Let the original speed of the train be $v$ km/h.

The phrase "existing speed" means the speed of the train at that specific moment. Therefore, if the train suffers a second incident, the new speed will be calculated based on its current reduced speed (the effects compound).

The total journey is $800$ km. We can divide the journey into three parts: the first $200$ km, the middle $400$ km, and the final $200$ km.

Case 1: Minor accident first, then mechanical defect

• 0 to 200 km: The train travels at its original speed $v$. Time taken = $\frac{200}{v}$
• 200 to 600 km: After the minor accident, the speed becomes half of the existing speed, i.e., $\frac{v}{2}$. Time taken = $\frac{400}{v/2} = \frac{800}{v}$
• 600 to 800 km: After the mechanical defect, the speed becomes one-fourth of the existing speed ($\frac{v}{2}$), i.e., $\frac{v}{2} \times \frac{1}{4} = \frac{v}{8}$. Time taken = $\frac{200}{v/8} = \frac{1600}{v}$

Total Time ($T_1$) = $\frac{200}{v} + \frac{800}{v} + \frac{1600}{v} = \frac{2600}{v}$

Case 2: Mechanical defect first, then minor accident

• 0 to 200 km: The train travels at its original speed $v$. Time taken = $\frac{200}{v}$
• 200 to 600 km: After the mechanical defect, the speed becomes one-fourth of the existing speed, i.e., $\frac{v}{4}$. Time taken = $\frac{400}{v/4} = \frac{1600}{v}$
• 600 to 800 km: After the minor accident, the speed becomes half of the existing speed ($\frac{v}{4}$), i.e., $\frac{v}{4} \times \frac{1}{2} = \frac{v}{8}$. Time taken = $\frac{200}{v/8} = \frac{1600}{v}$

Total Time ($T_2$) = $\frac{200}{v} + \frac{1600}{v} + \frac{1600}{v} = \frac{3400}{v}$

Notice that the time taken for the first 200 km and the last 200 km is identical in both cases. The difference in time comes entirely from the middle 400 km.

Calculating the Original Speed: The problem states that Case 2 takes $4$ more hours than Case 1. $T_2 - T_1 = 4$ $\frac{3400}{v} - \frac{2600}{v} = 4$ $\frac{800}{v} = 4$ $v = \frac{800}{4} = 200 \text{ km/h}$

The original speed of the train was $200$ km/h.