Distance = speed × time when speed is constant. The formula is simple; most exam mistakes come from units, relative motion or choosing the wrong distance. A train crossing a platform travels its own length plus the platform length, whereas a person crossing a road may be modelled as a point.
1. Units and conversion
Kilometres per hour and metres per second are related by 1 km/h = 5/18 m/s. Multiply by 5/18 to convert km/h to m/s; multiply by 18/5 for the reverse. A speed of 72 km/h is 20 m/s. If a question gives a train's length in metres and time in seconds, convert speed to m/s before dividing.
Worked example 1. A vehicle covers 150 km in 2.5 hours. Its average speed is 60 km/h. That is 60×5/18=16⅔ m/s. Dividing 150 by 2.5 uses hours, so the initial unit is km/h, not m/s.
2. Average speed is total distance over total time
When speeds change, average speed is total distance divided by total time. For equal distances at speeds u and v, average speed is 2uv/(u+v), the harmonic mean. For equal times at speeds u and v, the average is (u+v)/2. These formulas have different conditions and should not be swapped.
Worked example 2. A person travels 60 km at 30 km/h and returns 60 km at 60 km/h. The times are 2 hours and 1 hour. Total 120 km over 3 hours gives 40 km/h. The simple average 45 km/h would incorrectly treat the unequal travel times as equal.
3. Relative speed
For two objects moving towards one another on the same line, their separation closes at the sum of their speeds. Moving in the same direction, the faster gains at the difference of speeds. Use matching units. Relative motion converts a two-body situation into one speed applied to the distance to be closed.
Worked example 3. Two cyclists are 90 km apart and ride towards one another at 20 and 25 km/h. Closing speed is 45 km/h, so they meet after 90/45=2 hours. If both travel east and the faster starts 15 km behind, catch-up time is 15/(25−20)=3 hours.
4. Train crossing distances
A train of length L passing a stationary pole travels L during the crossing. Passing a platform of length P requires L+P. Two trains completely crossing each other cover the sum of their lengths at relative speed. An exam question may refer to passing a person in another moving train; identify which lengths must clear which points.
Worked example 4. A 180 m train moves at 54 km/h, or 15 m/s. It passes a pole in 180/15=12 seconds. To pass a 120 m platform, it travels 300 m and takes 300/15=20 seconds.
Worked example 5. Trains 150 m and 210 m long travel in opposite directions at 45 and 63 km/h. Combined length is 360 m. Relative speed is 108 km/h=30 m/s. They completely cross in 360/30=12 seconds.
5. Delays, gains and speed changes
For a fixed route, time is inversely proportional to speed. If speed rises 25%, travel time becomes 1/1.25=0.8 of the old time, a 20% reduction. A late departure or rest stop adds to elapsed time but does not change distance. For a planned arrival question, write scheduled and actual travel-time equations rather than combining percentages mentally.
If an object covers a distance at u and then at v, set out each segment explicitly. “Half the journey” can mean half the distance, while “half the time” means a different split. Underline which quantity is halved.
6. Error checks
A train-platform crossing must take longer than the same train passing a pole at the same speed. An average speed for positive-speed segments must lie between the slowest and fastest speeds. A catch-up problem with equal speeds has no finite catch-up time unless the objects start together.
Recall before practice
1. Convert 90 km/h into m/s. 2. Distinguish average speed for equal distances from equal times. 3. State the crossing distance for a train and a platform. 4. Explain why same-direction relative speed is a difference.
Chapter 13 practice — 50 questions
Choose one option for each item. Keep a separate answer list; explanations follow this set.
1. A car covers 7.5 metres every second. What is its speed in km/h?
A. 450 km/h B. 27 km/h C. 22.5 km/h D. 2.08 km/h
2. An athlete runs two laps of a track. She runs the first lap at 6 km/h and the second lap at 12 km/h. What is her average speed for the two laps?
A. 6 km/h B. 8 km/h C. 7 km/h D. 9 km/h
3. Two trains running in opposite directions at 25 m/s and 18 m/s cross each other completely in 8 seconds. What is the sum of their lengths?
