UPSC Prelims PYQ › Quantitative Aptitude › Time, Speed & Distance
Time, Speed & Distance — UPSC CSAT Paper II Quantitative Aptitude previous year questions
20 free solved previous year questions on Time, Speed & Distance, papers from 2015 to 2026, drawn from UPSC CSAT, covering questions of the kind UPSC asks in CSAT Paper II. Every answer key has been checked again against published sources and solved independently, and each explanation is written fresh.
Read all 20 questions with answers
- #1UPSC CSAT 2026The 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?
- A13.51
- B13.65
- C13.85
- D14.05
Answer: (C) 13.85Explanation
Train speed is 100 km/h.
When P walks in the same direction, relative speed = 100 – 4 = 96 km/h = 96 × 5/18 = 26.67 m/s.
Train length = 26.67 × 15 = 400 m.
When P walks in the opposite direction, relative speed = 100 + 4 = 104 km/h = 104 × 5/18 = 28.89 m/s.
Time = 400 ÷ 28.89 = 13.85 s (approximately).
Answer: option (c).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #2UPSC CSAT 2026X 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?
- A35
- B34
- C31
- D29
Answer: (A) 35Explanation
Count time in units where X runs 7 rounds and Y runs 5 rounds in 1 unit. X runs 21 rounds in 3 units.
First 1 unit, same direction: relative speed is 7 – 5 = 2 rounds per unit. They meet 2 times (at 0.5 and 1).
After 1 unit Y turns back with speed 10 rounds per unit. Now they run opposite ways: relative speed 7 + 10 = 17 rounds per unit.
In the next 2 units they close 34 rounds, so 34 meetings. The last one is at the end, when they stop, so it is not counted.
Total meetings = 2 + 34 – 1 = 35.
Answer: option (a).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #3UPSC CSAT 2026An 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.
- A4:59:00 PM
- B4:59:02 PM
- C4:58:42 PM
- D4:56:32 PM
Answer: (C) 4:58:42 PMExplanation
Let the explosion spot be d metres from the camp.
Sound reaches the camp in d/300 seconds. Drone 1 then flies d/30 seconds. Its photo is received at 5:02 PM.
Drone 2 starts at 5:02 PM and flies d/30 seconds. Its photo comes at 5:05 PM.
So d/30 = 3 minutes = 180 s, and d = 5400 m.
Sound time = 5400 ÷ 300 = 18 s.
Explosion time = 5:02:00 – (18 + 180) s = 5:02:00 – 3 min 18 s = 4:58:42 PM.
Answer: option (c).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #4UPSC CSAT 2026X 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?
- A3
- B4
- C5
- D6
Answer: (D) 6Explanation
Time for X: 1/5 + 2/10 + 3/4 hours = 0.2 + 0.2 + 0.75 = 1.15 hours.
Time for Y: 1/4 + 2/10 + 3/5 hours = 0.25 + 0.2 + 0.6 = 1.05 hours.
Y is faster by 1.15 – 1.05 = 0.1 hour.
0.1 hour = 0.1 × 60 = 6 minutes.
Answer: 6 minutes, option (d).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #5UPSC CSAT 2026A 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?
- A200
- B190
- C150
- D100
Answer: (A) 200Explanation
Let the original speed be v.
Case 1: 200 km at v, then accident: speed v/2 for 400 km, then defect: speed one-fourth of v/2 = v/8 for the last 200 km.
Time = 200/v + 800/v + 1600/v = 2600/v.
Case 2: 200 km at v, then defect: speed v/4 for 400 km, then accident: speed half of v/4 = v/8 for 200 km.
Time = 200/v + 1600/v + 1600/v = 3400/v.
Difference = 800/v = 4 hours, so v = 200 km/h.
Answer: option (a).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #6UPSC CSAT 2025P and Q walk along a circular track. They start at 5:00 am from the same point in opposite directions. P walks at an average speed of 5 rounds per hour and Q walks at an average speed of 3 rounds per hour. How many times will they cross each other between 5:20 a.m. and 7:00 a.m.?
- A12
- B13
- C14
- D15
Answer: (B) 13Explanation
In opposite directions they meet 5 + 3 = 8 times per hour, at times k/8 hours after 5:00 (k = 1, 2, ...).
Between 5:20 (1/3 h) and 7:00 (2 h) the meetings are k = 3 to 15, i.e. 13 meetings; the 16th meeting falls exactly at 7:00, the boundary.
Taking 'between' as exclusive, the count is 13.
Answer: option (b).Source: UPSC CSAT 2025 question; answer key checked by independent computation
- #7UPSC CSAT 2025A tram overtakes 2 persons X and Y walking at an average speed of 3 km/hr and 4 km/hr in the same direction and completely passes them in 8 seconds and 9 seconds respectively. What is the length of the tram?
