Mathematics — Chapter 3

Trains, Time & Distance — Relative Speed and Platform Problems

Why this chapter matters: Train questions are the most formula-stable block in one-paper arithmetic: every question is the same three quantities — speed, distance, time — plus two special rules (a point object takes only the train’s length; a platform, bridge or tunnel takes train length + object length) and one relative-speed rule (subtract for the same direction, add for the opposite direction). Our verified banks hold 44 topic-relevant items for this chapter: set 135’s 23 questions plus 21 tightly matched paper items. The set is the anchor — it contains every shape the papers ask, and one of its questions (the 360 m bridge at 45 km/h) reappears verbatim in FPSC papers of 2016 and 2017.

1. The three formulas — and the one conversion that unlocks everything

Speed, distance and time are one triangle:

• Speed = distance ÷ time · Distance = speed × time · Time = distance ÷ speed

The papers mix km/h with metres and seconds, so convert before calculating:

• km/h → m/s: multiply by 5/18 · m/s → km/h: multiply by 18/5

Why 5/18: 1 km = 1,000 m and 1 hour = 3,600 s, so 1 km/h = 1,000/3,600 = 5/18 m/s. Learn the two multipliers, not the fraction arithmetic — every set question uses them.

• Q1: 108 km/h × 5/18 = 30 m/s.
• Q2: 14 m/s × 18/5 = 50.4 km/h.
• Q7 (reverse use): 132 m in 6 s = 22 m/s = 22 × 18/5 = 79.2 km/h.

Conversion shortcut for the common speeds: 36 km/h = 10 m/s, 54 km/h = 15 m/s, 72 km/h = 20 m/s, 90 km/h = 25 m/s, 108 km/h = 30 m/s, 144 km/h = 40 m/s. Spot these and skip the multiplication: Q3’s 144 km/h is 40 m/s on sight.

2. Crossing a point object — pole, tree, man, signal, km stone

A pole, tree, standing man, signal or km stone has no length of its own, so the distance the train must cover is just the train’s own length:

• Time = train length ÷ speed · Train length = speed × time

• Q3: 100 m train at 144 km/h (40 m/s) → 100 ÷ 40 = 2.5 seconds.

• Q4: 280 m train at 63 km/h (17.5 m/s) → 280 ÷ 17.5 = 16 seconds.

• Q8 (two-step): 12 km in 10 minutes = 72 km/h = 20 m/s; length = 20 × 6 = 120 m. Convert the speed first, then use the pole time.

• Paper item (FPSC 2016, 2017): 60 km/h = 50/3 m/s; length = (50/3) × 9 = 150 m. The same question in two papers — a verbatim repeater.

• Paper item (NTS 2012): 150 m train at 30 km/h (25/3 m/s) passes a standing man in 150 ÷ (25/3) = 18 seconds.

Trap: a “man standing on the platform” is a point object (Q14’s first time, 20 seconds, gives the train’s length directly); a “man walking” is not — see Section 5.

3. Crossing a long object — platform, bridge, tunnel, station

A platform, bridge, tunnel or station has length, so the distance the train must cover is train length + object length:

• Time = (train length + object length) ÷ speed

• Object length = speed × time − train length

• Train length = speed × time − object length

• Q5: (110 + 132) ÷ 20 = 242 ÷ 20 = 12.1 seconds (72 km/h = 20 m/s).

• Q6: (360 + 140) ÷ 12.5 = 500 ÷ 12.5 = 40 seconds (45 km/h = 12.5 m/s). This is the question FPSC repeated in 2016 and 2017 — learn it, don’t just solve it.

• Q9: speed from the pole first: 240 ÷ 24 = 10 m/s; then (240 + 650) ÷ 10 = 89 seconds.

• Q10 (find the bridge): 45 km/h = 12.5 m/s × 30 s = 375 m covered; bridge = 375 − 130 = 245 m.

• Q11 (find the tunnel): 78 km/h = 65/3 m/s × 60 s = 1,300 m; tunnel = 1,300 − 800 = 500 m.

• Q12 (find the train): 72 km/h = 20 m/s × 26 s = 520 m; train = 520 − 250 = 270 m.

• Q13 (find the speed): (150 + 300) ÷ 40.5 = 100/9 m/s = (100/9) × 18/5 = 40 km/h.