A. 56 m B. 344 m C. 200 m D. 144 m
4. A motorcyclist covers 54 km at 54 km/h and 96 km at 48 km/h. What is the average speed of the motorcyclist for the whole journey?
A. 50 km/h B. 51 km/h C. 75 km/h D. 55 km/h
5. Two trains running in opposite directions at 15 m/s and 30 m/s cross each other completely in 19 seconds. What is the sum of their lengths?
A. 285 m B. 905 m C. 570 m D. 855 m
6. Two trains, each 175 m long, are running towards each other on parallel tracks at 72 km/h and 54 km/h. In how many seconds will they cross each other?
A. 17.5 seconds B. 10 seconds C. 5 seconds D. 70 seconds
7. A train 120 m long crosses a man standing on the platform in 12 seconds. What is the speed of the train in km/h?
A. 10 km/h B. 36 km/h C. 54 km/h D. 2.78 km/h
8. Two trains, each 220 m long, are running towards each other on parallel tracks at 18 km/h and 54 km/h. In how many seconds will they cross each other?
A. 29.33 seconds B. 22 seconds C. 44 seconds D. 11 seconds
9. An athlete runs two laps of a track. She runs the first lap at 6 km/h and the second lap at 18 km/h. What is her average speed for the two laps?
A. 10 km/h B. 8 km/h C. 9 km/h D. 12 km/h
10. Asha goes from home to office at 17 km/h and returns by the same route at 51 km/h. What is the average speed for the whole journey?
A. 29.44 km/h B. 25.5 km/h C. 34 km/h D. 22.67 km/h
11. A train takes 10 hours to cover a certain distance at 64 km/h. At what speed must it travel to cover the same distance in 5 hours?
A. 128 km/h B. 32 km/h C. 59 km/h D. 69 km/h
12. A 300 m long train running at 72 km/h crosses a platform 500 m long. How much time does it take?
A. 43 seconds B. 40 seconds C. 25 seconds D. 15 seconds
13. Two trains, each 200 m long, are running towards each other on parallel tracks at 18 km/h and 72 km/h. In how many seconds will they cross each other?
A. 16 seconds B. 20 seconds C. 26.67 seconds D. 8 seconds
14. A bike covers 22.5 metres every second. What is its speed in km/h?
A. 1350 km/h B. 6.25 km/h C. 67.5 km/h D. 81 km/h
15. A train 150 m long running at 54 km/h crosses a bridge in 50 seconds. What is the length of the bridge?
A. 150 m B. 600 m C. 450 m D. 750 m
16. Train A, 320 m long, is running at 81 km/h and train B, 220 m long, at 27 km/h on parallel tracks in the same direction. How long will train A take to pass train B completely?
A. 24 seconds B. 21 seconds C. 36 seconds D. 18 seconds
17. A bus covers 60 km at 60 km/h, 144 km at 36 km/h and 300 km at 100 km/h. What is the average speed of the bus for the whole journey?
A. 68 km/h B. 63 km/h C. 72 km/h D. 56 km/h
18. A train crosses a tree in 14 seconds and a 420 m long bridge in 42 seconds. What is the speed of the train in km/h?
A. 81 km/h B. 15 km/h C. 54 km/h D. 36 km/h
19. Two cyclists start together from the same point on a circular track 900 m long and run at 15 m/s and 10 m/s in opposite directions. After how many seconds will they meet for the first time?
A. 180 seconds B. 36 seconds C. 60 seconds D. 90 seconds
20. Two trains of equal length are running on parallel tracks in the same direction at 63 km/h and 27 km/h. The faster train passes the slower one in 20 seconds. What is the length of each train?
A. 350 m B. 200 m C. 250 m D. 100 m
21. A motorcyclist covers 270 km at 90 km/h, 400 km at 100 km/h and 30 km at 30 km/h. What is the average speed of the motorcyclist for the whole journey?
