Speed, Distance, and Time (SDT) is the backbone of a cluster of topics in SSC CHSL Quant — trains, boats and streams, circular tracks, and race problems all reduce to SDT at their core. Get this chapter cold and you are actually solving four chapters simultaneously.
The fundamental relationship is the simplest equation in arithmetic:
Think of it as a triangle. Cover the quantity you want to find, and multiply or divide the other two. Cover D — multiply S and T. Cover S — divide D by T. Cover T — divide D by S.
The analogy that locks this in: imagine a car odometer. The odometer measures distance. The speedometer measures speed. The clock measures time. None of the three values changes independently — shift one, and at least one of the others must respond. That inverse and direct relationship between the three is the entire chapter.
Here is where students go wrong at the concept level: average speed is not the arithmetic mean of speeds. If you drive 60 km/h to the office and 40 km/h back, your average speed for the round trip is not 50 km/h. It is the total distance divided by the total time, which gives you the harmonic mean. Every year, SSC sets at least one question designed to catch candidates who reach for (60 + 40) / 2 = 50. Don't be that candidate.
Unit consistency is the other silent killer. Speed in km/h, distance in metres, time in seconds — mix those up and your answer will look plausible but be wrong by a factor of 18. The conversion factors to memorise:
5/1818/5If a train problem gives speed in km/h and length in metres, convert before computing — every single time.
These three forms cover 80% of CHSL questions. The remaining 20% use one of three compound concepts below.
Case 1: Equal distances at different speeds
If a journey is split into equal halves (or thirds, etc.) at different speeds, use the harmonic mean:
For two equal halves:
Look at the PYQ below about a truck covering equal 100 km stretches at 200, 300, and 600 km/h. The instinct is (200 + 300 + 600)/3 = 366.67. The answer is 300. The harmonic mean formula catches this every time.
Case 2: Equal times at different speeds
If the traveller spends equal time at each speed, then average speed IS the arithmetic mean. This distinction between equal-distance and equal-time is what the setter is testing.
When two objects move:
|S₁ - S₂|S₁ + S₂The classic application: two people walking away from each other. Their combined speed is the rate at which the distance between them grows.
For the PYQ where Ram (3 km/h) and Shyam (5.25 km/h) walk in opposite directions: combined speed = 8.25 km/h, time to be 44 km apart = 44 / 8.25. Compute this as 44 × 4/33 = 176/33. Divide: 176 ÷ 33 = 5 remainder 11, which is 11/33 of an hour = 20 minutes. Answer: 5 hours 20 minutes.
When two objects start from the same point and move at right angles, find each distance separately using D = S × T, then apply Pythagoras. The 300 m–400 m–500 m configuration in the PYQ below is a 3-4-5 Pythagorean triple scaled by 100. Recognising the triple eliminates the square root computation entirely.
If speed changes by x%, time changes inversely. A 25% speed increase means new speed = 1.25 × old. For the same distance, new time = old distance / new speed. The key step candidates miss: when the distance itself also changes (e.g., "cover three-fourths of the distance"), multiply the two fractional effects together.
Original time 50 minutes × 3/5 = 30 minutes. No calculator needed.
For the cycling-back question: let distance = d. Going: d/8 hours. Returning: d/4 hours. Total: d/8 + 2d/8 = 3d/8 = 1.5. So d = 4 km. The trick is to get a common denominator immediately and not cross-multiply prematurely — that costs time.
For a round trip (or any equal-distance two-leg journey), average speed = 2S₁S₂ / (S₁ + S₂). Never add and halve.
Micro-example: 8 km/h onward, 4 km/h return. Avg = 2×8×4/(8+4) = 64/12 = 16/3 ≈ 5.33 km/h. The naive average gives 6 km/h — wrong, and it will appear as a trap option.
Standard method (plug in distance, compute each time, divide): 45 seconds. This formula: 12 seconds.
Two people walking apart (or trains approaching each other): gap at time T = (S₁ + S₂) × T. Don't set up two separate distance equations — add the speeds first, then multiply by time in one step.
Micro-example (Ram & Shyam PYQ): 3 + 5.25 = 8.25 km/h. Time for 44 km = 44/8.25 = 176/33. Recognise 176 = 5×33 + 11, so answer = 5 h 11/33 h = 5 h 20 min.
Standard two-equation setup: 6 steps. Direct addition method: 3 steps.
If speed becomes p/q of original, time becomes q/p of original (for the same distance). No formula to recall — just invert the fraction.
Speed increased by 25% → new speed = 5/4 of old → new time = 4/5 of old. If old time was 50 min, new time = 40 min. Then if distance is also 3/4: time = (4/5) × (3/4) × 50 = (3/5) × 50 = 30 min.
Standard algebra (compute distance first, then divide): 5 steps. Fraction-flip method: 2 steps.
When two objects travel at right angles, check if the distances form a Pythagorean triple before reaching for a calculator.
