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: Apply the given information.
- For (A): Using the principle of inclusion-exclusion for sets, we have: \[ n(X \cap Y) = n(X) + n(Y) - n(X \cup Y) = 17 + 23 - 38 = 2 \] Hence, \( n(X \cap Y) = 2 \), which corresponds to List-II option IV (2). - For (B): Since \( n(X) = 28 \) and \( n(Y) = 32 \), the union is: \[ n(X \cup Y) = n(X) + n(Y) - n(X \cap Y) = 28 + 32 - 10 = 50 \] Hence, \( n(X \cup Y) = 50 \), which corresponds to List-II option III (50). - For (C): Since \( n(X) = 10 \), we directly have \( n(X) = 10 \), which corresponds to List-II option I (10). - For (D): From part (A), we already calculated that \( n(X \cap Y) = 2 \), which corresponds to List-II option II (2).

Step 2: Conclusion.
Thus, the correct matching is: (A) - (IV), (B) - (III), (C) - (I), (D) - (II).