UPSC Prelims 2026 · CSAT
Question 69: Basic Numeracy
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?
AnswerUPSC’s answer is (a).
Let the speed be v. Actual journey: 200 km at v, 400 km at v/2, 200 km at v/8, because the second fault cuts the already reduced speed again. Time = 200/v + 800/v + 1600/v = 2600/v. Swapped order: 200 km at v, 400 km at v/4, 200 km at v/8, giving 200/v + 1600/v + 1600/v = 3400/v. The extra time is 800/v = 4 hours, so v = 200 km/h.
Why each option is right or wrong
- (a) Right: 800/v = 4 gives v = 200.
- (b) Wrong: 190 gives an extra time of about 4.2 hours.
- (c) Wrong: 150 gives an extra time of about 5.3 hours.
- (d) Wrong: 100 gives an extra time of 8 hours.
Only the provisional key (27 May 2026) was found; the final key replaces it when found.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2026, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.