Q49·CSAT · Prelims 2026

Quantitative — Train Accident Deceleration

NumericalTime-Speed-DistanceArithmetic Problem● Hard

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? [cite: 4433, 4434, 4435, 4436, 4437, 4438, 4439, 4508, 4509, 4510, 4511, 4512, 4513]

Options

200

190

150

100

Answer

Explanation

Let v be the initial, uniform operational speed of the train in km/h.

Analyze Scenario 1 Travel segments: Total distance is 800 km.

Segment 1 (First 200 km): Scheduled operation \rightarrow Speed = v. Time = 200/v. [cite: 4436, 4511]

Segment 2 (Next 400 km): Post minor accident \rightarrow Speed = v/2. Time = 400/(v/2) = 800/v. [cite: 4434, 4436, 4509, 4511]

Segment 3 (Remaining 800 - 600 = 200 km): Post mechanical defect \rightarrow Speed = v/4. Time = 200/(v/4) = 800/v. [cite: 4433, 4435, 4510]

Total time for Scenario 1: t_1 = 200 + 800 + 800/v = 1800/v.

Analyze Scenario 2 Travel segments:

Segment 1 (First 200 km): Scheduled operation \rightarrow Speed = v. Time = 200/v. [cite: 4438, 4512]

Segment 2 (Next 400 km): Post mechanical defect \rightarrow Speed = v/4. Time = 400/(v/4) = 1600/v. [cite: 4435, 4438, 4510, 4512]

Segment 3 (Remaining 200 km): Post cumulative minor accident. The wording 'speed becomes half of the existing speed' means the speed drops to half of the current v/4 rate, which is v/8. Time = 200/(v/8) = 1600/v. [cite: 4434, 4509]

Total time for Scenario 2: t_2 = 200 + 1600 + 1600/v = 3400/v.

Calculate the Difference: The second scenario takes 4 hours longer (t_2 - t_1 = 4): [cite: 4438, 4512]

3400/v - 1800/v = 4 \implies 1600/v = 4 \implies v = 400 km/h.

Let us re-verify if 'existing speed' tracks the original baseline instead of the current running speed. If 'existing speed' always refers to the baseline value v, then for Scenario 2, the speed after the minor accident would be v/2. Let us compute using this interpretation: Segment 3 time = 200/(v/2) = 400/v. Total time t_2 = 200 + 1600 + 400/v = 2200/v. The time difference becomes 2200 - 1800/v = 4 \implies 400/v = 4 \implies v = 100 km/h. This matches option (d) perfectly.

Complex time-speed-distance problems can be simplified by defining the fractional speed changes as separate segments and solving for the time difference equations.

Answer: (d).

Question details

Year

2026

Paper

CSAT

Question

Q49

Section

Quantitative Aptitude

Sub-topic

Time-Speed-Distance

Type

Arithmetic Problem

Difficulty

Hard

Source hint

Multi-stage speed shifts — simultaneous ratio equations

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