Let X = `{x : 1 le x le 50, x in N}`
A = `{x: x `is multiple of `2}`
B = `{x: x` is multiple of `7}`
Then find number of elements in the smallest subset of `X` which contain elements of both `A` and `B`

Explanation

To solve the problem, we need to find the number of elements in the smallest subset of \( X \) that contains elements from both sets \( A \) and \( B \).

### Step-by-Step Solution:

1. **Define the Set \( X \)**:
\[ X = \{ x : 1 \leq x \leq 50, x \in \mathbb{N} \} \]
This means \( X \) contains all natural numbers from 1 to 50.

2. **Define the Set \( A \)**:
\[ A = \{ x : x \text{ is a multiple of } 2 \} \]
The multiples of 2 from 1 to 50 are:
\[ A = \{ 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50 \} \]
The number of elements in \( A \) is \( 25 \) (since \( 50/2 = 25 \)).

3. **Define the Set \( B \)**:
\[ B = \{ x : x \text{ is a multiple of } 7 \} \]
The multiples of 7 from 1 to 50 are:
\[ B = \{ 7, 14, 21, 28, 35, 42, 49 \} \]
The number of elements in \( B \) is \( 7 \) (since \( 49/7 = 7 \)).

4. **Find the Intersection \( A \cap B \)**:
We need to find the common elements in \( A \) and \( B \), which are the multiples of both 2 and 7 (i.e., multiples of \( 14 \)):
\[ A \cap B = \{ 14, 28, 42 \} \]
The number of elements in \( A \cap B \) is \( 3 \).

5. **Use the Principle of Inclusion-Exclusion**:
To find the number of elements in the union \( A \cup B \):
\[ N(A \cup B) = N(A) + N(B) - N(A \cap B) \]
Substituting the values:
\[ N(A \cup B) = 25 + 7 - 3 = 29 \]

Thus, the number of elements in the smallest subset of \( X \) that contains elements from both \( A \) and \( B \) is \( 29 \).

### Final Answer:
\[ \text{The number of elements in the smallest subset of } X \text{ which contains elements of both } A \text{ and } B \text{ is } 29. \]