The L.C.M. of two different numbers are 30. Which of the following cannot be their H.C.F.?

Options

A. 12
B. 15
C. 6
D. 10

Answer

To solve the problem, we need to determine which of the given options cannot be the H.C.F. (Highest Common Factor) of two different numbers whose L.C.M. (Lowest Common Multiple) is 30.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The relationship between L.C.M. and H.C.F. of two numbers \( a \) and \( b \) is given by the formula:
\[ \text{L.C.M.}(a, b) \times \text{H.C.F.}(a, b) = a \times b \]
This means that the H.C.F. must be a divisor of the L.C.M.

2. **Identifying Divisors of 30**:
Since the L.C.M. is 30, we need to find all the divisors of 30. The divisors of 30 are:
\[ 1, 2, 3, 5, 6, 10, 15, 30 \]

3. **Evaluating the Options**:
We need to check which of the given options can be the H.C.F. by seeing if they divide 30 completely.
- **Option 1: 10**
- \( 30 \div 10 = 3 \) (Divides completely)
- **Option 2: 6**
- \( 30 \div 6 = 5 \) (Divides completely)
- **Option 3: 15**
- \( 30 \div 15 = 2 \) (Divides completely)
- **Option 4: 12**
- \( 30 \div 12 = 2.5 \) (Does not divide completely)

4. **Conclusion**:
The only option that does not divide 30 completely is 12. Therefore, 12 cannot be the H.C.F. of two numbers whose L.C.M. is 30.

### Final Answer:
The H.C.F. that cannot be of the two numbers is **12**.