In a party, the number of men, women and children guests are 72, 84 and 48 respectively. Find the minimum number of rooms required, if in each room, the same number of guests are to be seated all of them being of the same category.

Options

A. 20
B. 14
C. 17
D. 18

Answer

To solve the problem, we need to follow these steps:

1. **Identify the number of guests in each category**:
- Men: 72
- Women: 84
- Children: 48

2. **Calculate the total number of guests**:
\[ \text{Total guests} = \text{Men} + \text{Women} + \text{Children} = 72 + 84 + 48 \]
\[ \text{Total guests} = 204 \]

3. **Find the highest common factor (HCF) of the three numbers (72, 84, and 48)**:
- First, we can find the prime factorization of each number:
- \(72 = 2^3 \times 3^2\)
- \(84 = 2^2 \times 3^1 \times 7^1\)
- \(48 = 2^4 \times 3^1\)

- Now, we take the lowest power of each common prime factor:
- For \(2\): The lowest power is \(2^2\) (from 84).
- For \(3\): The lowest power is \(3^1\) (from 84 and 48).
- The prime \(7\) is not common to all three numbers.

- Therefore, the HCF is:
\[ \text{HCF} = 2^2 \times 3^1 = 4 \times 3 = 12 \]

4. **Calculate the minimum number of rooms required**:
\[ \text{Minimum number of rooms} = \frac{\text{Total guests}}{\text{HCF}} = \frac{204}{12} \]
\[ \text{Minimum number of rooms} = 17 \]

Thus, the minimum number of rooms required is **17**.