Step 1: Understanding the Question:
We have two scenarios for a train journey with different speeds and corresponding deviations from the scheduled time. We need to find the scheduled time of the journey. The distance is constant in both scenarios.
Step 2: Key Formula or Approach:
The fundamental relationship is Distance = Speed \( \times \) Time. We will set up two equations based on the two scenarios and solve them simultaneously.
Step 3: Detailed Explanation:
Let the distance be \(D\) km and the scheduled time be \(T\) hours.
Unit Conversion: First, convert all units to be consistent (km and hours). Speed in the second case is 8.33 m/s. Let's convert this to km/hr. Note that \(8.33 \approx 8\frac{1}{3} = \frac{25}{3}\). \[ \text{Speed}_2 = \frac{25}{3} \text{ m/s} = \frac{25}{3} \times \frac{18}{5} \text{ km/hr} = 5 \times 6 = 30 \text{ km/hr} \] Time deviations: 60 minutes = 1 hour. 300 minutes = 5 hours.
Scenario 1: Speed \(S_1 = 50\) km/hr. Time taken \(T_1 = T - 1\) hours (since it's 1 hour earlier). Distance \(D = S_1 \times T_1 = 50(T - 1)\). (Equation 1)
Scenario 2: Speed \(S_2 = 30\) km/hr. Time taken \(T_2 = T + 5\) hours (since it's 5 hours late). Distance \(D = S_2 \times T_2 = 30(T + 5)\). (Equation 2) Since the distance \(D\) is the same in both cases, we can equate the two expressions for \(D\). \[ 50(T - 1) = 30(T + 5) \] \[ 50T - 50 = 30T + 150 \] \[ 50T - 30T = 150 + 50 \] \[ 20T = 200 \] \[ T = \frac{200}{20} = 10 \text{ hours} \]
Step 4: Final Answer:
The correct (scheduled) time for the train to complete its journey is 10 hours.