At its core, Time-Speed-Distance (TSD) is one formula: Distance = Speed × Time. Everything else — relative speed, trains, circular tracks, boats — is just this formula applied in cleverly disguised situations.
Think of it this way: speed is how hungry the problem is for distance per unit time. If two cars are moving toward each other, the "combined hunger" is the sum of their speeds — they are eating up the gap between them from both ends simultaneously. If they move in the same direction, one is running away from the other, so only the difference matters.
The classic analogy that works in every classroom: imagine two trains on a track. Each train is like a ruler laid out on the track. The length of each train is the "extra distance" that needs to be covered during a crossing. The engine of train A does not clear train B the moment they touch noses — it has to travel until its tail passes B's tail. That extra distance is the sum of both lengths. This one insight clears roughly 40% of train-related questions.
For circular tracks, the concept shifts slightly. Two runners starting from the same point on a circle "meet" when one runner has lapped the other by exactly one full circle. Going in the same direction, the faster runner must gain exactly one full track length on the slower one. Going in opposite directions, they are collectively covering the track length — meeting is just the moment their combined travel equals one full lap.
Average speed deserves special attention because it is consistently mishandled. Average speed is never the arithmetic mean of two speeds unless time taken at each speed is equal. The correct formula when equal distances are covered at speeds u and v is: Average speed = 2uv/(u+v). If the time periods are equal (not distances), then the arithmetic mean works. SSC CGL tests this distinction explicitly — do not mix them up.
Unit conversion is where marks leak silently. km/h to m/s: multiply by 5/18. m/s to km/h: multiply by 18/5. Get this reflex before anything else.
D = S × T, and therefore S = D/T and T = D/S.
In exam settings, the question always gives you two of the three and asks for the third — or it gives you a relationship between them. Always write down what you know before computing.
Opposite directions: When two objects move toward each other or away from each other in opposite directions, relative speed = S₁ + S₂.
Same direction: Relative speed = |S₁ − S₂| (faster minus slower).
This applies everywhere: two cars on a road, two runners on a track, a boat in a stream (where the stream's speed is added or subtracted), trains crossing each other.
For boats and streams specifically:
u + vu − v(downstream + upstream)/2(downstream − upstream)/2When two trains cross each other:
When a train crosses a platform of length L at speed S in time T: length of train = S×T − L.
This is where most students lose time because they try to reason it out from scratch every time. Don't. Lock in these two rules:
Same direction: Time to meet = Track length ÷ (faster speed − slower speed)
Opposite directions: Time to meet = Track length ÷ (speed₁ + speed₂)
For the first meeting at the starting point (not just any meeting), you need the LCM of individual times to complete one round.
A thief-and-policeman or any head-start pursuit problem: the chaser needs to cover the head-start distance using only the relative speed advantage.
Time = Head start distance ÷ (chaser's speed − fleeing person's speed)
Be careful with units. If the head start is in meters and speeds are in km/h, convert before dividing.
A large family of SSC CGL questions follows this pattern: a train or person travels at two different speeds and arrives late by t₁ or early by t₂. The setup gives you:
D/(S₁) − D/(S₂) = (t₁ + t₂) — when one produces lateness and the other produces earliness.
Or more generally, set up two expressions for actual travel time and compare with scheduled time. The trick is to avoid solving for D separately — instead, look for a ratio or direct cancellation.
Case 1 — Equal distances at speeds u and v:
Average speed = 2uv/(u+v) (Harmonic Mean)
Case 2 — Equal time at speeds u and v:
Average speed = (u+v)/2 (Arithmetic Mean)
If a question gives you two legs of a journey at different speeds without specifying whether the distances or times are equal, read carefully. "Goes to a place at speed u and returns at speed v" means equal distances — use the harmonic mean.
When a question says "arrives x minutes late at speed s₁ and y minutes early at speed s₂", two equations describe the same distance D:
D = s₁ × (T + x/60) = s₂ × (T − y/60)
where T is the scheduled time. Divide the two equations to eliminate D, then solve for T, and back-substitute to find D if needed.
To convert km/h to m/s, multiply by 5 and divide by 18. But here's the faster way: 36 km/h = 10 m/s, 54 km/h = 15 m/s, 72 km/h = 20 m/s, 90 km/h = 25 m/s. Memorize these four anchor pairs. Most exam speeds are multiples or simple derivatives of these. When you see 240 km/h, spot it as 4 × 60 km/h — you already know 60 km/h = 50/3 m/s, so 240 km/h = 200/3 m/s. Standard method (multiply, divide): 6 steps. Pattern anchor: 2 steps.
For any two-object problem, immediately classify: same direction → subtract; opposite → add. Do this before reading the rest of the question. This locks your setup and prevents the most common sign error. In circular track problems, "same direction" means the faster one is chasing the slower one — that's subtraction. "Opposite directions" means they are running into each other — that's addition. Misclassifying here cascades into a wrong answer with a completely plausible-looking number. Standard method (reasoning it out): 30s. This reflex: 3s.
