1. Basics
Speed = distance/time. km/h × 5/18 = m/s; m/s × 18/5 = km/h. Same distance: speed ∝ 1/time.
2. Key results
| Case | Result |
|---|---|
| Average speed, equal distances at x and y | 2xy/(x + y) |
| Relative speed — same direction | x − y |
| Opposite direction | x + y |
| Late by t₁ at speed x, early by t₂ at y | D = xy(t₁ + t₂)/(y − x) |
| Speed reduced to a/b of usual, late by t | Usual time = a × t/(b − a) |
| Two people meet after starting together from A and B; times after meeting t₁, t₂ | Speed ratio = √t₂ : √t₁ |
| Stoppage per hour | (fast − slow)/fast × 60 min |
3. Solved examples
72 km/h = 72 × 5/18 = 20 m/s.
Goes at 40, returns at 60 → average 48 km/h.
Walking at 3/4 of usual speed, a man is 20 min late → usual time = 3 × 20 = 60 min.
Bus speed without stops 60, with stops 45 → stops 15/60 × 60 = 15 min per hour.
Thief starts; police follow 1 h later at 60 vs 40 km/h → gap 40 km closes at 20 km/h → 2 h.
(ASO 2021) Speeds 7 : 8; second runs 400 km in 4 h → first 87.5 km/h.
Solved examples by question type
Each type that comes in the ASO paper, worked step by step. Try it yourself first, then read the solution.
Basic level
Example 1. A speed of 72 km/h is equal to:
(a) 25 m/s (b) 259.2 m/s (c) 40 m/s (d) 20 m/s
Answer: (d) 20 m/s
Solution: km/h × 5/18 = m/s. 72 × 5/18 = 20 m/s.
Example 2. A speed of 72 km/h is equal to:
(a) 259.2 m/s (b) 20 m/s (c) 40 m/s (d) 25 m/s
Answer: (b) 20 m/s
Solution: km/h × 5/18 = m/s. 72 × 5/18 = 20 m/s.
Example 3. A speed of 10 m/s is equal to:
(a) 18 km/h (b) 36 km/h (c) 46 km/h (d) 2.78 km/h
Answer: (b) 36 km/h
Solution: m/s × 18/5 = km/h. 10 × 18/5 = 36 km/h.
Example 4. A bus moves at 72 km/h. How far does it go in 150 minutes?
(a) 180 km (b) 90 km (c) 360 km (d) 185 km
Answer: (a) 180 km
Solution: Time = 150/60 h. Distance = 72 × 150/60 = 180 km.
Example 5. A man goes to a place at 50 km/h and comes back by the same road at 75 km/h. His average speed for the whole journey is:
(a) 63 km/h (b) 62.5 km/h (c) 60 km/h (d) 58 km/h
Answer: (c) 60 km/h
Solution: For equal distances, average speed = 2xy/(x + y) = 2×50×75/125 = 60 km/h. Not (50+75)/2.
Example 6. Walking at 3/4 of his usual speed, a man reaches his office late. If his usual time is 40 minutes, how late is he?
(a) 18.33 min (b) 13.33 min (c) 10 min (d) 30 min
Answer: (b) 13.33 min
Solution: Speed × 3/4 → time × 4/3 = 53.33 min. Late by 53.33 − 40 = 13.33 min.
ASO exam level
Example 7. Walking at 3 km/h a student reaches school 12 minutes late; at 4 km/h he reaches 8 minutes early. The distance to the school is:
(a) 0.85 km (b) 8 km (c) 4 km (d) 5 km
Answer: (c) 4 km
Solution: Time difference = 20 min = 0.33 h. d/3 − d/4 = 0.33 → d = 3×4×0.33/(4−3) = 4 km.
Example 8. Walking at 3 km/h a student reaches school 8 minutes late; at 4 km/h he reaches 7 minutes early. The distance to the school is:
(a) 6 km (b) 4 km (c) 0.25 km (d) 3 km
Answer: (d) 3 km
Solution: Time difference = 15 min = 0.25 h. d/3 − d/4 = 0.25 → d = 3×4×0.25/(4−3) = 3 km.
