Time, speed and distance; trains, boats and streams; races: previous-year questions
46 real UPSC Prelims questions (2011–2026) with the official answer and a clear explanation. 16 of them came in the last five papers.
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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?
Answer (a) & explanation
1. Phase 1, same direction, speeds 7 : 5. X gains 2 rounds, so they meet 2 times. The second meeting is at the start point, when Y turns.
2. Phase 2, opposite directions, speeds 7 and 10. X runs 14 more rounds, which takes 2 units of time. Together they cover 17 × 2 = 34 rounds, so they meet 34 times.
3. The 34th meeting is at the moment they stop, and that one is excluded. Total = 2 + 33 = 35.
read "before they finally stopped". The meeting at the stopping moment does not count.
Traps like this come back in new clothes. The PYQ Book collects all of them, topic by topic, so none catch you twice. Try the free sample
Answer as per the official UPSC answer key.
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.
Answer (c) & explanation
The explosion happened at 4:58:42 PM (c). Step 1: drone 2 left at 5:02 (when picture 1 arrived) and reached the spot at 5:05. So one flight takes 180 seconds, and the distance = 30 × 180 = 5400 m. Step 2: drone 1 also flew 180 seconds and reached at 5:02, so it left at 4:59:00, the moment the sound reached the camp. Step 3: sound took 5400 ÷ 300 = 18 seconds. So the explosion was at 4:59:00 − 18 s = 4:58:42 PM. Option (a) forgets the sound delay. Options (b) and (d) miscalculate.
Tip: work backwards one step at a time. Pictures arrive instantly, so only flight time and sound travel time matter.
don't stop at 4:59:00. That is when the sound reached the camp, not when the blast happened.
Answer as per the official UPSC answer key.
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?
Answer (d) & explanation
Y finishes 6 minutes earlier (d). Step 1: time = distance ÷ speed. X: 1 km at 5 km/h = 12 min; 2 km at 10 km/h = 12 min; 3 km at 4 km/h = 45 min. Total 69 min. Step 2: Y: 1 km at 4 km/h = 15 min; 2 km at 10 km/h = 12 min; 3 km at 5 km/h = 36 min. Total 63 min. Step 3: 69 − 63 = 6 minutes. Options (a), (b) and (c) fail.
Tip: work in minutes: time (min) = distance ÷ speed × 60. Y wins because Y is faster over the longer last stretch.
Answer as per the official UPSC answer key.
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?
Answer (a) & explanation
The original speed was 200 km/h (a). Step 1: actual journey: 200 km at v; then a minor accident halves speed: 400 km at v/2; then a defect quarters it: 200 km at v/8. Time = 200/v + 800/v + 1600/v = 2600/v. Step 2: other order: 200 km at v; defect: 400 km at v/4; accident halves again: 200 km at v/8. Time = 200/v + 1600/v + 1600/v = 3400/v. Step 3: the difference is 4 hours: 800/v = 4, so v = 200 km/h. Options (b), (c) and (d) fail.
Tip: speed changes multiply: half then a quarter gives one-eighth, in either order. Only the middle stretch differs between the two cases.
the cuts apply to the existing speed. After both events the speed is v/8 in either order. Only the middle 400 km differs.
Answer as per the official UPSC answer key.
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?
Answer (c) & explanation
The train crosses him in about 13.85 seconds (c). Step 1: same direction: relative speed = 100 − 4 = 96 km/h. The train takes 15 seconds to pass him. Step 2: opposite direction: relative speed = 100 + 4 = 104 km/h. The train's length stays the same, so time falls in proportion. Step 3: time = 15 × 96 ÷ 104 = 1440 ÷ 104 ≈ 13.85 seconds. Options (a), (b) and (d) are close numbers from rounding or arithmetic slips.
Tip: when the distance (train length) is the same, time is inversely proportional to relative speed. No need to find the length itself.
Answer as per the official UPSC answer key.
Team X scored a total of N runs in 20 overs. Team Y tied the score in 10% less overs. Had team Y's average run rate (runs per over) been 50% higher, the scores would have been tied in 12 overs. How many runs were scored by team X?
