CAT 2025

Slot 1

QA

Question & SolutionQues & Sol

In a class, there were more than 10 boys and a certain number of girls. After 40% of the girls and 60% of the boys left the class, the remaining number of girls was 8 more than the remaining number of boys. Then, the minimum possible number of students initially in the class was

1. Concept Used

• Topic: Percentage, Linear Equations in Two Variables, Integer Constraints
• Formula: $$\text{Remaining Girls} - \text{Remaining Boys} = 8 \implies 0.6g - 0.4b = 8$$

2. Calculation

Let the number of girls be $g$ and the number of boys be $b$, where $b > 10$.

After 40% of the girls leave, the remaining girls $= 0.6g$.

After 60% of the boys leave, the remaining boys $= 0.4b$.

Using the given condition:

$0.6g - 0.4b = 8$

Multiplying both sides by 10:

$6g - 4b = 80$

Dividing by 2:

$3g - 2b = 40 \implies 3g = 2b + 40$

Now, since $0.6g$ must be a whole number (you can't have a fraction of a person), $g$ must be a multiple of 5. Similarly, since $0.4b$ must be a whole number, $b$ must also be a multiple of 5.

So we restrict our search to values where both $b$ and $g$ are multiples of 5, and $b > 10$.

From $3g = 2b + 40$, substituting $b = 5k$ (where $k$ is a positive integer and $5k > 10 \implies k \geq 3$):

$3g = 2(5k) + 40 = 10k + 40$

$g = \frac{10k + 40}{3}$

For $g$ to be an integer (and a multiple of 5), $10k + 40$ must be divisible by 3:

$10k + 40 \equiv k + 1 \pmod{3} \equiv 0 \pmod{3}$

$\implies k \equiv 2 \pmod{3}$

The smallest $k \geq 3$ satisfying $k \equiv 2 \pmod{3}$ is $k = 5$.

So $b = 5 \times 5 = 25$ and $g = \frac{10(5) + 40}{3} = \frac{90}{3} = 30$.

Verification: Remaining girls $= 0.6 \times 30 = 18$, Remaining boys $= 0.4 \times 25 = 10$, Difference $= 18 - 10 = 8$ ✅

Minimum total students $= b + g = 25 + 30 = 55$.

3. Solution

Answer = 55 ✅

The minimum possible number of students initially in the class is 55 (25 boys + 30 girls).