At out training institute we have `p - 1, p - 2, p - 3 and p - 4` processors in the ratio of `1/6, 1/5, 1/3 and 1/2` respectively. Minimum number of processors in out institute is :

Options

A. 16
B. 30
C. 32
D. 36

Answer

To solve the problem, we need to find the minimum number of processors in the institute given the ratios of processors \( p-1, p-2, p-3, \) and \( p-4 \).

### Step-by-Step Solution:

1. **Identify the Ratios**:
The processors are given in the following ratios:
- \( p-1 : \frac{1}{6} \)
- \( p-2 : \frac{1}{5} \)
- \( p-3 : \frac{1}{3} \)
- \( p-4 : \frac{1}{2} \)

2. **Convert Ratios to Whole Numbers**:
To convert these fractions into whole numbers, we will find the Least Common Multiple (LCM) of the denominators (6, 5, 3, and 2).

3. **Calculate the LCM**:
The LCM of 6, 5, 3, and 2 is 30.
- Factors of 6: \( 2 \times 3 \)
- Factors of 5: \( 5 \)
- Factors of 3: \( 3 \)
- Factors of 2: \( 2 \)
The LCM is \( 2 \times 3 \times 5 = 30 \).

4. **Multiply Each Ratio by the LCM**:
Now, we will multiply each ratio by 30 to convert them into whole numbers:
- \( p-1 = 30 \times \frac{1}{6} = 5 \)
- \( p-2 = 30 \times \frac{1}{5} = 6 \)
- \( p-3 = 30 \times \frac{1}{3} = 10 \)
- \( p-4 = 30 \times \frac{1}{2} = 15 \)

5. **Sum the Whole Numbers**:
Now, we will add these whole numbers to find the total number of processors:
\[ Total = p-1 + p-2 + p-3 + p-4 = 5 + 6 + 10 + 15 = 36 \]

6. **Conclusion**:
Therefore, the minimum number of processors in the institute is **36**.