CAT 2023 Slot 3 · Quant · Question 10

The 2021 population is 100000\left( 1-\frac{y}{100}\right), so making it as low as possible means making y as large as possible.

The 2022 population is 100000\left( 1-\frac{y}{100}\right)\left( 1+\frac{x}{100}\right), and it must exceed 100000.

A fall followed by a rise only recovers if the rise is the bigger percentage, so x>y, and with a difference of 10 that means x=y+10.

The condition becomes \left( 1-\frac{y}{100}\right)\left( 1+\frac{y+10}{100}\right)>1.

Clearing the denominators by multiplying through by 10000, \left( 100-y \right)\left( 110+y \right)>10000.

Expanding the left side, 11000+100y-110y-{y}^{2}=11000-10y-{y}^{2}.

So the condition is 11000-10y-{y}^{2}>10000, that is {y}^{2}+10y-1000<0.

The positive root of {y}^{2}+10y-1000=0 is \frac{-10+\sqrt{100+4000}}{2}=\frac{-10+\sqrt{4100}}{2}\approx 27.02.

So y<27.02, and since y is a natural number the largest allowed value is y=27.

The 2021 population is then 100000\times 0.73=73000 — the fourth option.

Checking: a 27% fall gives 73000, and a 37% rise on that gives 73000\times 1.37=100010, just above the 2020 figure.

Shortcut. Rather than solving the quadratic, test the options directly. At y=27 the 2022 figure is 73000\times 1.37=100010>100000, which works; at y=28 it is 72000\times 1.38=99360, which fails. So 73000 is the floor.