A. 73.33 km/h B. 100 km/h C. 87.5 km/h D. 77.78 km/h
22. A train 180 m long running at 36 km/h crosses a platform in 33 seconds. What is the length of the platform?
A. 30 m B. 330 m C. 180 m D. 150 m
23. An athlete runs at 12 km/h. How many seconds will he take to go once round a rectangular park 120 m long and 20 m wide?
A. 42 seconds B. 6 seconds C. 23 seconds D. 84 seconds
24. If a cyclist reduces the speed by 40%, by what per cent does the time taken to cover the same distance increase?
A. 20% B. 28 4/7% C. 66⅔% D. 40%
25. A train crosses a bridge 150 m long in 24 seconds and a platform 300 m long in 30 seconds at the same uniform speed. What is the length of the train?
A. 400 m B. 500 m C. 450 m D. 150 m
26. A truck covers the first one-third of a journey at 60 km/h, the next one-third at 36 km/h and the last one-third at 15 km/h. What is the average speed for the whole journey?
A. 37 km/h B. 27 km/h C. 45 km/h D. 30 km/h
27. A car runs for 3 hours at 45 km/h and then for 2 hours at p km/h. If its average speed for the whole trip is 51 km/h, find p.
A. 60 km/h B. 40 km/h C. 57 km/h D. 65 km/h
28. A train crosses an electric pole in 10 seconds and a 600 m long bridge in 60 seconds. What is the speed of the train in km/h?
A. 12 km/h B. 36 km/h C. 51.84 km/h D. 43.2 km/h
29. A car covers 20 km at 20 km/h, 120 km at 40 km/h and 480 km at 120 km/h. What is the average speed of the car for the whole journey?
A. 77.5 km/h B. 88.57 km/h C. 60 km/h D. 68.89 km/h
30. A boy runs at 12 km/h. How many seconds will he take to go once round a rectangular park 40 m long and 25 m wide?
A. 3 seconds B. 39 seconds C. 20 seconds D. 11 seconds
31. Train A, 140 m long, is running at 108 km/h and train B, 500 m long, at 36 km/h on parallel tracks in the same direction. How long will train A take to pass train B completely?
A. 32 seconds B. 7 seconds C. 16 seconds D. 21 seconds
32. An athlete runs two laps of a track. She runs the first lap at 3 km/h and the second lap at 15 km/h. What is her average speed for the two laps?
A. 12 km/h B. 9 km/h C. 6 km/h D. 5 km/h
33. Two trains 175 m and 320 m long are running towards each other on parallel tracks at 63 km/h and 18 km/h respectively. In how many seconds will they cross each other?
A. 7.78 seconds B. 39.6 seconds C. 22 seconds D. 28.29 seconds
34. A thief is spotted by a policeman from a distance of 1800 m. When the policeman starts the chase, the thief also starts running. The thief runs at 20 km/h and the policeman at 38 km/h. In how many minutes will the policeman catch the thief?
A. 8 minutes B. 6 minutes C. 3 minutes D. 2 minutes
35. A bus covers the first one-third of a journey at 48 km/h, the next one-third at 60 km/h and the last one-third at 40 km/h. What is the average speed for the whole journey?
A. 45 km/h B. 33 km/h C. 48 km/h D. 51 km/h
36. A train crosses a pole in 18 seconds and a 120 m long tunnel in 30 seconds. What is the speed of the train in km/h?
A. 36 km/h B. 14.4 km/h C. 10 km/h D. 90 km/h
37. A train crosses a platform 270 m long in 17 seconds and a tunnel 420 m long in 22 seconds at the same uniform speed. What is the length of the train?
A. 190 m B. 290 m C. 240 m D. 150 m
38. A train 200 m long crosses a man walking at 3 km/h in the same direction in 20 seconds. What is the speed of the train?
A. 10 km/h B. 39 km/h C. 33 km/h D. 36 km/h
39. Two trains of equal length are running on parallel tracks in opposite directions at 81 km/h and 54 km/h. They cross each other in 20 seconds. What is the length of each train?
A. 375 m B. 75 m C. 450 m D. 750 m
40. An athlete runs at 12 km/h. How many seconds will he take to go once round a rectangular park 80 m long and 60 m wide?
