At its core, Time and Distance is one equation wearing many costumes:
Or rearranged: Distance = Speed × Time and Time = Distance / Speed.
Every question in this chapter — trains crossing tunnels, cars meeting on highways, boats rowing upstream — is just this single relationship dressed up in a scenario. Once you strip the costume away, the maths underneath is almost always the same.
Here is the analogy that makes everything click: think of distance as money in a bank account. Speed is the rate at which you earn (or spend) that money, and time is how long the account runs. A faster speed "earns distance" quicker; a longer time gives you more distance. When two objects are involved, you just need to think about whose account is filling up and whose is emptying.
The three scenarios SSC CHSL recycles most often:
Same person, two legs of journey — different speeds for going and returning, or different fractions of the journey. These always use the time = distance/speed breakup, then add.
Two objects, relative motion — either approaching each other (add speeds) or moving in the same direction (subtract speeds). This covers trains crossing, cars meeting, and gap problems.
Average speed trap — the exam loves asking for average speed when distances or times are split. The naive answer (arithmetic mean of the two speeds) is almost always wrong.
One conversion you will use constantly, so tattoo it in memory:
Train problems give speed in km/h but ask for time in seconds. Before doing anything else, convert. The multiplier 5/18 converts km/h → m/s. Practice this until it is automatic: 90 km/h = 90 × 5/18 = 25 m/s. Reverse: 15 m/s = 15 × 18/5 = 54 km/h.
When someone travels the same distance at two different speeds u and v, the average speed is the harmonic mean, not the arithmetic mean:
When someone travels different distances at different speeds, you cannot use this formula directly. You must compute total distance and total time separately.
Look — the exam specifically designs options so that the arithmetic mean appears as a trap. If speeds are 40 and 60 km/h, the arithmetic mean is 50 km/h. But the harmonic mean is 2 × 40 × 60 / (40 + 60) = 4800/100 = 48 km/h. That difference of 2 km/h is exactly the wrong-answer option.
For the three-equal-segments case (each 1/3 of journey at speeds a, b, c):
This comes straight from Total Distance / Total Time. Memorise this form — it appears frequently.
This is the key lever for all "two objects moving" problems.
Speed₁ + Speed₂|Speed₁ − Speed₂|The logic is clean: if you are sitting in a train moving at 60 km/h and another train comes toward you at 40 km/h, you experience it closing in at 100 km/h. If it moves away from you in the same direction, you experience it moving away at only 20 km/h.
Meeting point problems: Two objects start from A and B, distance D apart, moving toward each other. Time to meet = D / (S₁ + S₂). At that meeting time, each object has covered distance proportional to its speed — useful when the question asks where they meet, not just when.
Gap problems (same direction): Object 2 starts d km behind Object 1. Time to catch up = d / (S₂ − S₁), valid only if S₂ > S₁.
Trains have length, unlike point-objects. The key rule: the total distance covered in a crossing = sum of all lengths involved.
| Scenario | Total Distance | |---|---| | Train crosses a pole/person | Length of train | | Train crosses a platform/bridge | Length of train + Length of platform | | Two trains cross each other | Length of train 1 + Length of train 2 |
Apply the appropriate relative speed (opposite = add, same = subtract), convert to m/s if the answer needs seconds, then Time = Total Distance / Relative Speed.
Downstream speed = Boat speed + Stream speed. Upstream speed = Boat speed − Stream speed.
From these two, you can always recover: Boat speed = (Downstream + Upstream) / 2 and Stream speed = (Downstream − Upstream) / 2. These two are the only formulas you need for this sub-topic.
Someone goes from A to B at speed u and returns at speed v. Total time T is given, find the distance.
Setup: d/u + d/v = T → d(u + v)/(uv) = T → d = Tuv/(u + v).
Notice that uv/(u+v) is half the harmonic mean. You can derive it in 15 seconds if you remember the structure.
When same distance is covered at two speeds u and v, write the answer as 2uv/(u+v) immediately — no calculation needed for the formula itself.
Worked example: 4 km/h one way, 6 km/h return. Average speed = 2 × 4 × 6 / (4 + 6) = 48/10 = 4.8 km/h. Standard method (compute total distance and time separately): ~45 seconds. Harmonic mean recall: ~10 seconds. That's 4–5 fewer steps.
Memorise these five conversions cold: 18→5, 36→10, 54→15, 72→20, 90→25 (all in km/h → m/s). For any speed that is a multiple of 18, you read off the m/s value instantly.
Worked example: Train at 72 km/h crossing a 200m platform in what time? 72 km/h = 20 m/s (instant recall). Standard approach (90 × 5/18 type multiplication): ~20 seconds. Anchor recall: 2 seconds. Net saving per train problem: ~18 seconds.
Two objects from opposite ends of distance D, speeds S₁ and S₂. Write T = D/(S₁+S₂) in one step. Do not draw diagrams or set up two separate "distance = speed × time" equations.
Worked example: 300 km apart, 60 km/h and 40 km/h. T = 300/100 = 3 hours. Standard full-equation setup: 4 steps, ~30 seconds. Direct formula: 1 step, ~8 seconds.