Answer (d) & explanation
Team X's runs cannot be determined (d). Step 1: Y tied the score in 10% fewer overs: 18 overs. So Y's run rate = N/18. Step 2: 50% higher rate = 1.5 × N/18 = N/12. At that rate, N runs take N ÷ (N/12) = 12 overs, whatever N is. Step 3: so the second fact is automatically true and gives no new information. N can be any number. Options (a) 72, (b) 144 and (c) 216 all fit equally well.
Tip: if a "condition" turns out to be true for every value, it adds no information. Check whether each fact gives a new equation before solving.
a condition that is always true adds no information. Test it before you start solving.
Answer as per the official UPSC answer key.
P and Q walk along a circular track. They start at 5:00 a.m. 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.?
Answer (b) & explanation
They cross 13 times (b). Step 1: walking in opposite directions on a circle, their speeds add: 5 + 3 = 8 rounds per hour. So they cross every 1/8 hour = 7.5 minutes. Step 2: crossings happen at 7.5, 15, 22.5, 30, … minutes after 5:00. The first one after 5:20 is at 22.5 minutes (the 3rd crossing). Step 3: 7:00 is 120 minutes, which is the 16th crossing. Between 5:20 and 7:00 (not counting 7:00), crossings 3 to 15 occur: 13 times. Options (a), (c) and (d) miscount the ends.
Tip: opposite directions add speeds; same direction subtracts them. Count crossings carefully at the start and end of the time window.
Answer as per the official UPSC answer key.
A 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?
Answer (c) & explanation
The tram is 20 m long (c). Step 1: the tram and walkers move the same way, so use relative speed = tram speed − walker speed. Length = (v − 3) × 8 = (v − 4) × 9, with speeds in km/hr and time in seconds. Step 2: 8v − 24 = 9v − 36, so v = 12 km/hr. Step 3: relative speed past X = 12 − 3 = 9 km/hr = 9 × 5/18 = 2.5 m/s. Length = 2.5 × 8 = 20 m. Check with Y: 8 km/hr = 20/9 m/s × 9 s = 20 m ✓. Options (a), (b) and (d) fail.
Tip: to change km/hr to m/s, multiply by 5/18. Passing a person (a point) means covering only the tram's own length.
Answer as per the official UPSC answer key.
A set (X) of 20 pipes can fill 70% of a tank in 14 minutes. Another set (Y) of 10 pipes fills 3/8th of the tank in 6 minutes. A third set (Z) of 16 pipes can empty half of the tank in 20 minutes. If half of the pipes of set X are closed and only half of the pipes of set Y are open, and all pipes of the set (Z) are open, then how long will it take to fill 50% of the tank?
Answer (d) & explanation
It takes 16 minutes to fill half the tank (d). Step 1: work out the rate of each full set per minute. X fills 70% in 14 min: 5% a minute; half of X = 2.5%. Y fills 37.5% in 6 min: 6.25%; half of Y = 3.125%. Z empties 50% in 20 min: 2.5% a minute (all of Z is open). Step 2: net filling = 2.5 + 3.125 − 2.5 = 3.125% a minute. Step 3: time for 50% = 50 ÷ 3.125 = 16 minutes. Options (a), (b) and (c) forget to halve a set or ignore the emptying pipes.
Tip: convert everything into "percent of tank per minute". Add filling pipes, subtract emptying ones, then divide.
Answer as per the official UPSC answer key.
X can complete one-third of a certain work in 6 days, Y can complete one-third of the same work in 8 days and Z can complete three-fourth of the same work in 12 days. All of them work together for n days and then X and Z quit and Y alone finishes the remaining work in 8 2/3 days. What is n equal to?
Answer (b) & explanation
n is 4 days (b). Step 1: find each person's full time. X does 1/3 in 6 days, so the whole work in 18 days. Y: 1/3 in 8 days, so 24 days. Z: 3/4 in 12 days, so 16 days. Step 2: together per day: 1/18 + 1/24 + 1/16 = 8/144 + 6/144 + 9/144 = 23/144. Step 3: Y alone works 8⅔ = 26/3 days: (26/3) × (1/24) = 26/72 = 52/144. The rest, 92/144, was done together: 23n/144 = 92/144, so n = 4. Options (a), (c) and (d) fail.
