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

Let's solve this problem step-by-step. We know:

• The population in 2020 was 100000.
• The population decreased by y% in 2021.
• The population increased by x% in 2022.
• Population in 2022 was greater than in 2020, i.e., greater than 100000.
• The difference between x and y is 10. Therefore, x = y + 10.

First, express the population changes in terms of equations:

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

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

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

Cancel out 100000 from both sides:

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

Expand and simplify:

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

The linear and constant terms cancel out, so keep:

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

Since \(-y(y+10)\) must be positive, solve for this condition. You'll find:

\(y = 20\)

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

\(x = 30\)

Now calculate the population in 2021:

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

Verify populations match all conditions. Minimizing \(P_{2021}\) requires checking other solutions, leading the lowest adequate population to adjust further:

Hence, the lowest possible population within given options is:

73000