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: Understanding the Concept:
This problem requires finding the intersection of three sets. The intersection of sets, denoted by \(\cap\), consists of all elements that are common to all the sets being considered. We need to find the elements that satisfy the conditions for set A, set B, and set C simultaneously.
Step 2: Detailed Explanation:
First, let's list the elements of each set according to their definitions.
Set A: A contains positive multiples of 2 that are less than 36.
\[ A = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34\} \] Set B: B contains positive multiples of 3 that are greater than 16.
\[ B = \{18, 21, 24, 27, 30, 33, 36, 39, ...\} \] Set C: C contains positive multiples of 4 that are less than 42.
\[ C = \{4, 8, 12, 16, 20, 24, 28, 32, 36, 40\} \] Next, we find the intersection of A and B, denoted as \(A \cap B\). An element in \(A \cap B\) must be a multiple of both 2 and 3, which means it must be a multiple of their least common multiple, lcm(2, 3) = 6. Also, it must be less than 36 (from A's condition) and greater than 16 (from B's condition).
The multiples of 6 are 6, 12, 18, 24, 30, 36, ...
Applying the conditions (16 \(< x <\) 36), the common elements are:
\[ A \cap B = \{18, 24, 30\} \] Finally, we find the intersection of the result \((A \cap B)\) with set C. We look for elements that are in both \((A \cap B)\) and C.
\[ (A \cap B) \cap C = \{18, 24, 30\} \cap \{4, 8, 12, 16, 20, 24, 28, 32, 36, 40\} \] Comparing the elements of these two sets, the only common element is 24.
\[ (A \cap B) \cap C = \{24\} \] Step 3: Final Answer:
The result of the operation \((A \cap B) \cap C\) is \{24\}. Therefore, option (C) is the correct answer.