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
- 74000
- 75000
- 73000
- 72000
The Correct Option is C
Solution and Explanation
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
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