Question

Match List-I with List-II.\[\begin{array}{|c|c|} \hline \textbf{List-I (Set operations)} & \textbf{List-II (Missing term)} \\ \hline \text{(A) If \(n(X)=17,\, n(Y)=23,\, n(X \cup Y)=38\), then \(n(X \cap Y)\) is} & \text{(I) 20} \\ \hline \text{(B) If \(n(X)=28,\, n(Y)=32,\, n(X \cap Y)=10\), then \(n(X \cup Y)\) is} & \text{(II) 10} \\ \hline \text{(C) If \(n(X)=10\), then \(n(X')\) is} & \text{(III) 50} \\ \hline \text{(D) If \(n(Y)=20\), then \(n\left(\tfrac{Y}{2}\right)\) is} & \text{(IV) 2} \\ \hline \end{array}\]

• (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
• (A) - (II), (B) - (I), (C) - (III), (D) - (IV)
• (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
• (A) - (IV), (B) - (II), (C) - (III), (D) - (I)

Hint:

For set theory problems, remember to use the inclusion-exclusion principle to solve for intersections and unions of sets.

Answer 1

The Correct Option is C

Step 1: Information Application.
- For (A): Using the inclusion-exclusion principle: \[ n(X \cap Y) = n(X) + n(Y) - n(X \cup Y) = 17 + 23 - 38 = 2 \] Therefore, \( n(X \cap Y) = 2 \), matching List-II option IV.- For (B): Given \( n(X) = 28 \) and \( n(Y) = 32 \), the union is: \[ n(X \cup Y) = n(X) + n(Y) - n(X \cap Y) = 28 + 32 - 10 = 50 \] Thus, \( n(X \cup Y) = 50 \), matching List-II option III.- For (C): Given \( n(X) = 10 \), this directly corresponds to \( n(X) = 10 \), matching List-II option I.- For (D): From part (A), \( n(X \cap Y) = 2 \) was calculated, matching List-II option II.

Step 2: Final Matching.
The correct pairings are: (A) - (IV), (B) - (III), (C) - (I), (D) - (II).