All 27 Time Speed & Distance questions the Punjab PSC asked between 2015 and 2026, each with a worked answer and the exam it came from. First 10 free to read.
3 of these 27 were set in more than one exam — the Commission reuses this topic, which is the argument for working through it rather than skimming it.
Free to read, with the worked answer and the exam each one came from. The book prints this topic in the order the Commission favours, so these are the ones to know first.
Still-water speed is the average of the upstream and downstream speeds. > Upstream speed = 4 ÷ 2 = 2 km/h > Downstream speed = 3 ÷ 1/2 = 6 km/h > Still-water speed = (2 + 6) ÷ 2 = 4 km/h > Time for 9 km = 9 ÷ 4 = 9/4 h = 2 hours 15 minutes (The stream's speed would be (6 − 2) ÷ 2 = 2 km/h.)
For equal distances covered at two speeds the average speed is the harmonic mean 2xy/(x + y), not the simple mean. > Average = 2 × 60 × 45 / (60 + 45) = 5400 / 105 > = 51.43 km/h So about 51.4 km/h. The simple mean 52.5 km/h is a common trap because the train spends longer on the slower leg.
Riding both ways saves 2 hours, so walking one way takes 2 hours more than riding one way. > Walk + Ride = 5 h 45 min > Walk = Ride + 2 h > Walk + Walk = 5 h 45 min + 2 h = 7 h 45 min Walking both ways takes 7 hours 45 minutes.
Speed = 60 km/h = 60 × 5/18 = 50/3 m/s. Crossing a pole, the train covers its own length: L = (50/3) × 30 = 500 m.
For the same distance, time is inversely proportional to speed, so the times are in the ratio of the reciprocals of the speeds. > Times = 1/2 : 1/3 : 1/4 > Multiply by 12: 6 : 4 : 3 Option (b), 4 : 3 : 2, is just the reversed speed ratio, a common trap.
Let the speed be s and the time t, so the distance is st. > st = (s + 216)(t − 1) gives s = 216(t − 1) > st = (s − 108)(t + 1) gives s = 108(t + 1) > 216(t − 1) = 108(t + 1) → t = 3, so s = 432 > Distance = 432 × 3 The plane covered 1296 km.
Convert the speed to metres per minute and divide. > 20 km/h = 20000 ÷ 60 = 333 1/3 m/min > Time = 1500 ÷ 333 1/3 = 4.5 min The man takes 4 1/2 minutes (4 minutes 30 seconds).
Let the trip last t hours. > Distance = 40 × 6 + 60 × (t − 6) = 60t − 120 > Average speed = (60t − 120)/t = 55 → 60t − 120 = 55t → 5t = 120 → t = 24 The whole trip takes 24 hours.
Let x km be covered at 40 km/h and (560 − x) km at 160 km/h; the two times add to 9.5 h. > x/40 + (560 − x)/160 = 9.5 > 4x + 560 − x = 1520 → 3x = 960 → x = 320 > Distance at 160 km/h = 560 − 320 = 240 km
B catches A when 4(t − 1) = 3t, that is t = 4 hours after 1 o'clock. So they meet at 5 o'clock, 12 km from Poona. C has been walking since 3 o'clock at 5 km/h, so C has covered 10 km. A and C are then 2 km apart. They walk towards each other at 3 + 5 = 8 km/h. > B catches A: 3t = 4(t − 1) → t = 4 h → 5:00, 12 km out > C at 5:00 = 5 × 2 = 10 km → gap = 2 km > time to meet = 2/8 h = 15 min → 5:15
You have read the ten the Commission asks most. The rest of this topic — and all 4,239 questions across the book, each with a worked solution — come with either edition.
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536 pages, delivered. Five full papers and five OMR sheets at the back. Online access is included free, from the moment you pay.
See the printed book →The Commission set 27 Time Speed & Distance questions across 38 papers between 2015–2026. That is the whole set, not a sample — every one of them is in the book, and the 10 most-asked are on this page free.
Yes. 3 of these 27 were set in more than one exam, which is why they are worth working rather than reading. Each question here carries a badge showing how many sittings asked it.
Here. Every question on this page is a solved MCQ: the four options as printed, the correct answer, and a worked explanation in plain English — not just a letter. The rest of the 27 are in the book, in the same form.
Switch to Attempt mode at the top of the questions. You get one question at a time and it is marked the moment you pick an option, with the reason underneath — the same way the exam feels, on your phone.
The questions are on screen rather than in a downloadable file, so they stay current and correct. The printed book has all 27 of these plus the rest of the Reasoning & Aptitude section, and every buyer gets the whole thing online as well.
The exams behind these questions, most first. A paper here is a sitting: a question set in two papers of the same sitting counts once.
| Exam | Year | Questions |
|---|---|---|
| Analysts | 2022 | 2 |
| Technical Assistants | 2019 | 2 |
| Superintendent Home | 2019 | 2 |
| Legal Assistant | 2019 | 2 |
| Research Officer | 2019 | 2 |
| Network Engineer | 2019 | 2 |
| Punjab Civil Services (Prelims) — CSAT | 2015 | 2 |
| PPSC DDSWO | 2026 | 1 |
| Child Development Project Officer (Group-B) | 2026 | 1 |
| Conservator (Group-B) | 2025 | 1 |
| Punjab Civil Services (Prelims) — CSAT | 2025 | 1 |
| Assistant Archaeological Engineer (Group-A) | 2025 | 1 |