NinetynineCAT 2023 Slot 3 QA Q10 Percentages & Profit-Loss
Question 10 of the Quantitative Ability (QA) section, CAT 2023 Slot 3, with CAT's official answer and a worked solution.
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
- A74000
- B75000
- C72000
- D73000
Answer: (D) 73000 Percentages & Profit-Loss MCQ
Solution
- For 2022 to exceed 2020 we need \(x>y\), so \(x=y+10\).
- Condition: \((100-y)(110+y)>10000 \Rightarrow 11000-10y-y^2>10000 \Rightarrow y^2+10y<1000\).
- \(y=27\): \(729+270=999<1000\) works; \(y=28\): \(1064\) fails. Max \(y=27\).
- Lowest 2021 population \(= 100000\times0.73 = 73000\).