The number of possible pairs of numbers , whose product is 5400 and HCF is 30:

Options

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

Answer

To solve the problem of finding the number of possible pairs of numbers whose product is 5400 and HCF is 30, we can follow these steps:

### Step 1: Understand the relationship between the numbers
Let the two numbers be \( A \) and \( B \). Given that their HCF (Highest Common Factor) is 30, we can express the numbers as:
\[ A = 30x \quad \text{and} \quad B = 30y \]
where \( x \) and \( y \) are co-prime integers (i.e., HCF of \( x \) and \( y \) is 1).

### Step 2: Set up the equation based on the product
According to the problem, the product of the two numbers \( A \) and \( B \) is:
\[ A \times B = 5400 \]
Substituting the expressions for \( A \) and \( B \):
\[ (30x) \times (30y) = 5400 \]
This simplifies to:
\[ 900xy = 5400 \]

### Step 3: Solve for \( xy \)
To find \( xy \), we divide both sides of the equation by 900:
\[ xy = \frac{5400}{900} = 6 \]

### Step 4: Find pairs of co-prime integers \( (x, y) \)
Now, we need to find pairs of integers \( (x, y) \) such that \( xy = 6 \) and \( \text{HCF}(x, y) = 1 \). The pairs of factors of 6 are:
1. \( (1, 6) \)
2. \( (2, 3) \)

Both pairs are co-prime, so we have:
- Pair 1: \( (1, 6) \)
- Pair 2: \( (2, 3) \)

### Step 5: Count the valid pairs
Each valid pair \( (x, y) \) corresponds to a unique pair of numbers \( (A, B) \):
1. From \( (1, 6) \): \( (30 \times 1, 30 \times 6) = (30, 180) \)
2. From \( (2, 3) \): \( (30 \times 2, 30 \times 3) = (60, 90) \)

Thus, we have two pairs of numbers:
1. \( (30, 180) \)
2. \( (60, 90) \)

### Final Answer
The number of possible pairs of numbers whose product is 5400 and HCF is 30 is **2**.

---