When a TSD problem gives a complicated algebraic setup — like a train covering partial distances at different speeds — skip the algebra and substitute options directly. Start with the middle option (option B or C), check if it satisfies the given time difference. If the time comes out too high, the speed is too low (more time means less speed), move to the next higher option. This works because TSD equations are monotonic: increasing speed always decreases time. Standard algebraic method: 90s of equation-setting + solving. Substitution from middle option: under 30s in most cases.
For equal-distance two-leg journeys, the average speed formula 2uv/(u+v) always gives a value less than the arithmetic mean. If a question asks for average speed when going at 40 km/h and returning at 60 km/h, the arithmetic mean is 50 but the correct answer is 2×40×60/100 = 48. The moment you see "goes and returns" with two different speeds, your answer must be below the arithmetic mean — use this as a sanity check. Eliminates one or two wrong options instantly even before computing.
For chase problems: write relative speed in the same unit as the head-start distance before dividing. The single biggest error in these problems is mixing km and meters. Convert head start to km if speeds are in km/h (or convert speeds to m/s if head start is in meters). Once units match, it is pure division. For the classic thief-policeman setup: 600 m head start, 3 km/h relative speed → 600 m = 0.6 km, time = 0.6/3 hours = 0.2 hours = 12 minutes. Two-step unit-aware calculation vs. four-step unit-conversion algebra: saves roughly 20 seconds per problem.
When you see a TSD question in the exam, run through this decision tree in under 10 seconds:
Step 1 — How many objects?
Step 2 — Are they on a circular track?
Step 3 — Are they moving toward each other, away, or is one chasing?
Step 4 — Does the question involve a train length or platform?
Step 5 — Can you substitute options?
The entire tree takes 8-10 seconds. Do not start writing until you have classified the problem.
Why this question: Tests average speed when the same time period is used but distance changes — a subtle variant of the standard formula.
Solving path: In 30 minutes at 240 km/h, normal distance = 240 × (30/60) = 120 km. Diverted distance = 120 × 1.20 = 144 km. Total time is still 30 minutes = 0.5 hours. Average speed = 144/0.5 = 288 km/h. The trap here: students sometimes average 240 and the new speed — don't. You're given total distance and total time directly; just divide.
Why this question: Classic two-unknown relative speed setup. Tests whether you can translate "opposite directions / same direction" into two clean equations.
Solving path: Opposite directions → x + y = 360/4 = 90. Same direction → x − y = 360/12 = 30. Add: 2x = 120, so x = 60 km/h. No quadratics, no guessing — two-equation setup and solve. The 360 km is not the actual track; it is the combined distance covered in the meeting interval. Read carefully.
Why this question: Circular track, same direction. Tests the "gain one lap" principle, which many students forget to apply and instead try to compute when they are at the same position by listing multiples.
Solving path: Same direction → relative speed = 12 − 6 = 6 m/s. B must gain one full track on A → time = 840/6 = 140 s. Direct application, no listing required.
Why this question: Head-start chase with a unit conversion trap built in. The head start is given in meters while speeds are in km/h.
Solving path: Relative speed = 15 − 12 = 3 km/h. Convert head start: 600 m = 0.6 km. Time = 0.6/3 = 0.2 hours = 12 minutes. Alternatively: 3 km/h = 3 × 1000/3600 = 5/6 m/s, time = 600 ÷ (5/6) = 720 s = 12 min. Both routes work; pick whichever unit you are more comfortable with, but do not mix them.
Why this question: Circular track with decimal speeds requiring a unit conversion before applying the formula — tests both the principle and calculation hygiene.
Solving path: Same direction (implied — "meet again for the first time" on a circular track without specifying opposite direction defaults to same direction in this question context; relative speed = 13.5 − 9 = 4.5 km/h). Convert: 4.5 × 5/18 = 1.25 m/s. Time = 750/1.25 = 600 s. The decimal conversion is where time is lost — practice the × 5/18 step as a reflex.
Averaging speeds arithmetically for equal-distance journeys. If a car goes at 40 km/h and returns at 60 km/h, the average is not 50 km/h — it is 48 km/h. The return trip takes less time, so the slow leg dominates. Always use 2uv/(u+v) for equal-distance two-leg problems.
Using the wrong relative speed direction for circular tracks. Same direction means subtraction; opposite means addition. If you swap these, your answer is off by a factor that creates a plausible-looking wrong option in the choices.
Ignoring train length in crossing problems. When a train crosses a platform or another train, the engine reaching the end of the platform is not the finish — the tail of the train must clear it. The distance is train length + platform length (or sum of both train lengths). Skipping one of the lengths gives you a wrong answer that still passes a rough check.
Not converting units before dividing in chase problems. Head start in meters, speed in km/h, and you divide directly — this is the most frequent arithmetic catastrophe. Always make units consistent before the final division.
Setting up late/early equations with wrong sign conventions. If a train arrives 24 minutes late at lower speed, its actual travel time is scheduled time + 24 minutes — the extra time is positive. If it arrives early at higher speed, actual time is scheduled time − 24 minutes. Reversing this sign gives an equation with no real solution, and students then guess randomly.
Confusing "first meeting" (any point on track) with "first meeting at starting point". These are different questions. First meeting anywhere: divide track by relative speed. First meeting back at starting point: find individual lap times and take their LCM. SSC CGL uses both — read the exact phrasing.