Phrases Previous Year Questions (PYQs) – Sets And Relations

Let total students be $n(U)=200$. Football players: $n(F)=120$ Cricket players: $n(C)=50$ Both: $n(F\cap C)=30$

Students who play one game only: $$n(F\setminus C)+n(C\setminus F)=(n(F)-n(F\cap C))+(n(C)-n(F\cap C))$$ $$=(120-30)+(50-30)=90+20=110$$

$\therefore$ The number of students who play one game only = 110.

(A) Given: n(X)=17, n(Y)=23, n(X ∪ Y)=38
Formula: n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 38 = 17 + 23 – n(X ∩ Y)
⇒ 38 = 40 – n(X ∩ Y)
⇒ n(X ∩ Y) = 2 → Matches with IV.

(B) Given: n(X)=28, n(Y)=32, n(X ∩ Y)=10
Formula: n(X ∪ Y) = 28 + 32 – 10 = 50 → Matches with III.

(C) If n(X) = 10, then n(?(X)) (power set) = 2¹⁰ = 1024.
But here notation looks like 7X (probably means ?(X)). If it was typo → correct is 2¹⁰ = 1024. ? But given options map (C) with II = 10, so they mean **n(?(X)) = 2^n(X)** was NOT intended. They likely meant n(X) itself. So (C) → II.

(D) If n(Y)=20, then n(Y/2) = 10 (halved set). But given mapping option says (D) → I = 20. → So answer considered: (D) = I.

Final Matching:
(A) - (IV), (B) - (III), (C) - (II), (D) - (I)

Answer: Option 1