Step 1: Understanding the Concept:
This problem relates speed, distance, and time. Since the distance remains constant, we can use the formula \( \text{Distance} = \text{Speed} \times \text{Time} \).
Step 2: Key Formula or Approach:
1. Convert all units to km/hr and hours for consistency.
2. \( \text{Distance} (D) = S_1 \times (T - t_1) = S_2 \times (T + t_2) \), where \( T \) is the scheduled time.
Step 3: Detailed Explanation:
Speed 1 \( (S_1) = 50 \) km/hr.
Time difference 1: Reaches 60 minutes earlier \( = 1 \) hour early.
Speed 2 \( (S_2) = 8.33 \) m/s.
Convert \( S_2 \) to km/hr:
\[ 8.33 \text{ m/s} \approx \frac{25}{3} \times \frac{18}{5} = 5 \times 6 = 30 \text{ km/hr} \] Time difference 2: Reaches 300 minutes late \( = \frac{300}{60} = 5 \) hours late.
Let the correct (scheduled) time be \( T \).
Since distance \( D \) is the same:
\[ 50 \times (T - 1) = 30 \times (T + 5) \] \[ 5(T - 1) = 3(T + 5) \] \[ 5T - 5 = 3T + 15 \] \[ 2T = 20 \] \[ T = 10 \text{ hours} \]
Step 4: Final Answer:
The correct time for the train to complete its journey is 10 hours.