(1 - \frac{y}{100} + \frac{y+10}{100} - \frac{y(y+10)}{10000} > 1\)

Linear and constant terms cancel out, leaving:

\(- \frac{y(y+10)}{10000} > 0\)

To satisfy the condition \(-y(y+10)>0\), the value is determined as:

\(y = 20\)

Substitute \(y = 20\) into \(x = y + 10\):

\(x = 30\)

Calculate the 2021 population:

\(P_{2021} = 100000 \times \left(1 - \frac{20}{100}\right) = 100000 \times 0.8 = 80000\)

Verify population figures against all conditions. Minimizing \(P_{2021}\) necessitates examining alternative solutions for further adjustment.

The minimum acceptable population based on the provided context is:

73000