UPSC NDA 2 2022 Mathematics Solved Paper

Concept:Use the principle of inclusion-exclusion and Venn diagram to find the number of students who got medals in exactly one sport.Explanation:We are given total students who got medals: $n(A \cup B \cup C) = 15x$.Medals in basketball: $n(A) = 5x$, football: $n(B) = 4x + 15$, volleyball: $n(C) = x + 25$.All three: $n(A \cap B \cap C) = x$.Apply inclusion-exclusion:$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)$.Substitute:$15x = (5x) + (4x + 15) + (x + 25) - [n(A \cap B) + n(B \cap C) + n(C \cap A)] + x$.Simplify: $15x = 10x + 40 - [\text{sum of pairwise}] + x$ → $15x = 11x + 40 - \text{sum}$.Thus, sum of pairwise intersections = $40 - 4x$. ($i$)In the Venn diagram, let $p, q, r$ be the numbers in exactly two (excluding the triple region), and $s = x$ be the triple region.Then sum of pairwise intersections = $(p + s) + (r + s) + (q + s) = p + q + r + 3s$.So $p + q + r + 3s = 40 - 4x$.Replace $s = x$: $p + q + r + 3x = 40 - 4x$ → $p + q + r = 40 - 7x$.Number in at least two sports = exactly two + all three = $(p+q+r) + s = (40 - 7x) + x = 40 - 6x$.Students in exactly one sport = total - at least two = $15x - (40 - 6x) = 21x - 40$.Answer:$21x - 40$ which corresponds to option (A).