Match List-I with List-II. $$ \begin{array}{|c|l|} \hline \textbf{List-I} & \textbf{List-II} \\ \hline (A)\ \text{If } X \text{ and } Y \text{ are two sets such that} \\ \ \ n(X) = 17,\ n(Y) = 23,\ n(X \cup Y) = 38, \\ \ \ \text{then } n(X \cap Y) \text{ is} & (I)\ 20 \\ \hline (B)\ \text{If } n(X) = 28,\ n(Y) = 32,\ n(X \cap Y) = 10, \\ \ \ \text{then } n(X \cup Y) \text{ is} & (II)\ 10 \\ \hline (C)\ \text{If } n(\overline{X}) = 10, \text{ and } n(S) = 30\\ \ \ \text{then } n(X) \text{ is} & (III)\ 50 \\ \hline (D)\ \text{If } n(Y) = 20, \\ \ \ \text{then } n\!\left(\tfrac{Y}{2}\right) \text{ is} & (IV)\ 2 \\ \hline \end{array} $$
(A) Given: \( n(X) = 17, \; n(Y) = 23, \; n(X \cup Y) = 38 \). Using the formula: \[ n(X \cup Y) = n(X) + n(Y) - n(X \cap Y) \] \[ 38 = 17 + 23 - n(X \cap Y) \] \[ n(X \cap Y) = 2 \] Hence, matches with option (IV).
(B) Given: \( n(X) = 28, \; n(Y) = 32, \; n(X \cap Y) = 10 \). \[ n(X \cup Y) = 28 + 32 - 10 = 50 \] Hence, matches with option (III).
(C) Given: \( n(\overline{X}) = 10 \). Assuming universal set \( U \) has \( n(U) = 30 \), \[ n(X) = n(U) - n(\overline{X}) = 30 - 10 = 20 \] Hence, matches with option (I).
(D) Given: \( n(Y) = 20 \). If we divide the set \( Y \) into 2 equal parts, then: \[ n\!\left(\tfrac{Y}{2}\right) = \frac{20}{2} = 10 \] Hence, matches with option (II).
(A) - (IV), (B) - (III), (C) - (I), (D) - (II)
Correct Answer: Option A
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