UPSC Prelims 2026 · CSAT Paper 2 · Set DMathematicsQ 49 / 80

UPSC Prelims 2026 CSAT Paper 2 — Question 49

Question

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)200
(b)190
(c)150
(d)100

UnlockIAS answer: (a)200

Solution

Each event acts on the "existing" (already reduced) speed. Actual time = 200/v + 800/v + 1600/v = 2600/v; the hypothetical = 200/v + 1600/v + 1600/v = 3400/v. The difference 800/v = 4 h gives v = 200 km/h.

UnlockIAS expert key, prepared 24 May 2026 and cross-verified by re-deriving the answer — this is our key, not the official UPSC provisional key (awaited).

Section

Mathematics

Topic Reference

Numeracy — Speed, time and distance

Official UPSC key still awaited

UPSC has announced (PIB Release 2262441, 18 May 2026) that the provisional answer key for CSE Prelims 2026 will be released “soon after Exam”. Until then, use the CSAT score calculator for your qualifying-status estimate, and join the per-question discussion below.

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Official Notification

PIB Release 2262441

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