Tip: turn part-work into full-work days first. Use the LCM (here 144) as the total work to avoid fractions.
Answer as per the official UPSC answer key.
A certain number of men can complete a piece of work in 6k days, where k is a natural number. By what percent should the number of men be increased so that the work can be completed in 5k days?
Answer (c) & explanation
The men must increase by 20% (c). Step 1: total work = men × days. So if M men take 6k days, M × 6k = M′ × 5k, where M′ is the new number of men. Step 2: M′ = 6M/5 = 1.2M. Step 3: the increase is 0.2M, that is 20%. Note that k cancels out, so its value does not matter. Option (b), 16.67%, is the fall in days (1k out of 6k), a common mix-up. Options (a) and (d) are wrong.
Tip: men and days are inversely related. If days go from 6 to 5, men go from 5 to 6, an increase of 1 on 5 = 20%. Always take the percentage on the old value.
(b) is the trap: 50/3% is the cut in days for the reverse change, not the rise in men.
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Answer as per the official UPSC answer key.
X, Y and Z can complete a piece of work individually in 6 hours, 8 hours and 8 hours respectively. However, only one person at a time can work in each hour and nobody can work for two consecutive hours. All are engaged to finish the work. What is the minimum amount of time that they will take to finish the work?
Answer (c) & explanation
The minimum time is 6 hours 45 minutes (c). Step 1: work done per hour: X = 1/6, Y = 1/8, Z = 1/8. To finish fast, use X, the fastest, as often as possible: every other hour, since no one can work two hours in a row. Step 2: order X, Y, X, Z, X, Y for 6 hours: 3 × 1/6 + 3 × 1/8 = 1/2 + 3/8 = 7/8 done. Step 3: 1/8 is left. Hour 7 can be X again: (1/8) ÷ (1/6) = 3/4 hour = 45 minutes. Total 6 h 45 min. Options (a) and (b) are too short; (d) wastes time.
Tip: in "one worker at a time" questions, give the fastest worker every slot the rule allows, and let the fastest finish the leftover bit.
Answer as per the official UPSC answer key.
A, B, C working independently can do a piece of work in 8, 16 and 12 days respectively. A alone works on Monday, B alone works on Tuesday, C alone works on Wednesday; A alone, again works on Thursday and so on.
Consider the following statements :
- 1. The work will be finished on Thursday.
- 2. The work will be finished in 10 days.
Which of the above statements is/are correct ?
Answer (a) & explanation
Only statement 1 is correct (a). Step 1: take total work = 48 units (LCM of 8, 16, 12). A does 6, B does 3 and C does 4 units a day. One 3-day cycle = 13 units. Step 2: three cycles (9 days) = 39 units. Day 10 (A) brings it to 45. Day 11 (B) brings it to 48. The work ends on day 11. Step 3: day 1 is Monday, so day 8 is Monday again and day 11 is a Thursday. Statement 1 is true; statement 2 (10 days) is false.
Tip: in rotating-work problems, use LCM units and count full cycles first, then finish day by day.
the last day is a B day, and B needs a full day for the last 3 units. So it takes 11 days, not 10.
Answer as per the official UPSC answer key.
Aman 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
Answer (d) & explanation
The village clock is 5 minutes fast (d). Step 1: let the outward drive take t minutes. The return route is 1.25 times longer at twice the speed, so it takes 1.25 ÷ 2 = 0.625t. Step 2: by the home clock, he was away 14:30 to 16:00 = 90 minutes. So t + 25 + 0.625t = 90, giving 1.625t = 65 and t = 40. Step 3: he truly reached the village at 14:30 + 40 = 15:10. The village clock showed 15:15, so it is 5 minutes fast. Options (a), (b) and (c) fail.
Tip: use only the reliable clock for time intervals. Then compare the true arrival time with what the other clock showed.
Answer as per the official UPSC answer key.
24 men and 12 women can do a piece of work in 30 days. In how many days can 12 men and 24 women do the same piece of work?
Answer (d) & explanation
The data are inadequate (d). Step 1: we are not told how fast a man works compared with a woman. Step 2: if men work faster, the new team (fewer men, more women) is slower, taking more than 30 days. If women work faster, it is quicker. If both work at the same rate, it takes the same 30 days. Step 3: since the answer changes with that one missing fact, we cannot decide. Option (a) assumes equal rates. Option (b) assumes men are faster. Option (c) is vague and not a conclusion.