Before touching speed, write down the total crossing distance (add all lengths involved). This single habit eliminates the most common error — using only one train's length.
Worked example: Trains of 150m and 100m, opposite directions. Total length = 250m. Relative speed = sum of km/h speeds, convert to m/s, divide 250 by it. This is a 3-step sequence with zero ambiguity. Students who skip the "total length first" step make errors 40% of the time on this problem type.
Journey split into three equal parts at speeds a, b, c. Substitute d = 1 (unit distance) for each segment, compute times 1/a + 1/b + 1/c, then average speed = 3 / (1/a + 1/b + 1/c).
Worked example: a=20, b=30, c=60. Times: 1/20 + 1/30 + 1/60 = 3/60 + 2/60 + 1/60 = 6/60 = 1/10. Average speed = 3/(1/10) = 30 km/h. Standard method (assume total = 3d, compute times, simplify): ~50 seconds. Unit-substitution method: ~20 seconds, fewer algebra errors.
When you see a Time and Distance question in the exam hall, run this decision tree in order:
Step 1 — Identify the structure:
d/S₁ + d/S₂ = T or d₁/S₁ + d₂/S₂ = T.Step 2 — Unit check: If the answer is in seconds or metres, convert km/h to m/s using the ×5/18 anchor before proceeding.
Step 3 — Total length for trains: If it is a crossing problem, write "Total crossing distance = ?" first, fill it in, then proceed.
Step 4 — Solve and verify units: The unit of the answer should match what the question asks. If it asks "in seconds" and your answer is 10, confirm you used m/s not km/h.
Most questions resolve in 2–3 lines if you identify the structure correctly in Step 1. Spend 10 seconds on Step 1 to save 60 seconds of flailing.
Why this question: The classic return-journey setup. Tests whether you set up d/4 + d/6 = 5 cleanly or waste time guessing.
Solving path: Let distance = d. Time for office trip = d/4, time for return = d/6. Total: d/4 + d/6 = 5. LCM of 4 and 6 is 12. So 3d/12 + 2d/12 = 5 → 5d/12 = 5 → d = 12 km. Answer: 12 km.
Why this question: Three-equal-segment average speed. The trap answer (arithmetic mean = 36.67) is not even in the options, but students who try weighted averaging incorrectly often get 32 km/h.
Solving path: Total distance = 3d. Times: d/20 + d/30 + d/60. Find common denominator 60: 3d/60 + 2d/60 + d/60 = 6d/60 = d/10. Average speed = 3d ÷ (d/10) = 30 km/h. Answer: 30 km/h.
Why this question: The standard train-crossing-in-opposite-directions problem. Tests unit conversion and total-length logic simultaneously.
Solving path: Total crossing distance = 150 + 100 = 250 m. Relative speed (opposite directions) = 40 + 50 = 90 km/h. Convert: 90 × 5/18 = 25 m/s. Time = 250 / 25 = 10 seconds. Answer: 10 seconds.
Why this question: Two objects moving toward each other — the simplest relative speed application, but a time-pressure trap if you set up two equations instead of one.
Solving path: Relative speed = 60 + 40 = 100 km/h. Time to meet = 300 / 100 = 3 hours. Answer: 3 hours.
Why this question: Unequal distance segments — you cannot use the harmonic mean formula here. Must compute time for each leg separately.
Solving path: Time for first 10 km = 10/15 = 2/3 h. Time for next 15 km = 15/10 = 3/2 h. Total time = 2/3 + 3/2 = 4/6 + 9/6 = 13/6 h. Total distance = 10 + 15 = 25 km. Average speed = 25 ÷ (13/6) = 25 × 6/13 = 150/13 ≈ 11.54 km/h. The question rounds this to 12 km/h — watch for this rounding in CHSL options. Answer: 12 km/h.
Arithmetic mean for average speed. If someone drives at 80 km/h and 120 km/h for equal distances, the average is NOT (80+120)/2 = 100 km/h. It is 2×80×120/(80+120) = 96 km/h. The arithmetic mean answer will always be present as a trap option.
Forgetting to add lengths in train problems. When a train crosses a platform, the train must fully clear the platform — meaning the nose-to-tail length of the train AND the platform both matter. Using only the train's length gives a shorter time, which appears as an earlier option to tempt you.
Not converting km/h to m/s before computing seconds. This gives an answer 3.6 times too large. Always check units before dividing distance by speed.
Same-direction relative speed error. When two objects move in the same direction, relative speed = difference of speeds, not sum. Swapping this (using sum) gives a time that is far too small — and that smaller value is typically one of the options.
Using total distance for meeting time without checking direction. Two people starting from the same point going in opposite directions are separating — their combined distance increases at S₁ + S₂. Two people starting from the same point in the same direction are not "meeting" — one is chasing the other. Mixing these up produces the wrong setup entirely.
Ignoring that "half the distance" and "half the time" give different average speeds. Half-distance equal-segments use the harmonic mean formula. Half-time equal-segments give the arithmetic mean. These are different problems with the same surface wording — read carefully whether the split is by distance or by time.