- A15 m
- B18 m
- C20 m
- D24 m
Answer: (C) 20 mExplanation
Person speeds are 5/6 m/s and 10/9 m/s. With tram speed v and length L: L = 8(v - 5/6) = 9(v - 10/9).
This gives v = 10/3 m/s, so L = 8 x (10/3 - 5/6) = 20 m.
Answer: option (c).Source: UPSC CSAT 2025 question; answer key checked by independent computation
- #8UPSC CSAT 2022X and Y run a 3 km race along a circular course of length 300 m. Their speeds are in the ratio 3:2. If they start together in the same direction, how many times would the first one pass the other (the start-off is not counted as passing)?
- A2
- B3
- C4
- D5
Answer: (B) 3Explanation
Let speeds be 3 and 2 m/s. The race takes the faster runner 3000/3 = 1000 s.
He gains 1 m/s on the other, so he gains one full lap (300 m) every 300 s.
Passes occur at 300 s, 600 s and 900 s, so 3 times.
Answer: option (b).Source: UPSC CSAT 2022 question; answer key checked by independent computation
- #9UPSC CSAT 2022A man started from home at 14:30 hours and drove to village, arriving there when the village clock indicated 15:15 hours. After staying for 25 minutes, he drove back by a different route of length 1.25 times the first route at a rate twice as fast reaching home at 16:00 hours. As compared to the clock at home, the village clock is
- A10 minutes slow
- B5 minutes slow
- C10 minutes fast
- D5 minutes fast
Answer: (D) 5 minutes fastExplanation
Total 90 min minus 25 min stay = 65 min driving. Let the outward trip take t minutes.
Return trip: 1.25 times distance at twice speed takes 0.625t. So 1.625t = 65 and t = 40.
He reached the village at 15:10 home time, but the village clock showed 15:15, so it is 5 minutes fast.
Answer: option (d).Source: UPSC CSAT 2022 question; answer key checked by independent computation
- #10UPSC CSAT 2021A person X from a place A and another person Y from a place B set out at the same time to walk towards each other. The places are separated by a distance of 15 km. X walks with a uniform speed of 1.5 km/hr and Y walks with a uniform speed of 1 km/hr in the first hour, with a uniform speed of 1.25 km/hr in the second hour and with a uniform speed of 1.5 km/hr in the third hour and so on.Which of the following is/are correct?1. They take 5 hours to meet.2. They meet midway between A and B.Select the correct answer using the code given below:
- A1 only
- B2 only
- CBoth 1 and 2
- DNeither 1 nor 2
Answer: (C) Both 1 and 2Explanation
Y covers 1, 1.25, 1.5, 1.75, 2 km in successive hours; in 5 hours that is 7.5 km.
X covers 1.5 x 5 = 7.5 km in 5 hours.
Together 15 km at 5 hours (after 4 hours only 6 + 5.5 = 11.5 km), so they meet after 5 hours.
Each has walked 7.5 km, so the meeting point is the midpoint.
Answer: option (c).Source: UPSC CSAT 2021 question; answer key checked by independent computation
- #11UPSC CSAT 2020A car travels from a place X to place Y at an average speed of v km/hr, from Y to X at an average speed of 2v km/hr, again from X to Y at an average speed of 3v km/hr and again from Y to X at an average speed of 4v km/hr. Then the average speed of the car for the entire journey
- Ais less than v km/hr
- Blies between v and 2v km/hr
- Clies between 2v and 3v km/hr
- Dlies between 3v and 4v km/hr
Answer: (B) lies between v and 2v km/hrExplanation
Each leg is distance D, so total distance = 4D.
Total time = D/v (1 + 1/2 + 1/3 + 1/4) = 25D/(12v).
Average speed = 48v/25 = 1.92v, which lies between v and 2v.
Answer: option (b).Source: UPSC CSAT 2020 question; answer key checked by independent computation
- #12UPSC CSAT 2020A man takes half time in rowing a certain distance downstream than upstream. What is the ratio of the speed in still water to the speed of current?
- A1 : 2
- B2 : 1
- C1 : 3
- D3 : 1
Answer: (D) 3 : 1Explanation
Same distance, half the time means downstream speed is twice upstream speed.
u + c = 2(u - c) gives u = 3c.
Ratio of still-water speed to current = 3 : 1.
Answer: option (d).Source: UPSC CSAT 2020 question; answer key checked by independent computation
- #13UPSC CSAT 2019X, Y and Z are three contestants in a race of 1000 m. Assume that all run with different uniform speeds. X gives Y a start of 40 m and X gives Z a start of 64 m. If Y and Z were to compete in a race of 1000 m, how many metres start will Y give to Z?