• Paper item (FPSC 2015): 100 m train at 48 km/h (40/3 m/s) crosses a bridge in 30 s → covered 400 m → bridge = 300 m.

• Paper item (FPSC 2004): 100 m train at 72 km/h (20 m/s) passes a bridge in 25 s → covered 500 m → bridge = 400 m.

• Paper item (mental-ability set): 60 km/h for 2 minutes covers 2,000 m; tunnel 1,500 m → train = 500 m.

The one rule behind all of these: the train has “completely crossed” only when its tail clears the far end — that is why the lengths add. If your answer used only the train’s length for a bridge, you forgot the bridge.

4. Two times, one unknown — the difference method

When the same train crosses two objects (or the same object at two times), the speed comes from the difference of the two situations — the unknown length cancels:

• Speed = (longer object − shorter object) ÷ (longer time − shorter time)

• Train length = speed × shorter time − shorter object

• Q14: platform in 36 s, man in 20 s, speed 54 km/h = 15 m/s. Train = 15 × 20 = 300 m; platform = 15 × 36 − 300 = 240 m. Equivalently, platform = 15 × (36 − 20) = 240 m — the time difference (16 s) at the known speed is the platform itself.

• Q15: 300 m train, pole 18 s, platform 39 s. Speed = 300 ÷ 18 = 50/3 m/s; platform = (50/3) × (39 − 18) = (50/3) × 21 = 350 m. Same trick: platform = speed × time difference.

• Q16: pole 15 s, platform 100 m in 25 s. Let length = L: L/15 = (L + 100)/25 → 5L = 3L + 300 → L = 150 m. When speed is not given, write the speed two ways and equate them.

• Q17: station 162 m in 18 s, station 120 m in 15 s. Speed = (162 − 120) ÷ (18 − 15) = 42 ÷ 3 = 14 m/s; length = 14 × 15 − 120 = 90 m. The difference method in its purest form: the 42 m of extra station took the extra 3 seconds.

• Paper item (FPSC 2001, 2008): a 110 m train “runs through a station” at 36 km/h (10 m/s) and takes 110 ÷ 10 = 11 seconds to pass a given point — the station’s length is not in the question, so the point rule of Section 2 applies, not the platform rule.

5. Relative speed — same direction: subtract

When two bodies move in the same direction, their relative speed is the difference of their speeds. The faster one gains on the slower one at that difference, and the distance to be gained is the length(s) involved.

• Relative speed (same direction) = faster speed − slower speed

• Time to cross a moving person = train length ÷ relative speed (the person is a point, so only the train’s length counts)

• Time to overtake a jogger ahead = (gap + train length) ÷ relative speed (the train must cover the gap and its own length to pass completely)

• Q18: man walking at 3 km/h in the train’s direction, train 63 km/h. Relative speed = 60 km/h = 50/3 m/s; time = 500 ÷ (50/3) = 30 seconds.

• Q19: jogger at 9 km/h, 240 m ahead; train 120 m long at 45 km/h. Relative speed = 36 km/h = 10 m/s; distance to gain = 240 + 120 = 360 m; time = 360 ÷ 10 = 36 seconds. The gap is the part students forget: the train must first reach the jogger, then pass him.

• Paper item (FPSC 2015): two trains same direction, 40 and 22 km/h, pass each other in 1 minute. Relative speed = 18 km/h = 5 m/s × 60 s = 300 m combined length; first train 125 m → second = 175 m.

• Paper item (FPSC 2015): a man in the slower train (32 km/h) watches the faster train (50 km/h) pass him in 15 s. Relative speed = 18 km/h = 5 m/s; faster train’s length = 5 × 15 = 75 m. From inside a train, the other train’s length is what passes you.

• Paper item (NTS 2012): a train passes two men walking at 2 km/h and 4 km/h in 9 s and 10 s. With train speed S and length L: (S − 2) × 9 = (S − 4) × 10 in consistent units → S = 22 km/h, L = 50 m. Check: (22 − 2) km/h = 50/9 m/s × 9 = 50 m; (22 − 4) km/h = 5 m/s × 10 = 50 m. Two equations, two unknowns — the Q16 method with moving men.

