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:
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:
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:
Options
Answer: (4) 73000
Explanation:
Step 1: Set up inequality (1 − y/100)(1 + x/100) > 1 Since (1 − y/100)(1 + y/100) < 1, we must have x > y → x = y + 10 Step 2: Substitute x = y + 10 (1 − y/100)(1 + (y + 10)/100) > 1 Multiply by 100: (100 − y)(110 + y) > 10,000 Expand: 11,000 − 10y − y^2 > 10,000 → y^2 + 10y − 1,000 < 0 Step 3: Solve quadratic inequality Complete the square: (y + 5)^2 − 1,025 < 0 → (y + 5)^2 < 1,025 → y + 5 < √1,025 ≈ 32.015 Maximum integer y = 27 Step 4: Minimum population in 2021 P₁ = 100,000 × (1 − 27/100) = 100,000 × 0.73 = 73,000
Explanation:
Step 1: Set up inequality (1 − y/100)(1 + x/100) > 1 Since (1 − y/100)(1 + y/100) < 1, we must have x > y → x = y + 10 Step 2: Substitute x = y + 10 (1 − y/100)(1 + (y + 10)/100) > 1 Multiply by 100: (100 − y)(110 + y) > 10,000 Expand: 11,000 − 10y − y^2 > 10,000 → y^2 + 10y − 1,000 < 0 Step 3: Solve quadratic inequality Complete the square: (y + 5)^2 − 1,025 < 0 → (y + 5)^2 < 1,025 → y + 5 < √1,025 ≈ 32.015 Maximum integer y = 27 Step 4: Minimum population in 2021 P₁ = 100,000 × (1 − 27/100) = 100,000 × 0.73 = 73,000
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