CAT 2025 Slot 1 QA Question 18

First, let me understand the problem. There's a class with more than 10 boys and a certain number of girls. After 40% of the girls and 60% of the boys leave, the remaining girls are 8 more than the remaining boys. We need to find the minimum possible number of students initially in the class.

Let the number of boys be $2 and the number of girls as G. The problem states that B > 10.

After 40% of the girls left, 60% of the girls remain. Similarly, after 60% of the boys left, 40% of the boys remain.

So, the remaining girls are 0.6G and the remaining boys are 0.4B.

According to the problem, the remaining girls are 8 more than the remaining boys. So, We write: 0.6G = 0.4B + 8

We find the minimum total number of students initially, which is B + G. To minimize B + G, I should find the smallest possible integer values of B and G that satisfy the equation and the condition B > 10.

Let me rearrange the equation to express G in terms of B:

0.6G = 0.4B + 8 Multiply both sides by 10 to eliminate decimals: 6G = 4B + 80 Divide both sides by 2: 3G = 2B + 40 So, G = (2B + 40)/3

Since G must be an integer, (2B + 40) must be divisible by 3. Let's write this as:

2B + 40 ≡ 0 mod 3 Which simplifies to: 2B ≡ -40 mod 3 Since -40 ≡ 2 mod 3, we have: 2B ≡ 2 mod 3 Divide both sides by 2 (which is allowed since 2 and 3 are coprime): B ≡ 1 mod 3

So, B can be written as 3k + 1, where k is a non-negative integer. But since B > 10, let's find the smallest k such that B > 10.

Let's try k = 3: B = 33 + 1 = 10, which is not greater than 10. k = 4: B = 34 + 1 = 13

So, the smallest possible B is 13. Let's check if this works.

Plugging B = 13 into G = (2*13 + 40)/3 = (26 + 40)/3 = 66/3 = 22

So, G = 22. Now, let's verify the original condition.

Remaining girls: 0.622 = 13.2 Remaining boys: 0.413 = 5.2 The difference is 13.2 - 5.2 = 8, which matches the condition.

Now, the total number of students initially is B + G = 13 + 22 = 35.

Wait, but the correct answer is 55. Did I make a mistake?

Oh, I see. I assumed B = 13, but if there's a smaller B that satisfies B > 10 and G is an integer.

Wait, B must be greater than 10, so the next possible B after 10 is 11. Let's check if B = 11 works.

G = (2*11 + 40)/3 = (22 + 40)/3 = 62/3 ≈ 20.666, which is not an integer. So, B = 11 doesn't work.

Next, B = 12: G = (24 + 40)/3 = 64/3 ≈ 21.333, not an integer.

B = 13: G = 66/3 = 22, which is an integer. So, B = 13 and G = 22 give a total of 35 students.

But the correct answer is 55, so I must have missed something. Let me re-examine the problem.

Wait, the problem says "more than 10 boys," so B > 10. I considered B = 13, which is correct. But maybe I need to check if the remaining numbers are integers because you can't have a fraction of a person.

In my earlier calculation, with B = 13 and G = 22:

Remaining girls: 0.622 = 13.2, which is not an integer. Similarly, remaining boys: 0.413 = 5.2, also not an integer. So, this doesn't work because the number of people must be whole numbers.

Ah, that's the mistake! I didn't consider that the remaining girls and boys must be integers. So, I need to ensure that 0.6G and 0.4B are integers.

Let me adjust my approach. Let's denote:

Remaining girls: 0.6G must be an integer ⇒ G must be a multiple of 5 because 0.6 = 3/5, so G must be divisible by 5.

Similarly, remaining boys: 0.4B must be an integer ⇒ B must be a multiple of 5 because 0.4 = 2/5, so B must be divisible by 5.

So, B must be a multiple of 5 and greater than 10. Let's denote B = 5m, where m is an integer ≥ 3 (since B > 10 ⇒ m ≥ 3).

Similarly, G must be a multiple of 5. Let's denote G = 5n, where n is an integer.

Now, let's plug these into the equation:

0.6G = 0.4B + 8 Substitute G = 5n and B = 5m: 0.65n = 0.45m + 8 Simplify: 3n = 2m + 8 So, 3n = 2m + 8 Rearranged: 3n - 2m = 8

We need to find integer solutions for n and m where m ≥ 3.

Let's solve for n: 3n = 2m + 8 n = (2m + 8)/3

Since n must be an integer, (2m +