CAT 2023 Slot 3 QA Question 10

First, the population in 2020 is 100,000. Then, it decreases by y% from 2020 to 2021. After that, it increases by x% from 2021 to 2022. We're told that the population in 2022 is greater than in 2020, and the difference between x and y is 10. We need to find the lowest possible population in 2021, which is after the decrease. The options are A) 72000, B) 75000, C) 74000, D) 73000.

Alright, let's start by understanding the problem. The population starts at 100,000 in 2020. It decreases by y% in 2021, so the population becomes 100,000(1 - y/100). Then, from 2021 to 2022, it increases by x%, so the population becomes 100,000(1 - y/100)*(1 + x/100). We know that this final population is greater than 100,000, so:

100,000(1 - y/100)(1 + x/100) > 100,000

Dividing both sides by 100,000 simplifies to:

(1 - y/100)*(1 + x/100) > 1

Also, we're told that x and y are natural numbers, and x - y = 10. So, x = y + 10.

Substituting x in the inequality:

(1 - y/100)*(1 + (y + 10)/100) > 1

Simplifying this expression. Let's denote y as a variable for simplicity.

First, expand the left side:

(1 - y/100)(1 + y/100 + 10/100) = (1 - y/100)(1 + y/100 + 0.1)

Compute each part:

1 - y/100 is straightforward.

1 + y/100 + 0.1 = 1.1 + y/100

So, multiplying these together:

(1 - y/100)(1.1 + y/100) = 1(1.1) + 1(y/100) - y/100(1.1) - (y/100)*(y/100)

Simplify term by term:

1.1 + y/100 - 1.1y/100 - y²/10000

Combine like terms:

1.1 + (y/100 - 1.1y/100) - y²/10000

Which is:

1.1 - 0.1y/100 - y²/10000

Simplify 0.1y/100 to y/1000:

1.1 - y/1000 - y²/10000

So, the inequality becomes:

1.1 - y/1000 - y²/10000 > 1

Subtract 1 from both sides:

0.1 - y/1000 - y²/10000 > 0

Multiply both sides by 10000 to eliminate denominators:

1000 - 10y - y² > 0

Rearranging:

-y² -10y + 1000 > 0

Multiply both sides by -1 (remember to flip the inequality sign):

y² + 10y - 1000 < 0

So, we have a quadratic inequality: y² + 10y - 1000 < 0

Let me solve the equation y² + 10y - 1000 = 0 to find critical points.

Using the quadratic formula:

y = [-10 ± \] / 2

\ is approximately 64.031

So, y = [-10 + 64.031]/2 ≈ 54.031/2 ≈ 27.015

And y = [-10 - 64.031]/2 ≈ negative value, which we can ignore since y is a natural number.

So, the inequality y² + 10y - 1000 < 0 holds for y between the two roots, but since y is positive, we consider y < 27.015.

Since y is a natural number, the maximum possible value for y is 27.

But we also have x = y + 10, so x would be 37 when y is 27.

Wait, but let's check if y=27 satisfies the original condition.

Compute the population in 2022:

Population in 2021: 100,000(1 - 27/100) = 100,0000.73 = 73,000

Population in 2022: 73,000(1 + 37/100) = 73,0001.37

Calculate 73,000*1.37:

73,000 * 1 = 73,000

73,000 * 0.37 = 27,010

Total: 73,000 + 27,010 = 100,010

Which is indeed greater than 100,000. So, y=27 is acceptable.

But we need the lowest possible population in 2021, which is 73,000 when y=27. But let's check if lower y values give lower populations.

Wait, if y is smaller, say y=26, then x=36.

Population in 2021: 100,000*(1 - 26/100) = 74,000

Population in 2022: 74,000*(1 + 3