Given:
The town's population in 2020 was 100,000.
The population decreased by y% from 2020 to 2021 and increased by x% from 2021 to 2022, where x and y are natural numbers.
Population in 2021: \(100000 \left(\frac{100-y}{100}\right)\)
Population in 2022: \(100000 \left(\frac{100-y}{100}\right)\left(\frac{100+x}{100}\right)\)
Additional Information:
The population in 2022 was greater than in 2020, and the difference between x and y is 10.
Therefore:
\(100000\left(\frac{100-y}{100}\right)\left(\frac{100+x}{100}\right)>100000\), and \(x - y = 10\)
Simplifying the inequality with \(x = y + 10\):
\(100000\left(\frac{100-y}{100}\right)\left(\frac{100+(y+10)}{100}\right)>100000\)
\(\left(\frac{100-y}{100}\right)\left(\frac{110+y}{100}\right)>1\)
To maximize the population in 2021, we need to maximize y.
The inequality \( (100-y)(110+y)>10000 \) leads to:
\(11000 + 100y - 110y - y^2>10000\)
\(11000 - 10y - y^2>10000\)
\(1000>y^2 + 10y\)
\(y^2 + 10y<1000\)
Completing the square:
\((y + 5)^2 - 25<1000\)
\((y + 5)^2<1025\)
Since \(32^2 = 1024\), the largest integer value for \(y+5\) is 32.
\(y + 5 = 32\)
\(y = 27\)
Population in 2021:
\(100000 \times \left(\frac{100-27}{100}\right) = 100000 \times \frac{73}{100} = 73000\)
The population in 2021 was 73,000.
The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64.Then, the largest number in the original set of three numbers is