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

\(100,000(1−\frac {4}{100})(1+\frac {}{100})>100,000\)

\((1−\frac {4}{100})(1+\frac {x}{100})>1\)

\((100−y)(100+x)>10,000\)
Since, \(x−y=10\)
\(⇒x=y+10\)
\((100−y)(100+y+10)>10000\)
\(−100y+100y−y2+1000−10y>0\)
\(y^2+10y−1000<0\)
\(y2+2⋅5⋅y+25−1025<0\)
\(y2+2⋅5⋅y+25<1025\)
\((y+5)^2<1025\)
\(y+5<\sqrt {1025}\)
\(y+5≤32.015\)
\(y+5≃32\)
For the minimum population in 2021, we need y to be as large as possible.
largest value of \(y=32−5=27\)
Now, the minimum population in 2021,
\(=100,000(1−\frac {27}{100})\)

\(=100,000 \times \frac {27}{100}\)
\(=73,000\)

So, the correct option is (C): \(73000\)