6. Relative speed — opposite direction: add

When two trains move towards each other (opposite directions), their relative speed is the sum of their speeds, and the distance to be covered is the sum of their lengths:

• Relative speed (opposite direction) = speed₁ + speed₂

• Crossing time = (length₁ + length₂) ÷ (speed₁ + speed₂)

• Q20: 270 m train at 120 km/h meets a train at 80 km/h in 9 s. Relative speed = 200 km/h = 500/9 m/s; combined length = (500/9) × 9 = 500 m; other train = 500 − 270 = 230 m.

• Q21: equal 120 m trains, pole times 10 s and 15 s. Speeds = 12 m/s and 8 m/s; relative speed = 20 m/s; time = (120 + 120) ÷ 20 = 12 seconds. Find each speed from its pole time first, then add.

• Q22: 108 m at 50 km/h meets 112 m in 6 s. Combined length 220 m in 6 s → relative speed = 220/6 m/s = 132 km/h; second train = 132 − 50 = 82 km/h.

• Q23: 150 m train passes a km stone in 15 s → 10 m/s = 36 km/h. It meets an equal 150 m train in 8 s: combined 300 m in 8 s → relative speed 37.5 m/s = 135 km/h; second train = 135 − 36 = 99 km/h.

• Paper item (FPSC 2015): 90 m and 110 m trains at 54 and 90 km/h approach each other. Combined 200 m; relative speed 144 km/h = 40 m/s; time = 200 ÷ 40 = 5 seconds.

Same-direction vs opposite-direction, one line: same direction, the faster train gains slowly (subtract); opposite direction, they close fast (add). Every relative-speed error in the banks is this sign.

7. General time & distance — average speed and the catch-up

The same triangle serves non-train questions, and the papers ask two of them repeatedly:

7.1 Average speed is not the average of the speeds

For equal distances at two speeds, the average is the harmonic mean: 2ab ÷ (a + b) — because more time is spent at the slower speed.

• Paper item (FPSC 2016): Nagpur → Allahabad at 100 km/h, back at 150 km/h. Average = 2 × 100 × 150 ÷ (100 + 150) = 30,000 ÷ 250 = 120 km/h — not 125. If the distances were unequal, average = total distance ÷ total time; the harmonic mean is only the equal-distance shortcut.

7.2 The catch-up (later train, same direction)

A later, faster train catches the earlier one when it has made up the head start:

• Head start = earlier speed × time gap · Catch-up time = head start ÷ (faster speed − slower speed) · Meeting distance = faster speed × catch-up time

• Paper item (FPSC 2015): trains leave Rawalpindi at 8:30 AM (60 km/h) and 9:00 AM (70 km/h). Head start = 60 × 0.5 = 30 km; closing speed = 10 km/h; catch-up in 3 hours (at 12:00 noon); meeting distance = 70 × 3 = 210 km from Rawalpindi. Check: the first train has run 3.5 hours × 60 = 210 km — the same point, as it must be.

• Paper item (FPSC 2002): speeds in ratio 7 : 8; the second covers 400 km in 4 hours = 100 km/h; the first = 100 × 7/8 = 87.5 km/h. Ratio questions are the same triangle with one speed found from distance ÷ time first.

• Paper item (PPSC 2026): 200 km/h from 12:24 to 14:12 = 1 hour 48 minutes = 1.8 hours → distance = 200 × 1.8 = 360 km. Time arithmetic (elapsed time) first, then the triangle.

7A. General distance & speed — the OPC drill (non-train questions)

The same triangle serves questions with no train in them. The OPC book’s Distance & Speed block (pp. 68–70) adds seven worked shapes that are not train-crossing questions — all answers recomputed independently in this run:

