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2026
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?

Explanation

Let the original speed be xx km/h.

In the original scenario, the distances covered are:

  • 200 km at xx km/h,
  • then 600 km at x2\frac{x}{2} km/h for the accident,
  • then 200 km at x4\frac{x}{4} km/h for the mechanical defect. The total travel time calculates to 200x+600x2+200x4=200x+1200x+800x=2200x\frac{200}{x} + \frac{600}{\frac{x}{2}} + \frac{200}{\frac{x}{4}} = \frac{200}{x} + \frac{1200}{x} + \frac{800}{x} = \frac{2200}{x}.

In the alternate scenario, the distances covered are:

  • 200 km at x4\frac{x}{4} km/h,
  • followed by 600 km at x2\frac{x}{2} km/h. The total travel time becomes 200x4+600x2=800x+1200x=2000x\frac{200}{\frac{x}{4}} + \frac{600}{\frac{x}{2}} = \frac{800}{x} + \frac{1200}{x} = \frac{2000}{x}.

The difference in time is: [\frac{2200}{x} - \frac{2000}{x} = 4 \Rightarrow \frac{200}{x} =

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