UPSC NDA 2 2022 Mathematics Solved Paper

Concept:Use set theory and a Venn diagram to find students who received medals in at least two sports.Explanation:Let $A$, $B$, $C$ represent students with medals in basketball, football, and volleyball.Given: $n(A)=5x$, $n(B)=4x+15$, $n(C)=x+25$, $n(A \cup B \cup C)=15x$, $n(A \cap B \cap C)=x$.Using the inclusion-exclusion principle:$n(A \cup B \cup C) = n(A)+n(B)+n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)$.Substitute values: $15x = 5x + 4x+15 + x+25 - [n(A \cap B)+n(B \cap C)+n(C \cap A)] + x$.Simplify: $15x = 11x + 40 - S + x$ where $S$ is the sum of three pairwise intersections.Thus $15x = 12x + 40 - S \implies S = 40 - 4x$. (Equation 1)From the Venn diagram, let $p$, $q$, $r$ be students in exactly two sports (each excluding the triple overlap), and $s = x$ be students in all three.Then $S = n(A \cap B)+n(B \cap C)+n(C \cap A) = (p+s)+(r+s)+(q+s) = p+q+r+3s$.Equating with Equation 1: $40-4x = p+q+r+3x$.Rearrange: $p+q+r = 40-7x$.Students in at least two sports = exactly two + all three = $(p+q+r)+s = (40-7x)+x = 40-6x$.Answer:$40-6x$ (Option D).