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CAT 2023 Slot 3 — QA Question 11

Successive Percentage ChangeEasy

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

Answer & solution

  • 73000

  • B

    72000

  • C

    75000

  • D

    74000

Solution

Easy

A drop of y%y\% followed by a rise of x%x\% is a successive percentage change. Use the formula a+b+ab100a+b+\dfrac{ab}{100} to get the net change, impose "2022 >> 2020" to force x>yx>y, then with x−y=10x-y=10 find the largest drop yy that still leaves a net gain. The biggest drop gives the lowest 2021 population.

1

Net two-year change (decrease y%y\%, then increase x%x\%) via a+b+ab100a+b+\dfrac{ab}{100} with a=−y, b=+xa=-y,\ b=+x:

net%=−y+x−xy100\text{net}\%=-y+x-\frac{xy}{100}
2

2022 exceeds 2020 ⇒\Rightarrow net >0⇒x>y>0\Rightarrow x>y. With x−y=10x-y=10, set x=y+10x=y+10:

net%=10−y(y+10)100\text{net}\%=10-\frac{y(y+10)}{100}
3

Require a genuine net gain and solve for the largest natural yy:

10−y(y+10)100>0⇒ y(y+10)<1000\begin{aligned} &10-\frac{y(y+10)}{100}\gt 0\\ &\Rightarrow\ y(y+10)\lt 1000 \end{aligned}

y=27: 27×37=999<1000 ✓y=27:\ 27\times 37=999\lt 1000\ \checkmark, while y=28: 28×38=1064>1000y=28:\ 28\times 38=1064\gt 1000. So ymax⁡=27y_{\max}=27.

4

Lowest 2021 population = biggest drop applied to 2020's value:

100000×(1−27100)=100000×0.73=73000100000\times\left(1-\frac{27}{100}\right)=100000\times 0.73=73000
Lowest 2021 population=73000\text{Lowest 2021 population}=\mathbf{73000}
CAT 2023 Slot 3 QA Q11: The population of a town in 2020 was 100000. The population decreased by y% from the year 2020 to 2021, and in — Solution | TheCATExam