Example 9. A car covers 550 km at a constant speed. Had its speed been 5 km/h more, it would have taken 1 hour less. Its speed is:
(a) 50 km/h (b) 45 km/h (c) 60 km/h (d) 55 km/h
Answer: (a) 50 km/h
Solution: 550/v − 550/(v + 5) = 1. Checking options is fastest: 550/50 − 550/55 = 11 − 10 = 1. So v = 50 km/h.
Example 10. A car covers 1,560 km at a constant speed. Had its speed been 5 km/h more, it would have taken 2 hours less. Its speed is:
(a) 60 km/h (b) 65 km/h (c) 70 km/h (d) 55 km/h
Answer: (a) 60 km/h
Solution: 1,560/v − 1,560/(v + 5) = 2. Checking options is fastest: 1,560/60 − 1,560/65 = 26 − 24 = 2. So v = 60 km/h.
Example 11. A cyclist reaches point A at 2 PM if he rides at 12 km/h, and at 12 noon if he rides at 18 km/h. At what speed must he ride to reach A at 1 PM?
(a) 15 km/h (b) 13.4 km/h (c) 14.4 km/h (d) 15.4 km/h
Answer: (c) 14.4 km/h
Solution: Distance = 12×18×2/(18−12) = 72 km. At 12 km/h he needs 6 h; to arrive 1 h earlier he has 5 h. Speed = 72/5 = 14.4 km/h.
Example 12. Without stoppages a train runs at 48 km/h; with stoppages its average is 36 km/h. For how many minutes per hour does it stop?
(a) 15 min (b) 20 min (c) 12 min (d) 17 min
Answer: (a) 15 min
Solution: It loses 12 km of running per hour. Time = 12/48 h = 15 min.
Example 13. Two cars start at the same time from places 300 km apart and travel towards each other at 35 km/h and 45 km/h. After how long will they meet?
(a) 30 h (b) 4 h 15 min (c) 3 h 45 min (d) 7 h 30 min
Answer: (c) 3 h 45 min
Solution: Speeds add when moving towards each other: 35 + 45 = 80 km/h. Time = 300/80 = 3.75 h = 3 h 45 min.
Example 14. A thief runs at 6 km/h. A policeman starts chasing him 3 hours later at 8 km/h from the same point. How far from the start will the policeman catch him?
(a) 80 km (b) 18 km (c) 72 km (d) 54 km
Answer: (c) 72 km
Solution: Head start = 6×3 = 18 km. Gap closes at 2 km/h → time 9 h. Distance = 8 × 9 = 72 km.
Advanced level
Example 15. Two trains start at the same time from A and B towards each other. After meeting, they take 16 hours and 25 hours to reach B and A respectively. The ratio of their speeds is:
(a) 4 : 5 (b) 5 : 4 (c) 25 : 16 (d) 16 : 25
Answer: (b) 5 : 4
Solution: Speed ratio = √(time of second) : √(time of first) = √25 : √16 = 5 : 4.
Example 16. Two trains start at the same time from A and B towards each other. After meeting, they take 4 hours and 9 hours to reach B and A respectively. The ratio of their speeds is:
(a) 3 : 2 (b) 4 : 9 (c) 9 : 4 (d) 2 : 3
Answer: (a) 3 : 2
Solution: Speed ratio = √(time of second) : √(time of first) = √9 : √4 = 3 : 2.
Topic-wise practice for Speed, time and distance (Quant): Average speed, relative speed, arriving late or early. Useful for OPSC ASO, OSSC, OSSSC and other Odisha exams.
Question types covered
- Conversion between km/h and m/s
- Basic speed, time, distance
- Average speed of a journey
- Relative speed: same and opposite direction
- Arriving late or early by changing speed
- Ratio of speeds and times
- Man on a bicycle or walking problems