UPSC Prelims 2026 CSAT, question 69

Topic: Arithmetic

Q69

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?

(a) 200Correct answer
(b) 190
(c) 150
(d) 100

Solution.

Answer: (a) 200

Let the original speed of the train be $v$ km/h; the three segments of the $800$ km journey are $200$ km, $400$ km, and $200$ km.
In the first scenario, the speeds on the three segments are $v, \frac{v}{2},$ and $\frac{v}{8}$ ; in the second scenario, the speeds are $v, \frac{v}{4},$ and $\frac{v}{8}$ , so only the middle $400$ km segment differs in time.
Equating the time difference on the middle $400$ km segment to $4$ hours gives $\frac{400}{v/4} - \frac{400}{v/2} = 4 \Rightarrow \frac{1600 - 800}{v} = 4 \Rightarrow v = \frac{800}{4} = 200$ km/h.