A. 6 seconds B. 84 seconds C. 42 seconds D. 23 seconds
41. Buses leave a depot every 30 minutes and run at 36 km/h. Vikas is walking away from the depot along the same road and is overtaken by a bus every 40 minutes. What is Vikas's walking speed?
A. 9 km/h B. 48 km/h C. 10 km/h D. 12 km/h
42. A cyclist wants to cover the same distance in 75% less time than usual. By what per cent must the speed be increased?
A. 75% B. 37½% C. 42 6/7% D. 300%
43. Asha walks from A to B at 200 m/min and hands a parcel to Sunita, who immediately runs from B to C at 2.5 times that speed. The total distance from A to C is 5 km and the parcel takes 16 minutes in all. Find the distance AB.
A. 2500 m B. 2000 m C. 1429 m D. 3000 m
44. Two cyclists start together from the same point on a circular track of length 600 m at 15 m/s and 5 m/s in the same direction. After how many seconds will they meet again at the starting point for the first time?
A. 4800 seconds B. 120 seconds C. 40 seconds D. 60 seconds
45. A train passes two persons walking in the same direction as the train at 3 km/h and 12 km/h in 24 seconds and 33 seconds respectively. What is the length of the train?
A. 270 m B. 220 m C. 160 m D. 240 m
46. A train passes two persons walking in the same direction as the train at 4 km/h and 9 km/h in 27 seconds and 32 seconds respectively. What is the length of the train?
A. 240 m B. 270 m C. 190 m D. 290 m
47. A train passes two persons walking in the same direction as the train at 3 km/h and 10 km/h in 15 seconds and 18 seconds respectively. What is the speed of the train?
A. 48 km/h B. 45 km/h C. 42 km/h D. 75 km/h
48. A train passes two persons walking in the same direction as the train at 6 km/h and 9 km/h in 24 seconds and 25 seconds respectively. What is the speed of the train?
A. 369 km/h B. 81 km/h C. 75 km/h D. 87 km/h
49. A train crosses a platform 350 m long in 18 seconds and a bridge 500 m long in 24 seconds at the same uniform speed. What is the speed of the train in km/h?
A. 75 km/h B. 108 km/h C. 90 km/h D. 25 km/h
50. Stations P and Q are 350 km apart. A train leaves P for Q at 6:00 a.m. at 36 km/h. Another train leaves Q for P at 7:30 a.m. at 75 km/h. At what time will the two trains meet?
A. 10:10 a.m. B. 9:09 a.m. C. 10:39 a.m. D. 10:40 a.m.
Chapter 13 — Answers and explanations
After checking the key, re-solve any miss without looking at the formula.
1. B. 1 m/s = 3600 m per hour = 3.6 km/h, i.e. multiply by 18/5. 7.5 × 18/5 = 27 km/h. (Multiplying by 5/18 gives 2.08, the reverse conversion.)
APAR26-13-03 | m/s to km/h | Easy
2. B. Equal distances at speeds u and v ⇒ average speed = 2uv/(u + v) = 2 × 6 × 12/(6 + 12) = 144/18 = 8 km/h. (The arithmetic mean 9 km/h is the standard trap — the slower leg takes longer, so the average is pulled below the mean.)
APAR26-13-02 | Athlete — two laps at two speeds | Easy
3. B. Relative speed (opposite directions) = 25 + 18 = 43 m/s. In 8 s the trains together cover 43 × 8 = 344 m, which is exactly the sum of the two lengths. (Using the difference of speeds gives 56 m.)
APAR26-13-28 | Combined length from crossing time | Easy
4. A. Average speed = total distance ÷ total time. Times: 54/54 = 1 h, 96/48 = 2 h; total time = 3 h, total distance = 150 km. Average = 150/3 = 50 km/h. (The simple mean of the speeds, 51 km/h, is the trap.)
APAR26-13-01 | Average speed over 2 distance segments | Easy
5. D. Relative speed (opposite directions) = 15 + 30 = 45 m/s. In 19 s the trains together cover 45 × 19 = 855 m, which is exactly the sum of the two lengths. (Using the difference of speeds gives 285 m.)
