GATE CS 2024 Set 2 — Question 3

Let $C$ be the set of students who like their core branches, and $O$ be the set of students who like other branches.
Total students $N = 10,000$.
Students who like neither: $n(C' \cap O') = 1,500$.Therefore, the number of students who like at least one branch is:$$n(C \cup O) = N - n(C' \cap O') = 10,000 - 1,500 = 8,500$$Let $n(C) = x$. The problem states that the number of students who like their core branches is $1/4^{\text{th}}$ of those who like other branches. Thus, $n(C) = \frac{1}{4} n(O)$, which implies $n(O) = 4x$.We are given the number of students who like both branches:$$n(C \cap O) = 500$$Using the principle of inclusion-exclusion:$$n(C \cup O) = n(C) + n(O) - n(C \cap O)$$$$8,500 = x + 4x - 500$$$$8,500 + 500 = 5x$$$$9,000 = 5x$$$$x = \frac{9,000}{5} = 1,800$$So, the number of students who like their core branches is $n(C) = 1,800$.