Set Theory and Cardinality Matching Solution

This question requires matching statements about the cardinalities of sets from List-I with their corresponding numerical values from List-II. We will analyze each item in List-I, applying relevant set theory formulas and concepts.

Analyzing List-I Item (A) for Intersection

Item (A) presents a scenario with two sets, X and Y, where $n(X) = 17$, $n(Y) = 23$, and $n(X \cup Y) = 38$. The goal is to find $n(X \cap Y)$.

Concept: The Principle of Inclusion-Exclusion for two sets is applied here. It states that $n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$.

Calculation: To determine $n(X \cap Y)$, we rearrange the formula:

\begin{equation*} n(X \cap Y) = n(X) + n(Y) - n(X \cup Y) \end{equation*}

Substituting the given values:

\begin{equation*} n(X \cap Y) = 17 + 23 - 38 \end{equation*}

\begin{equation*} n(X \cap Y) = 40 - 38 \end{equation*}

\begin{equation*} n(X \cap Y) = 2 \end{equation*}

The calculation yields a value of 2. The corresponding match for List-I item (A) from List-II is (II).

Analyzing List-I Item (B) for Union

Item (B) provides cardinalities for sets X and Y: $n(X) = 28$, $n(Y) = 32$, and $n(X \cap Y) = 10$. We need to find $n(X \cup Y)$.

Concept: The Principle of Inclusion-Exclusion is used again:

\begin{equation*} n(X \cup Y) = n(X) + n(Y) - n(X \cap Y) \end{equation*}

Calculation: Substitute the provided values into the formula:

\begin{equation*} n(X \cup Y) = 28 + 32 - 10 \end{equation*}

\begin{equation*} n(X \cup Y) = 60 - 10 \end{equation*}

\begin{equation*} n(X \cup Y) = 50 \end{equation*}

The calculation results in 50. The corresponding match for List-I item (B) from List-II is (I).

Analyzing List-I Item (C) for Scaled Cardinality

Item (C) gives the cardinality of set X as $n(X) = 10$, and asks for $n(7X)$.

Concept: The notation $n(kX)$ typically signifies scaling the cardinality of set X by a factor $k$, meaning $n(kX) = k \times n(X)$. This operation scales the count of elements.

Calculation: Applying this concept:

\begin{equation*} n(7X) = 7 \times n(X) \end{equation*}

\begin{equation*} n(7X) = 7 \times 10 \end{equation*}

\begin{equation*} n(7X) = 70 \end{equation*}

The calculation yields 70. The corresponding match for List-I item (C) from List-II is (IV).

Analyzing List-I Item (D) for Fractional Cardinality

Item (D) provides the cardinality of set Y as $n(Y) = 20$, and asks for $n(\frac{Y}{2})$.

Concept: Similarly, $n(\frac{Y}{k})$ indicates scaling the cardinality of set Y by $\frac{1}{k}$, i.e., $n(\frac{Y}{k}) = \frac{1}{k} \times n(Y)$. This represents a proportional reduction in the element count.

Calculation: Applying this concept:

\begin{equation*} n\left(\frac{Y}{2}\right) = \frac{1}{2} \times n(Y) \end{equation*}

\begin{equation*} n\left(\frac{Y}{2}\right) = \frac{1}{2} \times 20 \end{equation*}

\begin{equation*} n\left(\frac{Y}{2}\right) = 10 \end{equation*}

The calculation results in 10. The corresponding match for List-I item (D) from List-II is (III).

Summary of Matches

Based on the analysis, the matches between List-I and List-II are established as follows:

The correct option is 1: (A) - (II), (B) - (I), (C) - (IV), (D) - (III).