Let `x={1,2,3,….40},A={x:x"is divisible by 2 and 3"}` and `B={x:x=n^(2),x in N}` then `n(A)-n(B)` is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to find the number of elements in sets A and B, and then calculate \( n(A) - n(B) \).

### Step 1: Define the sets
We have:
- Set \( x = \{1, 2, 3, \ldots, 40\} \)
- Set \( A = \{ x : x \text{ is divisible by } 2 \text{ and } 3 \} \)
- Set \( B = \{ x : x = n^2, n \in \mathbb{N} \} \)

### Step 2: Find elements in set A
Since \( A \) consists of numbers that are divisible by both 2 and 3, we can conclude that these numbers must be divisible by \( 6 \) (since \( 6 \) is the least common multiple of \( 2 \) and \( 3 \)).

Now, we need to find how many multiples of \( 6 \) are there from \( 1 \) to \( 40 \):
- The multiples of \( 6 \) are: \( 6, 12, 18, 24, 30, 36 \).

Counting these, we find:
- \( 6, 12, 18, 24, 30, 36 \) (total of 6 elements).

Thus, \( n(A) = 6 \).

### Step 3: Find elements in set B
Set \( B \) consists of perfect squares from \( 1 \) to \( 40 \). We will list the perfect squares:
- \( 1^2 = 1 \)
- \( 2^2 = 4 \)
- \( 3^2 = 9 \)
- \( 4^2 = 16 \)
- \( 5^2 = 25 \)
- \( 6^2 = 36 \)

The next perfect square would be \( 7^2 = 49 \), which exceeds \( 40 \).

Thus, the elements in set \( B \) are:
- \( 1, 4, 9, 16, 25, 36 \) (total of 6 elements).

So, \( n(B) = 6 \).

### Step 4: Calculate \( n(A) - n(B) \)
Now we can find the difference:
\[ n(A) - n(B) = 6 - 6 = 0 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{0} \]