APAR26-13-32 | Combined length from crossing time | Easy
6. B. Opposite directions ⇒ relative speed = 72 + 54 = 126 km/h = 126 × 5/18 = 35 m/s. Distance = sum of lengths = 175 + 175 = 350 m. Time = 350/35 = 10 seconds. (Subtracting the speeds gives 70 s — that is for the same direction.)
APAR26-13-27 | Trains crossing in opposite directions | Easy
7. B. To cross a pole the train covers its own length. Speed = 120/12 = 10 m/s = 10 × 18/5 = 36 km/h. (Leaving the answer as 10 is m/s, not km/h.)
APAR26-13-30 | Speed from crossing a pole | Easy
8. B. Opposite directions ⇒ relative speed = 18 + 54 = 72 km/h = 72 × 5/18 = 20 m/s. Distance = sum of lengths = 220 + 220 = 440 m. Time = 440/20 = 22 seconds. (Subtracting the speeds gives 44 s — that is for the same direction.)
APAR26-13-29 | Trains crossing in opposite directions | Easy
9. C. Equal distances at speeds u and v ⇒ average speed = 2uv/(u + v) = 2 × 6 × 18/(6 + 18) = 216/24 = 9 km/h. (The arithmetic mean 12 km/h is the standard trap — the slower leg takes longer, so the average is pulled below the mean.)
APAR26-13-05 | Athlete — two laps at two speeds | Easy
10. B. Equal distances at speeds u and v ⇒ average speed = 2uv/(u + v) = 2 × 17 × 51/(17 + 51) = 1734/68 = 25.5 km/h. (The arithmetic mean 34 km/h is the standard trap — the slower leg takes longer, so the average is pulled below the mean.)
APAR26-13-06 | To-and-fro at two speeds | Easy
11. A. Distance = 64 × 10 = 640 km. Required speed = 640/5 = 128 km/h. (For a fixed distance speed ∝ 1/time: s2 = s1 × t1/t2 = 64 × 10/5; using 64 × 5/10 = 32 goes the wrong way.)
APAR26-13-04 | Speed for a new time over the same distance | Easy
12. B. Distance = train + platform = 300 + 500 = 800 m. Speed = 72 × 5/18 = 20 m/s. Time = 800/20 = 40 seconds. (Using only the platform length gives 25 s — the train's own length must be added.)
APAR26-13-26 | Time to cross a platform | Easy
13. A. Opposite directions ⇒ relative speed = 18 + 72 = 90 km/h = 90 × 5/18 = 25 m/s. Distance = sum of lengths = 200 + 200 = 400 m. Time = 400/25 = 16 seconds. (Subtracting the speeds gives 26.67 s — that is for the same direction.)
APAR26-13-31 | Trains crossing in opposite directions | Easy
14. D. 1 m/s = 3600 m per hour = 3.6 km/h, i.e. multiply by 18/5. 22.5 × 18/5 = 81 km/h. (Multiplying by 5/18 gives 6.25, the reverse conversion.)
APAR26-13-07 | m/s to km/h | Easy
15. B. Speed = 54 × 5/18 = 15 m/s. Distance covered in 50 s = 15 × 50 = 750 m = train + bridge. So bridge = 750 − 150 = 600 m. (Forgetting to subtract gives 750 m.)
APAR26-13-35 | Platform length from crossing time | Medium
16. C. Same direction ⇒ relative speed = 81 − 27 = 54 km/h = 15 m/s. Distance = 320 + 220 = 540 m. Time = 540/15 = 36 seconds. (Adding the speeds gives 18 s, which is for opposite directions.)
APAR26-13-44 | Faster train overtaking a slower one | Medium
17. B. Average speed = total distance ÷ total time. Times: 60/60 = 1 h, 144/36 = 4 h, 300/100 = 3 h; total time = 8 h, total distance = 504 km. Average = 504/8 = 63 km/h. (The simple mean of the speeds, 65.33 km/h, is the trap.)
