Q71 / 180 · COMEDK 2026 (9 May Shift 2) · Mathematics
Let A and B be two subsets of $\xi = \{1, 2, 3, -------, 44, 45\}$ such that
$A = \{x: x \text{ is divisible by } 3 \text{ and } 4\}$
$B = \{x: x \text{ is a perfect square number}\}$
Then $n(B - A)$ equals
$A = \{x: x \text{ is divisible by } 3 \text{ and } 4\}$
$B = \{x: x \text{ is a perfect square number}\}$
Then $n(B - A)$ equals
- A$9$
- B$2$
- Correct answerC$5$
- D$1$
Solution
The universal set is $\xi = \{1, 2, 3, \dots, 45\}$.
The set $A$ contains numbers divisible by both $3$ and $4$, which means they are divisible by $\text{LCM}(3, 4) = 12$.
$A = \{12, 24, 36\}$
The set $B$ contains perfect square numbers up to $45$.
$B = \{1^2, 2^2, 3^2, 4^2, 5^2, 6^2\} = \{1, 4, 9, 16, 25, 36\}$
The set $B - A$ contains elements that are in $B$ but not in $A$.
$B - A = \{1, 4, 9, 16, 25\}$
Therefore, $n(B - A) = 5$.
Answer: $5$
The set $A$ contains numbers divisible by both $3$ and $4$, which means they are divisible by $\text{LCM}(3, 4) = 12$.
$A = \{12, 24, 36\}$
The set $B$ contains perfect square numbers up to $45$.
$B = \{1^2, 2^2, 3^2, 4^2, 5^2, 6^2\} = \{1, 4, 9, 16, 25, 36\}$
The set $B - A$ contains elements that are in $B$ but not in $A$.
$B - A = \{1, 4, 9, 16, 25\}$
Therefore, $n(B - A) = 5$.
Answer: $5$
Hence, the correct answer is option: 'C'.