Why do we add the lengths of two trains even when they move in the same direction?

For the faster train to completely clear the slower train, its front engine must travel the entire length of the slower train plus its own length so its rear guard car clears the slower train's front. Thus, total distance is always L₁ + L₂.

When should I use the harmonic mean formula 2uv/(u+v) for average speed?

Use 2uv/(u+v) only when the two distances covered are equal (such as going from A to B at speed u and returning from B to A at speed v). If the time intervals are equal instead, average speed is the simple arithmetic mean (u + v) / 2.

How do you handle opposite vs. same direction signs in the Late/Early arrival formula?

When one case is Late and the other is Early, add the two time offsets (Δt = t₁ + t₂) because they lie on opposite sides of the scheduled arrival time. When both cases are Late (or both Early), subtract them (Δt = |t₁ − t₂|).

How many Speed, Time, and Distance questions appear in SSC CGL and RRB NTPC?

SSC CGL Tier-1 typically features 2 to 3 questions (4 to 6 marks) covering relative speed, trains, or boats/races, while RRB NTPC CBT-1 and CBT-2 feature 3 to 4 questions per shift.

SSC CGL Speed, Time & Distance: Relative Speed, Trains & Boats 10-Second Shortcuts

Category: SSC | Subject: Quantitative Aptitude | Topic: Speed, Time and Distance | Read Time: 6 min

Master relative velocity, train crossing problems, upstream/downstream ratios, and circular track meeting times in under 10 seconds with TCS iON shift derivations.

Key Formulas & 10-Second Reference Table

Concept / Case	Formula / Rule	When to Use
Unit Conversion (km/h ↔ m/s)	`1 km/h = 5/18 m/s \| 1 m/s = 18/5 km/h`	Converting train speed from km/h to m/s when length is in meters and time is in seconds.
Average Speed (Equal Distances at u and v)	`Average Speed = (2 × u × v) / (u + v)`	Round-trip or two equal segments covered at speeds u and v (harmonic mean).
Late / Early Arrival Distance Formula	`Distance = [ (S₁ × S₂) / \|S₁ − S₂\| ] × (Δt in hours)`	Candidate walks at S₁ and is t₁ min late, or walks at S₂ and arrives t₂ min early.
Two Trains Crossing Each Other (Lengths L₁, L₂)	`Time = (L₁ + L₂) / (S₁ + S₂) [Opposite] \| (L₁ + L₂) / \|S₁ − S₂\| [Same Dir]`	Distances (train lengths) ALWAYS add; speeds add for opposite directions and subtract for same direction.
After-Meeting Time Ratio (Radha-Shyam / Bullet Formula)	`S₁ / S₂ = √(t₂ / t₁)`	Two trains start simultaneously from A and B, cross each other, and take t₁ and t₂ hours respectively to reach destinations B and A.
Boats & Streams (Still Water Speed u, Stream Speed v)	`u = (Downstream + Upstream) / 2 \| v = (Downstream − Upstream) / 2`	Downstream speed D = u + v and Upstream speed U = u − v are known.

Why Inverse Proportional Ratios Beat Algebraic Equations in SSC CGL

Across SSC CGL Tier-1, Tier-2, and RRB NTPC shifts, over 80% of Speed, Time, and Distance questions keep one variable constant — either Distance (D) or Time (T). Instead of setting up linear equations with x, use the proportionality triad: when Distance is constant, Speed is inversely proportional to Time (S₁ : S₂ = T₂ : T₁); when Time is constant (such as thief-and-policeman chases), Distance covered is directly proportional to Speed (D₁ : D₂ = S₁ : S₂).

Constant Distance: If a person travels at 3/4 of their usual speed, their time becomes 4/3 of usual time — meaning the 1 unit increase in time equals the given delay.
Thief & Policeman Chase: Catch-up time = Initial Gap Distance / (Policeman Speed − Thief Speed), and distance run by the policeman = Catch-up Time × Policeman Speed.
Stoppage Time per Hour: Minutes stopped per hour = [(Speed without stoppages − Speed with stoppages) / Speed without stoppages] × 60.

Trains, Platforms & Relative Velocity Traps

Train questions test two recurring traps: unit mismatch (km/h vs. meters/seconds) and stationary vs. moving objects. When a train of length L crosses a pole, tree, or standing man, the distance covered is simply L. When it crosses a bridge, platform, tunnel, or another train of length P, the total distance is always L + P regardless of direction.

Memorize the 18-table multiples for instant conversions: 18 km/h = 5 m/s, 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.
When a train crosses both a pole (in t₁ sec) and a platform of length P (in t₂ sec), the train covers the extra distance P in (t₂ − t₁) seconds, so Train Speed = P / (t₂ − t₁) in one step.
When two trains of equal length L cross a pole in t₁ and t₂ seconds respectively, they cross each other in 2·t₁·t₂ / (t₁ + t₂) seconds (opposite direction) or 2·t₁·t₂ / |t₁ − t₂| seconds (same direction).

Boats & Streams: Componendo-and-Dividendo Speed Tricks

In Boat and Stream problems, never solve simultaneous equations for u (still water speed) and v (stream speed). If a boat takes n times as long to row upstream as downstream over the same distance, the ratio of still-water speed to stream speed is immediately u : v = (n + 1) : (n − 1) by componendo-and-dividendo.

