The LCM of two natural numbers is 380 and their HCF is 19. How many sets of values are possible?

Options

A. 1
B. 10
C. 4
D. 2

Answer

To solve the problem, we need to determine how many sets of values are possible for two natural numbers given their LCM (Least Common Multiple) and HCF (Highest Common Factor).

### Step-by-Step Solution:

1. **Understanding the Relationship between LCM and HCF**:
The relationship between two natural numbers \(a\) and \(b\) can be expressed as:
\[ a \times b = \text{LCM}(a, b) \times \text{HCF}(a, b) \]
In this case, we know:
- LCM = 380
- HCF = 19

2. **Setting Up the Equation**:
Using the relationship, we can substitute the known values:
\[ a \times b = 380 \times 19 \]

3. **Calculating the Product**:
First, we calculate \(380 \times 19\):
\[ 380 \times 19 = 7220 \]
So, we have:
\[ a \times b = 7220 \]

4. **Expressing the Numbers in Terms of HCF**:
Since the HCF is 19, we can express the two numbers as:
\[ a = 19x \quad \text{and} \quad b = 19y \]
where \(x\) and \(y\) are co-prime numbers (they have no common factors other than 1).

5. **Substituting into the Product**:
Substitute \(a\) and \(b\) into the product equation:
\[ (19x) \times (19y) = 7220 \]
Simplifying gives:
\[ 361xy = 7220 \]

6. **Dividing by 361**:
Now, we divide both sides by 361 to isolate \(xy\):
\[ xy = \frac{7220}{361} = 20 \]

7. **Finding Co-prime Pairs**:
Now we need to find pairs of natural numbers \(x\) and \(y\) such that:
\[ xy = 20 \]
The pairs of factors of 20 are:
- (1, 20)
- (2, 10)
- (4, 5)

8. **Checking for Co-primality**:
We need to check which of these pairs are co-prime:
- (1, 20): Co-prime
- (2, 10): Not co-prime (common factor is 2)
- (4, 5): Co-prime

Thus, the valid pairs are (1, 20) and (4, 5).

9. **Counting the Valid Sets**:
Therefore, the total number of sets of values for \( (x, y) \) is 2.

### Final Answer:
The number of sets of values possible is **2**.