Question

Let A and B denote the sets of students in a school who play cricket and football respectively. If \(n(A) = 45, n(B) = 35\) and \(n(A \cap B) = 13\), then \(n((A \cap B)' \cap (A \cup B))\) =

• A. 43
• B. 48
• C. 50
• D. 52
• E. 54

Hint:

The expression \((A \cap B)' \cap (A \cup B)\) is just a formal way of asking for the number of students who play exactly one sport. You can calculate this as [Students playing only Cricket] + [Students playing only Football] = \((45-13) + (35-13) = 32 + 22 = 54\).

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The problem asks for the number of elements in the set \((A \cap B)' \cap (A \cup B)\).
This set represents the elements that are in the union of A and B but are NOT in their intersection.
In set theory notation, this is equivalent to the symmetric difference, represented as \((A \cup B) \setminus (A \cap B)\).

Step 2: Key Formula or Approach:
1. The Principle of Inclusion-Exclusion for two sets: \(n(A \cup B) = n(A) + n(B) - n(A \cap B)\).
2. The definition of the relative complement: \(n(X \cap Y') = n(X) - n(X \cap Y)\).
Applying this to our expression: \(n((A \cup B) \cap (A \cap B)') = n(A \cup B) - n((A \cup B) \cap (A \cap B))\).
Since \((A \cap B) \subseteq (A \cup B)\), this simplifies to \(n(A \cup B) - n(A \cap B)\).

Step 3: Detailed Explanation:
First, we find the total number of students playing at least one of the two games:
\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \]
Substituting the given values:
\[ n(A \cup B) = 45 + 35 - 13 \]
\[ n(A \cup B) = 80 - 13 = 67 \]
Now, we find the number of elements in the required set \((A \cap B)' \cap (A \cup B)\):
\[ n((A \cap B)' \cap (A \cup B)) = n(A \cup B) - n(A \cap B) \]
\[ n((A \cap B)' \cap (A \cup B)) = 67 - 13 \]
\[ n((A \cap B)' \cap (A \cup B)) = 54 \]

Step 4: Final Answer:
The value of \(n((A \cap B)' \cap (A \cup B))\) is 54.