Find the length of Train A.
Statement I: Train A of length (x + 50) m can cross a man moving at the speed of 5 m/s in the opposite direction of that of train A in 10 seconds. It can cross the same man in 15 seconds if he is moving in the same direction.
Statement II: Train A of length (x + 50) m can cross another train B of length (x + 150) m moving in the opposite direction at a speed of 10 m/s in 20 seconds. It can cross train B in 28 seconds if train B is standing on the platform.

Options

A. The data in Statement I alone are sufficient to answer the question, while the data in Statement II alone are not sufficient to answer the question.
B. The data in Statement II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question
C. The data either in Statement I or in Statement II alone are sufficient to answer the question.
D. The data in both Statements I and II together are necessary to answer the question.

Explanation

To find the length of Train A, we will analyze both statements provided in the question systematically.

### Step-by-Step Solution:

**Statement I:**
1. **Given Information:**
- Length of Train A = \( x + 50 \) meters
- Speed of the man = 5 m/s
- Time taken to cross the man in the opposite direction = 10 seconds
- Time taken to cross the man in the same direction = 15 seconds

2. **Equation for crossing the man in the opposite direction:**
- The relative speed when crossing the man in the opposite direction is \( s_A + 5 \) m/s (where \( s_A \) is the speed of Train A).
- Distance = Speed × Time
- Therefore, we have:
\[ x + 50 = (s_A + 5) \times 10 \quad \text{(Equation 1)} \]

3. **Equation for crossing the man in the same direction:**
- The relative speed when crossing the man in the same direction is \( s_A - 5 \) m/s.
- Thus, we have:
\[ x + 50 = (s_A - 5) \times 15 \quad \text{(Equation 2)} \]

4. **Solving Equations 1 and 2:**
- From Equation 1:
\[ x + 50 = 10s_A + 50 \implies x = 10s_A \]
- From Equation 2:
\[ x + 50 = 15s_A - 75 \implies x = 15s_A - 125 \]
- Setting the two expressions for \( x \) equal:
\[ 10s_A = 15s_A - 125 \]
\[ 5s_A = 125 \implies s_A = 25 \text{ m/s} \]
- Substituting \( s_A \) back into \( x = 10s_A \):
\[ x = 10 \times 25 = 250 \]
- Therefore, the length of Train A:
\[ \text{Length of Train A} = x + 50 = 250 + 50 = 300 \text{ meters} \]

**Statement II:**
1. **Given Information:**
- Length of Train A = \( x + 50 \) meters
- Length of Train B = \( x + 150 \) meters
- Speed of Train B = 10 m/s
- Time taken to cross Train B in the opposite direction = 20 seconds
- Time taken to cross Train B when stationary = 28 seconds

2. **Equation for crossing Train B in the opposite direction:**
- The relative speed is \( s_A + 10 \) m/s.
- Therefore:
\[ (x + 50) + (x + 150) = (s_A + 10) \times 20 \quad \text{(Equation 3)} \]

3. **Equation for crossing Train B when stationary:**
- The relative speed is just \( s_A \) m/s.
- Thus:
\[ (x + 50) + (x + 150) = s_A \times 28 \quad \text{(Equation 4)} \]

4. **Solving Equations 3 and 4:**
- From Equation 3:
\[ 2x + 200 = 20s_A + 200 \implies 2x = 20s_A \]
\[ x = 10s_A \]
- From Equation 4:
\[ 2x + 200 = 28s_A \implies 2x = 28s_A - 200 \]
- Setting the two expressions for \( x \) equal:
\[ 10s_A = 14s_A - 100 \]
\[ 4s_A = 100 \implies s_A = 25 \text{ m/s} \]
- Substituting \( s_A \) back into \( x = 10s_A \):
\[ x = 10 \times 25 = 250 \]
- Therefore, the length of Train A:
\[ \text{Length of Train A} = x + 50 = 250 + 50 = 300 \text{ meters} \]

### Conclusion:
From both statements, we find that the length of Train A is \( \text{300 meters} \).