UPSC CSE Prelims
Time, Speed & Distance Previous Year Questions (PYQs)
Practice solved questions for Time, Speed & Distance with detailed step-by-step solutions, key insights, and trend analysis for UPSC CSE PRELIMS.
Solved Previous Year Questions
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The speed of a train $T$ is $100$ km per hour and the speed of a person $P$ is $4$ km per hour. $T$ crosses $P$ in $15$ seconds, if $P$ travels along the direction of motion of $T$. If $P$ travels along the opposite direction of $T$, then in how much time does $T$ cross $P$, in seconds, approximately?
Detailed Explanation:
Same-direction crossing: Relative speed = 100 - 4 = 96 km/h = 80/3 m/s. Train length L = (80/3) × 15 = 400 m.
Opposite-direction crossing: Relative speed = 100 + 4 = 104 km/h = 260/9 m/s. Time = 400 ÷ (260/9) = 3600/260 ≈ 13.846 seconds.
Therefore, the train crosses in approximately 13.85 seconds, making Option 3 the correct answer.
An explosion takes place at a certain distance from an army camp. As soon as the sensor in the camp receives the sound of the explosion, a drone starts flying towards the spot of explosion. The drone clicks a picture from the spot and the camp receives it at the same time. Immediately another drone starts flying to the spot and it also sends a picture as soon as it reaches the spot. The two pictures were received at 5:02 PM and 5:05 PM, respectively. If the speed of the drones is 30 m/s, at what time did the explosion take place? Assume that the speed of sound is 300 m/s.
Detailed Explanation:
The second drone leaves at 5:02 PM (when first picture arrives) and reaches at 5:05 PM, taking 180 seconds. At 30 m/s, distance d = 5400 m.
Working backward from 5:02 PM: Sound traveled 5400/300 = 18 s to reach camp, then first drone took 5400/30 = 180 s to reach spot. Total elapsed time = 198 s = 3 min 18 s.
Therefore, explosion occurred at 5:02:00 − 3:18 = 4:58:42 PM, making Option 3 the correct answer.
$X$ and $Y$ are two runners who run for the same duration of time on the same circular track. They started running at the same time in the same direction with uniform speeds. When $X$ completed $7$ rounds, $Y$ did exactly $5$. After completing $5$ rounds, $Y$ changed his direction and started running in the opposite direction with speed which is double of his earlier speed. On the other hand, $X$ continued to run with the same speed. They stopped running when $X$ completed exactly $21$ rounds. How many times did $X$ and $Y$ meet after they had started and before they finally stopped?
Detailed Explanation:
Phase 1 (Same direction): When X completes 7 rounds, Y completes 5 rounds in time T. Speed ratio is 7:5. Relative speed = 7 − 5 = 2 rounds/T. In time T, they meet 2 times (at t = 0.5T and t = T).
Phase 2 (Opposite directions): After Y completes 5 rounds, he reverses and doubles his speed to 10 rounds/T. X continues at 7 rounds/T. Relative speed = 7 + 10 = 17 rounds/T. X runs 21 − 7 = 14 more rounds in time 2T. Meetings in Phase 2 = 17 × 2 = 34 times.
Therefore, total meetings after start and before final stop = 2 + 33 = 35 (excluding the meeting at final stop), making Option 1 the correct answer.
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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?
Detailed Explanation:
Let original speed be v km/h. The journey is 800 km divided into segments: 0–200 km, 200–600 km, 600–800 km. Speed reductions compound on the existing (current) speed.
Case 1: Minor accident at 200 km (speed → v/2), mechanical defect at 600 km (speed → v/2 × 1/4 = v/8). Time = 200/v + 400/(v/2) + 200/(v/8) = 200/v + 800/v + 1600/v = 2600/v hours.
Case 2: Mechanical defect at 200 km (speed → v/4), minor accident at 600 km (speed → v/4 × 1/2 = v/8). Time = 200/v + 400/(v/4) + 200/(v/8) = 200/v + 1600/v + 1600/v = 3400/v hours.
Given: Case 2 takes 4 hours more. So, 3400/v - 2600/v = 4 → 800/v = 4 → v = 200 km/h.
Therefore, the original speed was 200 km/h, making Option 1 the correct answer.
X travels 6 km on a bicycle with average speeds of 5 km per hour, 10 km per hour and 4 km per hour during the first 1 km, the next 2 km and the remaining 3 km, respectively. Y travels the same distances with average speeds of 4 km per hour, 10 km per hour and 5 km per hour, respectively. How many minutes early will Y complete the journey if both X and Y start at the same time?
Detailed Explanation:
X's journey: Time = 1/5 + 2/10 + 3/4 = 0.2 + 0.2 + 0.75 = 1.15 hours = 69 minutes.
Y's journey: Time = 1/4 + 2/10 + 3/5 = 0.25 + 0.2 + 0.6 = 1.05 hours = 63 minutes.
Therefore, Y completes the journey 69 - 63 = 6 minutes earlier, making Option 4 the correct answer.
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