APAR26-13-14 | Average speed over 3 distance segments | Medium
18. C. Let length L m and speed v m/s. Pole: L = 14v; bridge: L + 420 = 42v. Subtracting, 420 = 28v ⇒ v = 15 m/s = 54 km/h, and L = 14 × 15 = 210 m. Answer: 54 km/h. (The extra 28 s is the time to cover the bridge alone — the key step.)
APAR26-13-33 | Train speed from pole and platform times | Medium
19. B. Opposite directions ⇒ relative speed = 15 + 10 = 25 m/s, and together they cover one lap (900 m) between meetings. First meeting after 900/25 = 36 s. (Using a − b gives 180 s.)
APAR26-13-17 | First meeting on a circular track | Medium
20. D. Relative speed = 63 − 27 = 36 km/h = 36 × 5/18 = 10 m/s. In 20 s the distance covered = 10 × 20 = 200 m = 2L (both lengths). So L = 200/2 = 100 m. (Forgetting the factor 2 gives 200 m.)
APAR26-13-40 | Length of equal trains from passing time | Medium
21. C. Average speed = total distance ÷ total time. Times: 270/90 = 3 h, 400/100 = 4 h, 30/30 = 1 h; total time = 8 h, total distance = 700 km. Average = 700/8 = 87.5 km/h. (The simple mean of the speeds, 73.33 km/h, is the trap.)
APAR26-13-08 | Average speed over 3 distance segments | Medium
22. D. Speed = 36 × 5/18 = 10 m/s. Distance covered in 33 s = 10 × 33 = 330 m = train + platform. So platform = 330 − 180 = 150 m. (Forgetting to subtract gives 330 m.)
APAR26-13-34 | Platform length from crossing time | Medium
23. D. Perimeter = 2 × (120 + 20) = 280 m. Speed = 12 km/h = 12 × 5/18 = 3.33 m/s. Time = 280/3.33 = 84 seconds. (Forgetting to convert km/h to m/s gives 23.33 — meaningless.)
APAR26-13-19 | Time to go round a rectangular field | Medium
24. C. For a fixed distance, speed × time is constant. New speed = 60% = 60/100 of the old ⇒ new time = 100/60 of the old. Change = |100/60 − 1| × 100% = 40/60 × 100% = 66⅔% increase. (Answering 40% assumes the change is symmetric — it is not.)
APAR26-13-10 | Percentage change in time from change in speed | Medium
25. C. Let length L m, speed v m/s. L + 150 = 24v and L + 300 = 30v. Subtracting: 150 = 6v ⇒ v = 25 m/s = 90 km/h. Then L = 24 × 25 − 150 = 450 m. Answer: 450 m. (The extra 6 s covers exactly the extra 150 m — that is the key.)
APAR26-13-42 | Train length from two crossings | Medium
26. B. Let each part be d km. Total time = d/60 + d/36 + d/15; average = 3d ÷ (d/60 + d/36 + d/15) = 3/(1/60 + 1/36 + 1/15) = 27 km/h (harmonic mean of the three speeds). The simple mean 37 km/h is wrong.
APAR26-13-09 | Three equal parts at three speeds | Medium
27. A. Total distance = average × total time = 51 × (3 + 2) = 255 km. First leg = 45 × 3 = 135 km, so the second leg = 255 − 135 = 120 km in 2 h ⇒ p = 120/2 = 60 km/h. (Guessing p = 2 × 51 − 45 = 57 works only when the two times are equal.)
APAR26-13-11 | Unknown leg speed from overall average | Medium
28. D. Let length L m and speed v m/s. Pole: L = 10v; bridge: L + 600 = 60v. Subtracting, 600 = 50v ⇒ v = 12 m/s = 43.2 km/h, and L = 10 × 12 = 120 m. Answer: 43.2 km/h. (The extra 50 s is the time to cover the bridge alone — the key step.)
APAR26-13-39 | Train speed from pole and platform times | Medium
29. A. Average speed = total distance ÷ total time. Times: 20/20 = 1 h, 120/40 = 3 h, 480/120 = 4 h; total time = 8 h, total distance = 620 km. Average = 620/8 = 77.5 km/h. (The simple mean of the speeds, 60 km/h, is the trap.)
