Question

Let \( A = \{x : x \text{ is a positive multiple of } 2 \text{ less than } 36\},\)
\(B = \{x : x \text{ is a positive multiple of } 3 \text{ greater than } 16\}, \text{ and }\)
\(C = \{x : x \text{ is a positive multiple of } 4 \text{ less than } 42\}. \text{ Then } (A \cap B) \cap C =\)

• A. \(\{12, 24, 36\}\)
• B. \(\{12, 24\}\)
• C. \(\{24\}\)
• D. \(\{24, 36\}\)
• E. \(\phi\)

Hint:

For intersection of sets, always list the elements of each set carefully and then pick only the common elements. Here, the final answer must belong to all three sets simultaneously.

Answer 1

The Correct Option is C

Step 1: Write the set \(A\).
Since \(A\) contains all positive multiples of \(2\) less than \(36\), we get:
\[ A = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34\} \]

Step 2: Write the set \(B\).
Since \(B\) contains all positive multiples of \(3\) greater than \(16\), the relevant elements are:
\[ B = \{18, 21, 24, 27, 30, 33, 36, 39, \dots\} \] Only the common values with \(A\) will matter later.

Step 3: Find \(A \cap B\).
The common elements of \(A\) and \(B\) are numbers that are multiples of both \(2\) and \(3\), that is, multiples of \(6\), subject to the restrictions of the sets. Hence:
\[ A \cap B = \{18, 24, 30\} \]

Step 4: Write the set \(C\).
Since \(C\) contains all positive multiples of \(4\) less than \(42\), we get:
\[ C = \{4, 8, 12, 16, 20, 24, 28, 32, 36, 40\} \]

Step 5: Find \((A \cap B) \cap C\).
Now compare:
\[ A \cap B = \{18, 24, 30\} \] with
\[ C = \{4, 8, 12, 16, 20, 24, 28, 32, 36, 40\} \] The only common element is \(24\).

Step 6: State the intersection clearly.
Therefore,
\[ (A \cap B) \cap C = \{24\} \]

Step 7: Match with the options.
The set \(\{24\}\) appears in option (3). Hence, the correct option is \((3)\).