A train has to complete a journey of 800 km. If it meets a minor accident, its speed becomes half of the existing speed. If there is a mechanical defect, the speed becomes one-fourth of the existing speed. On its way, the train meets with a minor accident after 200 km; and 400 km thereafter, it develops a mechanical defect. Had the train developed the mechanical defect after 200 km and met the minor accident 400 km thereafter, it would have taken 4 more hours to reach its destination. What was the original speed of the train in km per hour?

Correct Answer: Option A (200)

To solve this UPSC CSAT Time, Speed, and Distance problem, we must carefully interpret the phrase "existing speed," which dictates a cascading effect on the train's velocity rather than a standard reduction from the initial speed.

Let the original speed be v km/hr. The total distance is 800 km.

Scenario 1:

• First 200 km at speed v: Time = 200/v
• Next 400 km after a minor accident (speed becomes v/2): Time = 400/(v/2) = 800/v
• Remaining 200 km after a mechanical defect (speed becomes ¼ of the existing v/2, i.e., v/8): Time = 200/(v/8) = 1600/v
• Total Time T₁ = 200/v + 800/v + 1600/v = 2600/v

Scenario 2:

• First 200 km at speed v: Time = 200/v
• Next 400 km after mechanical defect (speed becomes v/4): Time = 400/(v/4) = 1600/v
• Remaining 200 km after minor accident (speed becomes ½ of the existing v/4, i.e., v/8): Time = 200/(v/8) = 1600/v
• Total Time T₂ = 200/v + 1600/v + 1600/v = 3400/v

We are given that Scenario 2 takes 4 hours longer (T₂ - T₁ = 4). 3400/v - 2600/v = 4 ⟹ 800/v = 4 ⟹ v = 200 km/hr.

Why other options are incorrect:

• Option B (190) & Option C (150): Mathematically incorrect; substituting these into the derived equation 800/v yields 4.21 hours and 5.33 hours respectively, not 4.
• Option D (100): This is a deliberate distractor. Aspirants arrive at 100 km/hr if they incorrectly assume the "existing speed" always refers to the initial starting speed v, which incorrectly sets up the equation as 400/v = 4.

Takeaway: In Quantitative Aptitude problems involving sequential events, modifiers like "existing" or "remaining" require cascaded calculations (compounding fractional changes) rather than applying deductions to the initial base value.