Tip: before calculating, check whether every rate is given. If one key comparison is missing, "data inadequate" is often right.
(a) assumes a man and a woman work equally fast. The question never says so.
Answer as per the official UPSC answer key.
X 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)?
Answer (b) & explanation
X passes Y 3 times (b). Step 1: the race is 3 km = 3000 m on a 300 m track. So X runs 10 rounds. Step 2: the speeds are 3 : 2, so in the same time Y runs 2/3 × 10 = 6⅔ rounds. Step 3: X gains 10 − 6⅔ = 3⅓ rounds over Y. Each full round gained is one pass. So X passes Y 3 times. Options (a), (c) and (d) miscount; the extra ⅓ round is not a full pass.
Tip: on a circular track, the number of passes = the number of full laps the faster runner gains. The start is not counted.
Answer as per the official UPSC answer key.
A 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 :
Answer (c) & explanation
Both statements are correct (c). Step 1: X walks at a steady 1.5 km/hr. In 5 hours X covers 1.5 × 5 = 7.5 km. Step 2: Y's speed rises by 0.25 each hour: 1, 1.25, 1.5, 1.75, 2. In 5 hours Y covers 1 + 1.25 + 1.5 + 1.75 + 2 = 7.5 km. Step 3: together they cover 7.5 + 7.5 = 15 km, the full distance. So they meet after 5 hours (statement 1). Each has walked 7.5 km, which is exactly midway (statement 2). So options (a), (b) and (d) fail.
Tip: Y's speeds average 1.5 km/hr over those 5 hours, the same as X. Equal average speeds for equal time mean equal distance.
Answer as per the official UPSC answer key.
A man completes 7/8 of a job in 21 days. How many more days will it take him to finish the job if quantum of work is further increased by 50% ?
Answer (d) & explanation
He needs 15 more days (d). Step 1: he does 7/8 of the job in 21 days. So the whole job takes 21 ÷ 7/8 = 24 days. Step 2: the work is raised by 50%, so the total is now 1.5 jobs. Work left = 1.5 − 7/8 = 12/8 − 7/8 = 5/8 of a job. Step 3: 5/8 of a job takes 5/8 × 24 = 15 days. Option (a), 24, is the time for one full job. Options (b) and (c) miscalculate the remaining work.
Tip: first find the time for one full job. Then any amount of work is just that fraction of the time.
the 50% is added to the whole job, not to the remaining 1/8.
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Answer as per the official UPSC answer key.
A person X can complete 20% of work in 8 days and another person Y can complete 25% of the same work in 6 days. If they work together, in how many days will 40% of the work be completed?
Answer (a) & explanation
40% of the work takes 6 days (a). Step 1: X does 20% in 8 days, so the full job takes 40 days. Y does 25% in 6 days, so the full job takes 24 days. Step 2: together, per day: 1/40 + 1/24 = 3/120 + 5/120 = 8/120 = 1/15. So together they finish in 15 days. Step 3: 40% of the job takes 0·4 × 15 = 6 days. Option (b), 8, and the others fail.
Tip: convert each person's partial work into full-job time. Then add daily rates (1/days), not days. Using LCM units (here 120 units of work) also makes it quick.
Answer as per the official UPSC answer key.
A 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
Answer (b) & explanation
The average speed lies between v and 2v (b). Step 1: each trip covers the same distance d. Average speed = total distance ÷ total time, not the average of the speeds. Step 2: total time = d/v + d/2v + d/3v + d/4v = (d/v)(1 + ½ + ⅓ + ¼) = 25d/12v. Step 3: average speed = 4d ÷ 25d/12v = 48v/25 = 1·92v. That is between v and 2v. Option (c) is the trap from averaging the speeds (2·5v).
Tip: for equal distances, the average speed is the harmonic mean, always less than the simple average. Slow legs take more time, so they pull the average down.
the plain average of v, 2v, 3v, 4v is 2·5v. The car spends more time at low speed.
Answer as per the official UPSC answer key.