Three numbers are in the ratio of `3: 5: 10` and their LCM is 630. Find their HCF.
A 21 B 42 C 63 D 36

Options

A. A
B. D
C. C
D. B

Answer

To solve the problem, we need to find the HCF of three numbers that are in the ratio of 3:5:10 and have an LCM of 630.

### Step-by-Step Solution:

1. **Understanding the Ratio**:
Let's denote the three numbers as \(3x\), \(5x\), and \(10x\) where \(x\) is a common multiplier.

2. **Finding the LCM**:
The LCM of the numbers \(3x\), \(5x\), and \(10x\) can be calculated using the formula for LCM based on the prime factors:
\[ \text{LCM}(3x, 5x, 10x) = \text{LCM}(3, 5, 10) \cdot x \]
The LCM of the coefficients \(3\), \(5\), and \(10\) is \(30\). Therefore:
\[ \text{LCM}(3x, 5x, 10x) = 30x \]

3. **Setting up the Equation**:
We know from the problem statement that the LCM is \(630\):
\[ 30x = 630 \]

4. **Solving for \(x\)**:
To find \(x\), we divide both sides of the equation by \(30\):
\[ x = \frac{630}{30} = 21 \]

5. **Finding the Actual Numbers**:
Now that we have \(x\), we can find the three numbers:
- First number: \(3x = 3 \times 21 = 63\)
- Second number: \(5x = 5 \times 21 = 105\)
- Third number: \(10x = 10 \times 21 = 210\)

6. **Finding the HCF**:
To find the HCF of \(63\), \(105\), and \(210\), we can use the prime factorization:
- \(63 = 3^2 \times 7\)
- \(105 = 3 \times 5 \times 7\)
- \(210 = 2 \times 3 \times 5 \times 7\)

The common prime factors are \(3\) and \(7\). Therefore, the HCF is:
\[ HCF = 3^1 \times 7^1 = 21 \]

### Final Answer:
The HCF of the three numbers is \(21\).