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

Options

A. a) 4
B. b) 3
C. c) 2
D. d) 1

Answer

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

### Step-by-Step Solution:

1. **Let the two numbers be expressed in terms of their HCF:**
\[ \text{Let the two numbers be } a = 15x \text{ and } b = 15y \]
where \(x\) and \(y\) are co-prime integers (since the HCF of \(x\) and \(y\) is 1).

2. **Set up the equation using the product of the two numbers:**
\[ a \times b = 6300 \]
Substituting \(a\) and \(b\):
\[ (15x) \times (15y) = 6300 \]
Simplifying this gives:
\[ 225xy = 6300 \]

3. **Solve for \(xy\):**
\[ xy = \frac{6300}{225} \]
Simplifying the fraction:
\[ xy = 28 \]

4. **Find pairs of integers \((x, y)\) such that \(xy = 28\):**
The pairs of integers that multiply to 28 are:
- \( (1, 28) \)
- \( (2, 14) \)
- \( (4, 7) \)
- \( (7, 4) \)
- \( (14, 2) \)
- \( (28, 1) \)

5. **Check the pairs for co-primality:**
We need to ensure that \(x\) and \(y\) are co-prime:
- For \( (1, 28) \): Co-prime
- For \( (2, 14) \): Not co-prime (HCF is 2)
- For \( (4, 7) \): Co-prime
- For \( (7, 4) \): Co-prime
- For \( (14, 2) \): Not co-prime (HCF is 2)
- For \( (28, 1) \): Co-prime

Valid pairs are:
- \( (1, 28) \)
- \( (4, 7) \)

6. **Count the valid pairs:**
The valid pairs of \((x, y)\) that are co-prime are:
- \( (1, 28) \)
- \( (4, 7) \)

Therefore, there are **2 valid pairs** of numbers.

### Conclusion:
The number of possible pairs of the two numbers is **2**.