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

Answer: (C)

73000

Answer 2

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

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

$(100−y)(100+x)>10,000$
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): $73000$