APAR26-13-20 | Average speed over 3 distance segments | Medium
30. B. Perimeter = 2 × (40 + 25) = 130 m. Speed = 12 km/h = 12 × 5/18 = 3.33 m/s. Time = 130/3.33 = 39 seconds. (Forgetting to convert km/h to m/s gives 10.83 — meaningless.)
APAR26-13-16 | Time to go round a rectangular field | Medium
31. A. Same direction ⇒ relative speed = 108 − 36 = 72 km/h = 20 m/s. Distance = 140 + 500 = 640 m. Time = 640/20 = 32 seconds. (Adding the speeds gives 16 s, which is for opposite directions.)
APAR26-13-36 | Faster train overtaking a slower one | Medium
32. D. Equal distances at speeds u and v ⇒ average speed = 2uv/(u + v) = 2 × 3 × 15/(3 + 15) = 90/18 = 5 km/h. (The arithmetic mean 9 km/h is the standard trap — the slower leg takes longer, so the average is pulled below the mean.)
APAR26-13-18 | Athlete — two laps at two speeds | Medium
33. C. Opposite directions ⇒ relative speed = 63 + 18 = 81 km/h = 81 × 5/18 = 22.5 m/s. Distance = sum of lengths = 175 + 320 = 495 m. Time = 495/22.5 = 22 seconds. (Subtracting the speeds gives 39.6 s — that is for the same direction.)
APAR26-13-41 | Trains crossing in opposite directions | Medium
34. B. Relative speed = 38 − 20 = 18 km/h = 18 × 5/18 = 5 m/s. Time to close the 1800 m gap = 1800/5 = 360 s = 6 minutes. (Using the policeman's own speed instead of the relative speed is the trap.)
APAR26-13-12 | Time to catch a thief | Medium
35. C. Let each part be d km. Total time = d/48 + d/60 + d/40; average = 3d ÷ (d/48 + d/60 + d/40) = 3/(1/48 + 1/60 + 1/40) = 48 km/h (harmonic mean of the three speeds). The simple mean 49.33 km/h is wrong.
APAR26-13-15 | Three equal parts at three speeds | Medium
36. A. Let length L m and speed v m/s. Pole: L = 18v; tunnel: L + 120 = 30v. Subtracting, 120 = 12v ⇒ v = 10 m/s = 36 km/h, and L = 18 × 10 = 180 m. Answer: 36 km/h. (The extra 12 s is the time to cover the tunnel alone — the key step.)
APAR26-13-38 | Train speed from pole and platform times | Medium
37. C. Let length L m, speed v m/s. L + 270 = 17v and L + 420 = 22v. Subtracting: 150 = 5v ⇒ v = 30 m/s = 108 km/h. Then L = 17 × 30 − 270 = 240 m. Answer: 240 m. (The extra 5 s covers exactly the extra 150 m — that is the key.)
APAR26-13-45 | Train length from two crossings | Medium
38. B. Relative speed = 200/20 = 10 m/s = 10 × 18/5 = 36 km/h. Same direction ⇒ relative speed = train − man, so train = 36 + 3 = 39 km/h. (Subtracting the man's speed instead gives 33 km/h.)
APAR26-13-37 | Train overtaking a man walking the same way | Medium
39. A. Relative speed = 81 + 54 = 135 km/h = 135 × 5/18 = 37.5 m/s. In 20 s the distance covered = 37.5 × 20 = 750 m = 2L (both lengths). So L = 750/2 = 375 m. (Forgetting the factor 2 gives 750 m.)
APAR26-13-43 | Length of equal trains from passing time | Medium
40. B. Perimeter = 2 × (80 + 60) = 280 m. Speed = 12 km/h = 12 × 5/18 = 3.33 m/s. Time = 280/3.33 = 84 seconds. (Forgetting to convert km/h to m/s gives 23.33 — meaningless.)
