The LCM of two numbers is 40 times its HCF. If the product of the numbers is 1440, find their HCF.

Options

A. 6
B. 12
C. 15
D. 8

Answer

To solve the problem, we need to find the HCF of two numbers given that the LCM of the two numbers is 40 times their HCF, and the product of the two numbers is 1440.

### Step-by-Step Solution:

1. **Let the HCF be denoted as H**.
We know from the problem that the LCM (Least Common Multiple) of the two numbers is 40 times the HCF. Therefore, we can express this as:
\[ \text{LCM} = 40H \]

2. **Let the two numbers be denoted as x and y**.
According to the relationship between LCM, HCF, and the product of two numbers, we have:
\[ \text{LCM} \times \text{HCF} = x \times y \]

3. **Substituting the LCM in the equation**:
Substitute the expression for LCM from step 1 into the equation from step 2:
\[ (40H) \times H = x \times y \]
This simplifies to:
\[ 40H^2 = x \times y \]

4. **Using the product of the two numbers**:
We know from the problem statement that the product of the two numbers is given as:
\[ x \times y = 1440 \]

5. **Setting the equations equal**:
Now we can set the two expressions for the product equal to each other:
\[ 40H^2 = 1440 \]

6. **Solving for H**:
To find H, we can divide both sides of the equation by 40:
\[ H^2 = \frac{1440}{40} \]
Simplifying the right side:
\[ H^2 = 36 \]

7. **Taking the square root**:
Now, take the square root of both sides to find H:
\[ H = \sqrt{36} = 6 \]

### Conclusion:
The HCF of the two numbers is **6**.