The HCF and product of two numbers are 15 and 6300 respectively. The number of possible pairs of the numbers is:

Options

A. 4
B. 3
C. 2
D. 1

Answer

To solve the problem, we need to find the number of possible pairs of two numbers whose Highest Common Factor (HCF) is 15 and whose product is 6300.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
We know that for any two numbers \( a \) and \( b \), the relationship between HCF and LCM is given by:
\[ \text{HCF} \times \text{LCM} = a \times b \]
Here, we have the HCF as 15 and the product \( a \times b = 6300 \).

2. **Expressing the Numbers**:
Since the HCF of the two numbers is 15, we can express the two numbers as:
\[ a = 15x \quad \text{and} \quad b = 15y \]
where \( x \) and \( y \) are coprime (i.e., HCF of \( x \) and \( y \) is 1).

3. **Setting Up the Equation**:
Substituting \( a \) and \( b \) into the product equation:
\[ (15x) \times (15y) = 6300 \]
Simplifying this gives:
\[ 225xy = 6300 \]

4. **Solving for \( xy \)**:
Dividing both sides by 225:
\[ xy = \frac{6300}{225} \]
Calculating the right side:
\[ xy = 28 \]

5. **Finding Pairs of \( x \) and \( y \)**:
Now, we need to find pairs of integers \( (x, y) \) such that their product is 28. The pairs of factors of 28 are:
- \( (1, 28) \)
- \( (2, 14) \)
- \( (4, 7) \)

6. **Counting the Coprime Pairs**:
We need to ensure that \( x \) and \( y \) are coprime:
- \( (1, 28) \) are coprime.
- \( (2, 14) \) are not coprime (common factor is 2).
- \( (4, 7) \) are coprime.

Therefore, the valid pairs of \( (x, y) \) are \( (1, 28) \) and \( (4, 7) \).

7. **Counting the Possible Pairs**:
Each pair \( (x, y) \) can be arranged in two ways: \( (x, y) \) and \( (y, x) \). Thus, the pairs are:
- From \( (1, 28) \): \( (15 \times 1, 15 \times 28) \) and \( (15 \times 28, 15 \times 1) \)
- From \( (4, 7) \): \( (15 \times 4, 15 \times 7) \) and \( (15 \times 7, 15 \times 4) \)

This gives us a total of 4 pairs.

### Final Answer:
The number of possible pairs of the numbers is **4**.