APAR26-13-13 | Time to go round a rectangular field | Medium
41. A. Consecutive buses are 36 × 30/60 = 18 km apart. Walking away, a bus gains the gap at (36 − w) km/h in 40 min: (36 − w) × 40/60 = 18 ⇒ 36 − w = 27 ⇒ w = 9 km/h. (Key relation: bus speed × T = relative speed × t.)
APAR26-13-23 | Walker meeting buses at fixed intervals | Difficult
42. D. For a fixed distance, speed × time is constant. New time = 25% = 25/100 of the old ⇒ new speed = 100/25 of the old. Change = |100/25 − 1| × 100% = 75/25 × 100% = 300% increase. (Answering 75% assumes the change is symmetric — it is not.)
APAR26-13-25 | Percentage change in time from change in speed | Difficult
43. B. Let AB = x m; then BC = 5000 − x. Second speed = 2.5 × 200 = 500 m/min. x/200 + (5000 − x)/500 = 16. Multiply by 500: 2.5x + 5000 − x = 8000 ⇒ 1.5x = 3000 ⇒ x = 2000 m. (Check: 2000/200 + 3000/500 = 10 + 6 = 16 min.)
APAR26-13-24 | Two travellers in sequence with speeds in ratio | Difficult
44. B. Each returns to the start after every lap: lap times are 600/15 = 40 s and 600/5 = 120 s. Both are at the start together after LCM(40, 120) = 120 seconds — direction does not matter. (60 s is when they first meet ANYWHERE on the track, not at the start.)
APAR26-13-22 | Meeting again at the starting point | Difficult
45. B. Let train speed v km/h, length L m. L = (v − 3) × 5/18 × 24 = (v − 12) × 5/18 × 33. So 24(v − 3) = 33(v − 12) ⇒ 9v = 324 ⇒ v = 36 km/h. Then L = (36 − 3) × 24 × 5/18 = 792 × 5/18 = 220 m. Answer: 220 m.
APAR26-13-49 | Train length from passing two walkers | Difficult
46. A. Let train speed v km/h, length L m. L = (v − 4) × 5/18 × 27 = (v − 9) × 5/18 × 32. So 27(v − 4) = 32(v − 9) ⇒ 5v = 180 ⇒ v = 36 km/h. Then L = (36 − 4) × 27 × 5/18 = 864 × 5/18 = 240 m. Answer: 240 m.
APAR26-13-46 | Train length from passing two walkers | Difficult
47. B. Let train speed v km/h, length L m. L = (v − 3) × 5/18 × 15 = (v − 10) × 5/18 × 18. So 15(v − 3) = 18(v − 10) ⇒ 3v = 135 ⇒ v = 45 km/h. Then L = (45 − 3) × 15 × 5/18 = 630 × 5/18 = 175 m. Answer: 45 km/h.
APAR26-13-50 | Train speed from passing two walkers | Difficult
48. B. Let train speed v km/h, length L m. L = (v − 6) × 5/18 × 24 = (v − 9) × 5/18 × 25. So 24(v − 6) = 25(v − 9) ⇒ 1v = 81 ⇒ v = 81 km/h. Then L = (81 − 6) × 24 × 5/18 = 1800 × 5/18 = 500 m. Answer: 81 km/h.
APAR26-13-47 | Train speed from passing two walkers | Difficult
49. C. Let length L m, speed v m/s. L + 350 = 18v and L + 500 = 24v. Subtracting: 150 = 6v ⇒ v = 25 m/s = 90 km/h. Then L = 18 × 25 − 350 = 100 m. Answer: 90 km/h. (The extra 6 s covers exactly the extra 150 m — that is the key.)
APAR26-13-48 | Train speed from two crossings | Difficult
50. A. By 7:30 a.m. the first train has covered 36 × 1.5 = 54 km; gap left = 350 − 54 = 296 km. Closing speed = 36 + 75 = 111 km/h, so they meet 296/111 h = 2 hours 40 minutes after 7:30 a.m., i.e. at 10:10 a.m. (dividing the full 350 km by 111 ignores the head start).
APAR26-13-21 | Meeting time with different start times | Difficult