In the PYQ: 300 m and 400 m — that is 3×100 and 4×100. Hypotenuse = 5×100 = 500 m. Done. No square root computation needed.
Common triples to spot: (3,4,5), (5,12,13), (8,15,17), (7,24,25). If the distances are multiples of any of these, the answer is immediate.
Computing √(300² + 400²) from scratch: 30 seconds. Triple recognition: 5 seconds.
Three equal distances at speeds S₁, S₂, S₃: average speed = 3 / (1/S₁ + 1/S₂ + 1/S₃).
Micro-example (truck PYQ): 3 / (1/200 + 1/300 + 1/600) = 3 / ((3+2+1)/600) = 3 × 600/6 = 300 km/h. The key step: find the LCM of the denominators (600 here) and convert all reciprocals to that base before summing.
Naively computing each time and then dividing total distance by total time: 6 arithmetic steps. Harmonic mean with LCM shortcut: 3 steps.
In the exam hall, the first 10 seconds on any SDT question should run this check:
If none of these flags triggers, it is a plain D = S × T substitution — solve in under 20 seconds.
Why this question: Tests whether you know the harmonic mean formula for same-speed averages — the most common CHSL trap in this chapter.
Solving path: Three equal distances → harmonic mean applies. LCM of 200, 300, 600 is 600. Reciprocals: 3/600 + 2/600 + 1/600 = 6/600 = 1/100. Average speed = 3 / (1/100) = 300 km/h. The trap option 366.67 is the arithmetic mean — spot it and move on.
Why this question: Tests opposite-direction relative speed with a mixed-number speed (5¼ km/h) designed to slow you down with fraction arithmetic.
Solving path: Convert 5¼ to 21/4. Combined speed = 21/4 + 3 = 21/4 + 12/4 = 33/4 km/h. Time = 44 ÷ (33/4) = 44 × 4/33 = 176/33. Divide: 33 × 5 = 165, remainder 11. So 5 hours and 11/33 hours = 5 hours 20 minutes (since 11/33 × 60 = 20 min).
Why this question: Tests the combined effect of a speed percentage change and a distance fraction change — two multipliers at once.
Solving path: Original time = 50 min. Speed increases by 25% → new speed = 5/4 of old → time for same distance = 4/5 of old = 40 min. Distance is now 3/4 of original → time = 3/4 × 40 = 30 min. No need to compute the actual distance in km.
Why this question: Pythagoras in disguise. Tests whether you treat a right-angle motion problem as a geometry problem rather than an algebra problem.
Solving path: Convert speeds to m/s: 54 km/h = 15 m/s; 72 km/h = 20 m/s. In 20 seconds: car 1 covers 300 m, car 2 covers 400 m. Right-angle motion → 300–400–500 triple. Distance = 500 m.
Why this question: Tests percentage decrease in speed — candidates who compute "new speed minus old speed" often forget to express the decrease as a percentage of the original speed.
Solving path: Original speed = 1200/2 = 600 km/h. New speed = 1200/3 = 400 km/h. Decrease = 200 km/h. Percentage decrease = 200/600 × 100 = 33.33%. The trap is dividing by 400 (new speed) instead of 600 (original) — that gives a different wrong answer which will appear as an option.
Why this question: Classic round-trip question where one speed is double the other — designed to tempt the arithmetic mean answer (6 km/h) as a distractor.
Solving path: Let distance = d. Equation: d/8 + d/4 = 1.5. Common denominator 8: d/8 + 2d/8 = 3d/8 = 1.5. So d = 4 km. Verify: 4/8 + 4/4 = 0.5 + 1 = 1.5 hours. Correct.
Taking arithmetic mean as average speed for equal-distance legs. The question "covers equal distances at S₁ and S₂" is a fixed signal for the harmonic mean formula 2S₁S₂/(S₁+S₂). The arithmetic mean (S₁+S₂)/2 will always be a trap option.
Forgetting to convert units before computing. If speed is in km/h and a length (train, platform, person's height — anything) is in metres, convert speed to m/s first. Skipping this step gives an answer that is off by a factor of 3.6 (= 18/5).
Applying opposite-direction formula when objects move in the same direction. "They start from the same point" does not mean opposite directions. Read whether the directions are stated explicitly. Same direction → subtract; opposite → add.
Computing percentage decrease off the wrong base. Percentage decrease in speed = (old − new) / old × 100. Dividing by the new speed is a common error, and the resulting wrong number is often placed as an option.
In right-angle motion, computing the arithmetic sum of distances instead of the Pythagorean hypotenuse. 300 + 400 = 700 is not the distance between the two cars. That will also be an option. Pythagoras — always.
Treating "half the journey" as half the time rather than half the distance. When the problem says "first half of the journey at 60 km/h and rest at 45 km/h," it means the first half of the total distance. Setting up equal times instead of equal distances will give a wrong total and one of the distractors will match it.