• Length units first: 1 km = 100,000 cm, so 2 km = 200,000 cm. Papers that ask “how many centimetres” want the unit ladder, not the speed triangle.
• Street crossing: a person crosses a 600 m street in 5 minutes (300 s). Speed = 600 ÷ 300 = 2 m/s = 2 × 18/5 = 7.2 km/h. Minutes → seconds first, then the conversion.
• Elapsed time, then the triangle: a car leaves Lahore at 7:12 AM, covers 180 miles, arrives in Multan at 10:57 AM. Elapsed time = 3 hours 45 minutes = 3.75 hours. Average speed = 180 ÷ 3.75 = 48 mph. The units stay as asked (miles, mph) — do not convert miles to kilometres when the question asks mph.
• Plain division: a truck covers 360 km in 5 hours → 360 ÷ 5 = 72 km/h; a car travels 700 km in 3 hours 30 minutes (3.5 h) → 700 ÷ 3.5 = 200 km/h.
• Segment average (time-weighted, not harmonic): a ship averages 32 km/h for the first hour and 41 km/h for the next two hours. Total distance = 32 + (41 × 2) = 114 km; total time = 3 hours; average = 114 ÷ 3 = 38 km/h. Contrast with Section 7.1: the harmonic mean 2ab ÷ (a + b) is only for equal distances; when the question gives times per segment, average = total distance ÷ total time. Using the harmonic mean here (2 × 32 × 41 ÷ 73 = 35.9) would be wrong.
• Speed → distance in one second: at 30 km/h, one second covers 30 × 5/18 = 25/3 ≈ 8.3 metres. “In one second” means the speed itself, converted.
• Minutes as a fraction of an hour: at 40 km/h from 9:48 AM to 10:15 AM (27 minutes = 27/60 hours), distance = 40 × 27/60 = 18 km.

Two of the OPC block’s conversion examples are already this chapter’s Section 1 (108 km/h = 30 m/s is Q1; 14 m/s = 50.4 km/h is Q2) and are not repeated here. The train-crossing formulas of the OPC block (pole = L only; stationary object = L + b; same direction u − v; opposite direction u + v; crossing time = (a + b) ÷ relative speed) are Sections 2, 3, 5 and 6 of this chapter — the one OPC formula that is not already here is the post-crossing ratio, in Section 7B.

7B. After crossing: the square-root speed ratio (OPC formula corrected)

When two trains start from A and B towards each other, cross, and then take a seconds (train from A) and b seconds (train from B) to reach their destinations:

• Speed ratio (train from A : train from B) = √b : √a

Why the roots: at the crossing point, each train’s remaining distance is the other’s already-covered distance, so the speed ratio squared equals b/a — the ratio is the square root, not the raw times.

• Worked illustration: after crossing, the train from A takes 16 seconds to reach B and the train from B takes 9 seconds to reach A. Speed ratio = √9 : √16 = 3 : 4 (the train from A is slower — it needed longer after the crossing).

Trap — the OPC book prints this rule wrong. It gives “ratio of speeds = b : a”, which for the illustration above would say 9 : 16. That omits the square roots and is not taught here: always root both times before forming the ratio. No bank item in this chapter’s count tests this shape yet, so it is taught as method, with no bankCount change and no mark claimed.

8. Master table — the train rules at a glance

9. Rapid-fire one-liners

• 108 km/h = 30 m/s; 14 m/s = 50.4 km/h; 144 km/h = 40 m/s.
• Pole rule: distance = train length only. 100 m at 40 m/s → 2.5 s; 280 m at 17.5 m/s → 16 s.
• Platform rule: distance = train + object. (110 + 132) ÷ 20 = 12.1 s; (360 + 140) ÷ 12.5 = 40 s.
• Find the bridge: v × t − L. 12.5 × 30 − 130 = 245 m; 20 × 26 − 250 = train 270 m.
• Find the speed: (150 + 300) ÷ 40.5 = 100/9 m/s = 40 km/h.
• Two times: platform = v × Δt. 15 × (36 − 20) = 240 m; (50/3) × (39 − 18) = 350 m.
• Two stations: v = (162 − 120) ÷ (18 − 15) = 14 m/s; L = 14 × 15 − 120 = 90 m.
• Same direction: subtract. 63 − 3 = 60 km/h → 500 m in 30 s; jogger: (240 + 120) ÷ 10 = 36 s.
• Opposite direction: add. 120 + 80 = 200 km/h → other train 230 m; 12 + 8 = 20 m/s → 240 ÷ 20 = 12 s.
• Second train’s speed: relative − known. 132 − 50 = 82 km/h; 135 − 36 = 99 km/h.
• Opposite paper item: (90 + 110) ÷ 40 = 5 s.
• Average speed, equal distances: 2 × 100 × 150 ÷ 250 = 120 km/h, not 125.
• Segment average (times given): (32 + 82) ÷ 3 = 38 km/h — total distance ÷ total time, not the harmonic mean.
• Post-crossing ratio: 16 s and 9 s after crossing → speeds 3 : 4 (square roots first — OPC’s unrooted b : a is wrong).
• Length units: 2 km = 200,000 cm; 600 m in 5 min = 7.2 km/h; 180 miles in 3 h 45 min = 48 mph; 40 km/h for 27 min = 18 km.
• Catch-up: head start 30 km ÷ closing 10 km/h = 3 h → meeting 210 km out.
• Two walking men: train 22 km/h, length 50 m (both crossings give 50 m).

