There are two places X and Y, 200 km apart from each other. Initially two persons P and Q both are at 'X'. The speed of P is 20 km/h and speed of Q is 30 km/h. Later on they starts to move to and fro between X and Y.
After starting their race, they meet each other for the `n^(th)` time at point X, then what is the minimum possible value of n?

Options

A. 1
B. 2
C. 3
D. 5

Answer

To solve the problem, we need to determine how many times persons P and Q meet at point X after they start moving back and forth between points X and Y.

### Step-by-Step Solution:

1. **Identify the Speeds and Distance**:
- The distance between X and Y is 200 km.
- Speed of P = 20 km/h.
- Speed of Q = 30 km/h.

2. **Calculate the Time to Meet for the First Time**:
- When both start moving towards each other, their relative speed is the sum of their speeds:
\[ \text{Relative Speed} = 20 \text{ km/h} + 30 \text{ km/h} = 50 \text{ km/h} \]
- The time taken to meet for the first time can be calculated using the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{200 \text{ km}}{50 \text{ km/h}} = 4 \text{ hours} \]

3. **Calculate the Distance Covered by Each Person**:
- In 4 hours, P covers:
\[ \text{Distance covered by P} = 20 \text{ km/h} \times 4 \text{ hours} = 80 \text{ km} \]
- In 4 hours, Q covers:
\[ \text{Distance covered by Q} = 30 \text{ km/h} \times 4 \text{ hours} = 120 \text{ km} \]

4. **Determine the Meeting Points**:
- After meeting for the first time, P will continue to Y and Q will return to X.
- The total distance they cover before meeting again is 200 km (to Y) + 200 km (back to X) = 400 km.

5. **Calculate the Time for Subsequent Meetings**:
- The time taken for them to meet again after the first meeting is:
\[ \text{Time for next meetings} = \frac{400 \text{ km}}{50 \text{ km/h}} = 8 \text{ hours} \]

6. **Calculate Total Time for n Meetings**:
- The first meeting takes 4 hours, and each subsequent meeting takes 8 hours.
- If they meet for the nth time at point X, the total time can be expressed as:
\[ \text{Total Time} = 4 + 8(n-1) \]
- We need to find the minimum value of n such that they meet at point X.

7. **Determine the Minimum Value of n**:
- The total distance covered by P in time T is:
\[ \text{Distance covered by P} = 20 \text{ km/h} \times T \]
- The total distance covered by Q in time T is:
\[ \text{Distance covered by Q} = 30 \text{ km/h} \times T \]
- For them to meet at point X, the total distance covered by both must be a multiple of 200 km.

8. **Calculate the Total Distance**:
- The total distance covered in time T is:
\[ \text{Total Distance} = 20T + 30T = 50T \]
- We need \(50T\) to be a multiple of 200 km:
\[ 50T = 200k \quad \text{for some integer } k \]
- Simplifying gives:
\[ T = 4k \]
- Since \(T = 4 + 8(n-1)\), equate:
\[ 4 + 8(n-1) = 4k \]
- Rearranging gives:
\[ 8(n-1) = 4k - 4 \]
\[ 2(n-1) = k - 1 \]
- This implies \(k = 2(n-1) + 1\), which must be an integer.

9. **Finding the Minimum n**:
- The smallest integer solution for \(k\) occurs when \(n = 5\).

### Conclusion:
Thus, the minimum possible value of n such that P and Q meet at point X for the nth time is **5**.