Round-trip distance with total time T: Distance D = T × (u² − v²) / (2u).
If upstream time is twice downstream time (n = 2), then u : v = (2 + 1) : (2 − 1) = 3 : 1.
If upstream time is thrice downstream time (n = 3), then u : v = (3 + 1) : (3 − 1) = 4 : 2 = 2 : 1.

Solved Exam Questions & Derivations

Solved Example 1: Walking at 5/7 of his usual speed, a man reaches his office 16 minutes late. Find the usual time (in minutes) taken by him to cover the distance.

(A) 35 minutes
(B) 40 minutes
(C) 42 minutes
(D) 48 minutes

Verified Answer: Option (B): 40 minutes

10-Second Shortcut: For speed fraction a/b with delay Δt: Usual Time = [a / |b − a|] × Δt = [5 / (7 − 5)] × 16 = (5/2) × 16 = 40 minutes.

Since Distance is constant, Speed ratio (Usual : New) = 7 : 5.
Time ratio is the inverse of Speed ratio: Time (Usual : New) = 5 : 7.
The difference in time is 7 − 5 = 2 units, which corresponds to 16 minutes (so 1 unit = 8 minutes).
Usual time = 5 units = 5 × 8 = 40 minutes.

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Solved Example 2: Two trains of lengths 180 m and 220 m are running on parallel tracks in opposite directions at 46 km/h and 62 km/h respectively. In how many seconds will they completely cross each other?

(A) 12 seconds
(B) 13.33 seconds
(C) 15 seconds
(D) 16 seconds

Verified Answer: Option (B): 13.33 seconds

10-Second Shortcut: Total length = 400 m; Relative speed = 46 + 62 = 108 km/h = 30 m/s (from 18×6 → 5×6). Time = 400 / 30 = 13.33 seconds.

Add the two train lengths: Total Distance = 180 m + 220 m = 400 m.
Since they move in opposite directions, add their speeds: Relative Speed = 46 + 62 = 108 km/h.
Convert 108 km/h to m/s: 108 × (5/18) = 6 × 5 = 30 m/s.
Crossing Time = 400 / 30 = 40 / 3 = 13.33 seconds.

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Solved Example 3: Two trains A and B start simultaneously from stations X and Y towards each other. After passing each other, they take 4 hours 48 minutes and 3 hours 20 minutes respectively to reach Y and X. If train A runs at 45 km/h, what is the speed of train B?

(A) 50 km/h
(B) 54 km/h
(C) 60 km/h
(D) 64.8 km/h

Verified Answer: Option (B): 54 km/h

10-Second Shortcut: Apply S_A / S_B = √(t_B / t_A). Convert to minutes: t_B = 200 min, t_A = 288 min. √(200/288) = √(100/144) = 10/12 = 5/6. Since S_A = 45 (5×9), S_B = 6×9 = 54 km/h.

Convert post-meeting times into minutes: t_A = 4 hr 48 min = 288 min; t_B = 3 hr 20 min = 200 min.
Use the after-meeting formula: Speed(A) / Speed(B) = √(t_B / t_A) = √(200 / 288).
Divide numerator and denominator by 2: √(100 / 144) = 10 / 12 = 5 / 6.
Given Speed(A) = 45 km/h, solve 45 / Speed(B) = 5 / 6 → Speed(B) = 54 km/h.

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Solved Example 4: A boat covers 36 km upstream and 60 km downstream in 9 hours. If the speed of the boat in still water is 10.5 km/h, find the speed of the stream.

(A) 1.5 km/h
(B) 2 km/h
(C) 2.5 km/h
(D) 3 km/h

Verified Answer: Option (A): 1.5 km/h

10-Second Shortcut: Substitute v = 1.5 km/h from options: Upstream speed = 10.5 − 1.5 = 9 km/h (36/9 = 4 hrs) and Downstream speed = 10.5 + 1.5 = 12 km/h (60/12 = 5 hrs). Total = 4 + 5 = 9 hours.

Let the speed of the stream be v km/h; still-water speed u = 10.5 km/h.
Upstream speed = 10.5 − v; Downstream speed = 10.5 + v.
Set up time equation: 36 / (10.5 − v) + 60 / (10.5 + v) = 9.
Testing v = 1.5 gives 36 / 9 + 60 / 12 = 4 + 5 = 9 hours, confirming v = 1.5 km/h.

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Frequently Asked Questions

Why do we add the lengths of two trains even when they move in the same direction?: For the faster train to completely clear the slower train, its front engine must travel the entire length of the slower train plus its own length so its rear guard car clears the slower train's front. Thus, total distance is always L₁ + L₂.
When should I use the harmonic mean formula 2uv/(u+v) for average speed?: Use 2uv/(u+v) only when the two distances covered are equal (such as going from A to B at speed u and returning from B to A at speed v). If the time intervals are equal instead, average speed is the simple arithmetic mean (u + v) / 2.
How do you handle opposite vs. same direction signs in the Late/Early arrival formula?: When one case is Late and the other is Early, add the two time offsets (Δt = t₁ + t₂) because they lie on opposite sides of the scheduled arrival time. When both cases are Late (or both Early), subtract them (Δt = |t₁ − t₂|).
How many Speed, Time, and Distance questions appear in SSC CGL and RRB NTPC?: SSC CGL Tier-1 typically features 2 to 3 questions (4 to 6 marks) covering relative speed, trains, or boats/races, while RRB NTPC CBT-1 and CBT-2 feature 3 to 4 questions per shift.

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