Question

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?

Hint:

When percentages of people leave, ensure that those percentages give integer counts. Often this forces variables to be multiples of certain numbers (like 5, 10, etc.), which helps in solving and minimizing or maximizing totals.

Answer 1

Correct Answer: 55

Step 1: Let \(x\) = number of boys remaining after 60% leave, so \(x = 0.4B\), meaning \(B = 2.5x\), which is an integer only when \(x\) is even.
Step 2: Remaining girls \(= x+8 = 0.6G\), so \(G = \dfrac{5(x+8)}{3}\), which is an integer only when \(x+8\) is a multiple of 3, i.e. \(x \equiv 1 \pmod 3\).
Step 3: Since \(B=2.5x>10\), we need \(x>4\); combined with \(x\) even and \(x\equiv1\pmod3\), the smallest valid \(x\) is \(10\) (checking \(x=6,8\) fail the mod-3 test, but \(x=10\) gives \(x+8=18\), divisible by 3). At \(x=10\): \(B=25\), \(G=\frac{5(18)}{3}=30\). Total \(=25+30=55\).
\[ \boxed{55} \]