10. Most-asked facts (from our verified past-paper banks)

• Set 135 is the topic’s anchor: 23 questions — conversion 2, pole/point 3, platform/bridge/tunnel 6, two-time unknowns 4, same-direction relative 2, opposite-direction relative 4, plus the speed-from-distance items (Q8, Q13) that open the two families.
• The paper items (21) are led by platform/bridge/tunnel and pole questions (10) and general speed-distance items (6) — the papers test the two basic rules more than the two-train relative-speed shapes, so master Sections 2–3 before Sections 5–6.
• The verbatim repeaters: set Q6 (360 m bridge, 45 km/h, 40 s) appears in FPSC 2016 and 2017, and the 60 km/h pole question (150 m) appears in the same two papers. A set question that appears in two papers is the highest-yield item in the chapter.
• The 2015 FPSC welfare paper is a train cluster: bridge-length, opposite-direction crossing (5 s) and same-direction passing (175 m) in one paper — the three rules of Sections 3, 5 and 6 side by side.
• The two-men question (NTS 2012) is the hardest shape in the topic: two moving men give two equations for speed and length. Solve it by the Q16 method (equate the two expressions for length), not by guessing the speed.

11. Traps & confusions

• Point vs long object. Pole, tree, standing man, signal, km stone: train length only. Platform, bridge, tunnel, station: train + object. This single distinction decides Sections 2 vs 3 — and Q14 uses both in one question (man 20 s, platform 36 s).
• “Passes a station” is ambiguous — read the question. A train “passing a given point” at a station (paper item, 110 m at 36 km/h) is a point crossing (11 s); a train “passing through a station 162 m long” (Q17) is a long-object crossing. The station’s length, when given, is the tell.
• Same direction: subtract, opposite: add. Q18’s man (63 − 3 = 60) vs Q20’s trains (120 + 80 = 200). Adding in the same direction, or subtracting in the opposite direction, is the most common relative-speed error.
• The jogger’s gap. Q19’s 240 m gap must be added to the train’s 120 m: 360 m, not 240 m and not 120 m. “Pass the jogger” means the train’s tail has cleared him.
• Convert before calculating. A speed in km/h divided into metres gives nonsense. Q3 only works because 144 km/h became 40 m/s first. Write the converted speed on the paper before touching the lengths.
• Average speed is not (a + b) ÷ 2. 100 and 150 km/h over equal distances average 120 km/h, not 125 — the slower leg takes longer, so it weighs more.
• Segment average is a different rule, not the harmonic mean. When the question gives times per segment (32 km/h for 1 hour, 41 km/h for 2 hours), average = total distance ÷ total time = 38 km/h. The harmonic mean 2ab ÷ (a + b) applies only when the distances are equal. Mixing the two rules is the OPC block’s likeliest error.
• Post-crossing ratio needs square roots. OPC’s “b : a” without roots is wrong: 16 s and 9 s give 3 : 4, not 9 : 16. If your ratio looks like the raw times, you forgot the roots.
• Elapsed time before the triangle. Lahore 7:12 → Multan 10:57 is 3 hours 45 minutes = 3.75 hours; 180 ÷ 3.75 = 48 mph. A time-arithmetic error here is not a speed error.
• Keep the asked units. A question asking mph is answered in mph (48 mph, not a kilometre conversion); a question asking centimetres wants the unit ladder (2 km = 200,000 cm).
• “Find the platform” vs “find the train”. Both are v × t minus the known length — subtract the other length. Q10 subtracts the train (130 m) to find the bridge; Q12 subtracts the platform (250 m) to find the train. Subtracting the wrong one swaps the answer.
• The man inside the train. In the FPSC 2015 slower-train item, the man is a point on a moving train, so the distance is the faster train’s length and the speed is the relative speed (5 m/s × 15 s = 75 m) — not the faster train’s own speed.
• Clock arithmetic before the triangle. The PPSC 2026 item needs 12:24 → 14:12 = 1 hour 48 minutes = 1.8 hours before any multiplication. A 108-minute error here is a time error, not a speed error.