- A20
- B25
- C30
- D35
Answer: (B) 25Explanation
When X runs 1000 m, Y runs 960 m and Z runs 936 m.
So Y : Z speeds = 960 : 936 = 40 : 39.
When Y runs 1000 m, Z runs 975 m, so Y gives a 25 m start.
Answer: option (b).Source: UPSC CSAT 2019 question; answer key checked by independent computation
- #14UPSC CSAT 2018A train 200 metres long is moving at the rate of 40 kmph. In how many seconds will it cross a man standing near the railway line?
- A12
- B15
- C16
- D18
Answer: (D) 18Explanation
Speed = 40 kmph = 100/9 m/s.
Time = 200 / (100/9) = 18 s.
Answer: option (d).Source: UPSC CSAT 2018 question; answer key checked by independent computation
- #15UPSC CSAT 2018Two persons, A and B are running on a circular track. At the start, B is ahead of A and their positions make an angle of 30° at the centre of the circle. When A reaches the point diametrically opposite to his starting point, he meets B. What is the ratio of speeds of A and B, if they are running with uniform speeds?
- A6 : 5
- B4 : 3
- C6 : 1
- D4 : 2
Answer: (A) 6 : 5Explanation
A covers 180 degrees of arc to reach the opposite point.
B starts 30 degrees ahead and meets A at that same point, so B covers 180 - 30 = 150 degrees.
Speeds are in the ratio of arcs in equal time: 180 : 150 = 6 : 5.
Answer: option (a).Source: UPSC CSAT 2018 question; answer key checked by independent computation
- #16UPSC CSAT 2017A freight train left Delhi for Mumbai at an average speed of 40 km/hr. Two hours later, an express train left Delhi for Mumbai, following the freight train on a parallel track at an average speed of 60 km/hr. How far from Delhi would the express train meet the freight train?
- A480 km
- B260 km
- C240 km
- D120 km
Answer: (C) 240 kmExplanation
The freight train has an 80 km head start.
The express gains 20 km per hour, so it catches up in 80/20 = 4 hours.
Distance from Delhi = 60 x 4 = 240 km.
Answer: option (c).Source: UPSC CSAT 2017 question; answer key checked by independent computation
- #17UPSC CSAT 2016A daily train is to be introduced between station A and station B starting from each end at 6 AM and the journey is to be completed in 42 hours. What is the number of trains needed in order to maintain the shuttle service?
- A2
- B3
- C4
- D7
Answer: (C) 4Explanation
A train leaving A at 6 AM reaches B at midnight 42 h later and can start back at the next 6 AM (day 2).
It returns to A at midnight and restarts at 6 AM on day 4, so one full cycle takes 4 days.
With one departure per day from each end, 4 trains are needed.
Answer: option (c).Source: UPSC CSAT 2016 question; answer key checked by independent computation
- #18UPSC CSAT 2016A and B walk around a circular park. They start at 8 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other after 8:00 a.m. and before 9:30 a.m.?
- A7
- B6
- C5
- D8
Answer: (A) 7Explanation
Opposite directions: relative speed = 5 rounds per hour, so they meet every 1/5 hour.
In 1.5 hours the relative distance is 7.5 rounds, giving meetings at 1, 2, ..., 7 rounds.
Seven crossings (the 8th would fall after 9:30).
Answer: option (a).Source: UPSC CSAT 2016 question; answer key checked by independent computation
- #19UPSC CSAT 2015In a 500 metres race, B starts 45 metres ahead of A, but A wins the race while B is still 35 metres behind. What is the ratio of the speeds of A to B assuming that both start at the same time?
- A25:21
- B25:20
- C5:3
- D5:7
Answer: (A) 25:21Explanation
When A finishes, A has run 500 m.
B started 45 m ahead and is 35 m short of the finish, so B has run 500 - 45 - 35 = 420 m.
Time is the same, so speed ratio = 500 : 420 = 25 : 21.
Answer: option (a).Source: UPSC CSAT 2015 question; answer key checked by independent computation
- #20UPSC CSAT 2015Two cities A and B are 360 km apart. A car goes from A to B with a speed of 40 km/h and returns to A with a speed of 60 km/h. What is the average speed of the car?
- A45 km/hr
- B48 km/hr
- C50 km/hr
- D55 km/hr
Answer: (B) 48 km/hrExplanation
Time going = 360/40 = 9 h; time returning = 360/60 = 6 h.
Total distance 720 km in 15 h.
Average speed = 720 / 15 = 48 km/h.
Answer: option (b).Source: UPSC CSAT 2015 question; answer key checked by independent computation
Want all the Time, Speed & Distance questions?
143 more questions on this topic, plus mock tests, revision and progress tracking, are in the full practice set. One payment of ₹499.
Get full access – ₹499 Already have a code? Unlock here