12. Memory tricks

• 5/18 down, 18/5 up: km/h → m/s uses the smaller fraction (5/18, speed number shrinks: 108 → 30); m/s → km/h uses the bigger one (18/5, number grows: 14 → 50.4). Down = shrink, up = grow.
• Point = train only; platform = train + platform. Say it before every question: “point or platform?” — the answer to that question is the distance.
• Same = subtract, opposite = add. Same direction, the gap closes slowly; opposite direction, they rush together. Slow = subtract, fast = add.
• Difference method: when two times are given, the difference of the times at the known speed is the difference of the objects — Q14’s platform is just 15 × 16 = 240 m.
• 36–54–72–90: the ladder 10–15–20–25 m/s. Recognise 36, 54, 72, 90 km/h on sight and the conversion step disappears.
• Harmonic, not arithmetic: equal distances, two speeds — “harmonic” sounds like “harmony”, and the average sits closer to the slower speed, in harmony with the longer time spent there.

13. Visuals (site build specs)

• DIAGRAM — Point vs platform: one train shown crossing a pole (only its own length marked) beside the same train crossing a bridge (train length + bridge length marked, tail clearing the far end) — placed at the head of Section 3, using Q3 (pole, 2.5 s) and Q5 (bridge, 12.1 s) as the pair.
• FLOWCHART — Which rule? “Is the object a point?” → yes: distance = L; no: distance = L + object. Then “Is anything else moving?” → same direction: subtract; opposite: add — placed in Section 1 as the chapter’s decision graphic, with Q18 and Q20 as the two relative-speed leaves.
• DIAGRAM — The jogger overtake: track diagram with the 240 m gap, the 120 m train and the 10 m/s closing speed, showing the 360 m total — placed in Section 5 beside Q19.
• DIAGRAM — Segment average vs harmonic mean: two bars for the ship example (32 km/h × 1 h + 41 km/h × 2 h = 114 km ÷ 3 h = 38 km/h) beside the Nagpur–Allahabad equal-distance case (120 km/h harmonic) — placed in Section 7A to stop the two averages being swapped.
• DIAGRAM — Post-crossing ratio: two trains meeting at a point, with post-crossing times 16 s and 9 s labelled and the √9 : √16 = 3 : 4 ratio shown — placed in Section 7B beside the corrected rule.

14. Practice this topic

• Practice set: 44 topic-relevant items in the verified banks serve this topic — set 135’s 23 problems-on-trains questions and 21 tightly matched paper items (platform/bridge/tunnel 5, pole/point 5, same-direction/relative 4, opposite-direction 1, speed-distance general 6). The site serves them as this chapter’s practice set; work set 135 first for every shape, then the papers’ platform and pole repeaters for speed. Vocabulary, analogy, grammar and general-knowledge items that merely contain the word “train” are excluded from this count and served by their own chapters. Sections 7A–7B (the OPC general-distance drill and the corrected post-crossing ratio) are teaching material: the bank script for this run (hidden_files/goldmine-notes-count-ma03-ma04-opc-dist-age.py, 199 files / 24,587 questions) found no new train-topic bank rows behind them, so bankCount stays 44; two general-speed rows (cyclist, planes) sit outside the train-scoped count and are not counted.
• Book mode: ← Previous: Problems on Numbers · Next: Problems on Ages →

Goldmine Notes · Mathematics · Chapter 3 · Sources: FPSC Edition 32, Arithmetic set 135 — Problems on Trains (23 questions); verified PPSC/FPSC past-paper banks (21 tightly matched paper items, script-counted — counting script hidden_files/goldmine-notes-count-ma03.py); speed–distance–time, relative-speed and platform-crossing methods as standard arithmetic. Standard answers only.