Speed, Distance and Time is one of the most reliable scoring areas in SSC MTS. The questions are almost always direct — no deep algebra, no multi-step traps. You need one iron-clad mental model and two or three formula variants.
Here is the mental model: imagine you are driving on a highway. The distance you cover is the road itself. The time you take is how long you sit in the car. Your speed is how efficiently you are converting time into distance. These three quantities are locked together:
Rearrange it two ways and you have everything:
Think of it as a triangle — cover the quantity you want to find, and multiply or divide the remaining two.
The inverse relationship is where most SSC MTS questions live. When distance is fixed, speed and time pull in opposite directions. Double the speed, half the time. Reduce speed to 4/5th, time becomes 5/4th. This "inverse" idea shows up in nearly every late-arrival or early-arrival question.
Units matter more than students realize. Speed in km/h, time in hours — distance comes out in km. If the question gives you minutes, convert first. The conversion you must know cold: to go from km/h to m/s, multiply by 5/18. To go from m/s to km/h, multiply by 18/5. These two come up whenever trains or runners are involved.
Average speed is the second trap. Most people instinctively add the two speeds and divide by 2. That works only if time is equal for both legs. If distance is equal (same route, go and come back), use the harmonic mean formula. If neither time nor distance is equal, go back to basics: total distance divided by total time.
Think of average speed like your "effective pace" for the whole journey — not the pace at any one moment, but what a single constant speed would have needed to be to complete the same journey in the same total time.
Everything flows from this. Memorize the three rearrangements as muscle memory, not as derivations.
| Want to find | Formula | |---|---| | Distance | | | Speed | | | Time | |
A quick check: 36 km/h = 36 × 5/18 = 10 m/s. That is a number worth remembering.
When distance is constant:
So if speed becomes a fraction of original, time becomes of original.
Example: Speed reduced to 4/5 of original (20% less).
This fraction-flip method beats setting up equations from scratch every time. See it as: numerator and denominator of the speed fraction swap to give you the time fraction.
Case 1: Unequal distances or unequal times (General formula)
Always compute total distance and total time separately. Never average the speeds directly unless the times are exactly equal.
Case 2: Same distance both ways (Harmonic mean)
This formula applies specifically when you travel the same route at two different speeds — like going to a place at 60 km/h and returning at 80 km/h. The harmonic mean always gives a result closer to the lower speed, not the midpoint. That is the intuition: you spend more time at the lower speed, so it pulls the average down.
When two objects move:
For two trains moving toward each other, the gap between them closes at the rate of their combined speed. So time to meet = gap / (S₁ + S₂).
For two trains moving in the same direction, the faster one closes the gap at (S₁ − S₂). The time for the faster to fully overtake depends on the combined length of both trains divided by relative speed.
Key pattern for "one covered X km more" questions:
If Train A travels at and Train B at (both moving toward each other, starting simultaneously), and they meet after time :
So if that difference is given, . Then total distance = .
For the same distance:
Ratios flip. Speed ratio 2:3:4 → time ratio 1/2 : 1/3 : 1/4. Multiply through by LCM (12) to get 6:4:3. This is a direct question type in SSC MTS.
When a vehicle stops for some time during a journey, the moving speed covers the actual distance, but the average speed uses total elapsed time including stoppages.
The driving time itself uses only the driving speed. This is the trap: students forget to add the stop time to the denominator.
When speed changes by a fraction, time changes by the reciprocal fraction. If speed becomes m/n of original, time becomes n/m of original. The extra or saved time is the difference between the new time fraction and 1, multiplied by original time T.
Worked example: Speed reduced by 25% → new speed = 3/4 original → new time = 4/3 original → extra time = 1/3 × T. If extra time = 20 min, T = 60 min.
Standard method (setting up S×T = S'×T' with two variables): ~50 seconds. Fraction flip: ~15 seconds — you just read off the fraction and solve one equation.
For any "go at speed A, return at speed B" question, skip total distance and total time. Apply directly: Average Speed = 2AB/(A+B).
Worked example: A = 60, B = 80. Average = 2×60×80/(60+80) = 9600/140 = 480/7 ≈ 68.57 km/h.
The formula takes 4 arithmetic steps. Setting up total distance = d + d, total time = d/60 + d/80, then solving: 8 steps minimum, higher chance of error. This saves at least 30 seconds.
When two trains start simultaneously toward each other and one covers X km more than the other at the meeting point: Time = X / (difference in speeds). Then total distance = sum of speeds × time.
Worked example: Speeds 72 and 84, one covers 108 km more. Time = 108 / (84 − 72) = 108/12 = 9 hours. Total distance = (72+84)×9 = 1404 km.
Building two simultaneous equations for this: ~60 seconds. The difference-over-difference pattern: ~10 seconds.
For the same distance, time ratios are the reciprocals of speed ratios. To convert, write the speeds as fractions with 1 in the numerator, then scale to whole numbers by multiplying through by the LCM.
Worked example: Speed ratio 2:3:4 → time ratio = 1/2 : 1/3 : 1/4. LCM of 2, 3, 4 is 12. Multiply: 6 : 4 : 3.
Writing out full equations for each person: 6 lines of algebra. Reciprocal swap + LCM scale: 2 lines.
Whenever a question mentions a break, stop, or halt — add it to total time before dividing. The single most common error in average speed questions is leaving out stop time.
Worked example: Drove 60 km/h for 3 hours, stopped 1 hour, drove 60 km/h for 4 hours. Total distance = 60×7 = 420 km. Total time = 3+1+4 = 8 hours (not 7). Average = 420/8 = 52.5 km/h. If you used 7 hours you get 60 — which is the wrong answer and the first option listed as a trap.
Recognizing the trap before computing: saves one full recalculation cycle.
When you see a Speed-Distance-Time question in the exam hall, run through this sequence:
What is fixed — distance, time, or speed? If distance is fixed across two scenarios, inverse relationship applies. If only one journey is described, it is a direct formula question.
Is it a same-route return trip? Yes → use 2ab/(a+b) for average speed. No → use total distance / total time.
Does it involve two objects moving? Same direction → subtract speeds. Opposite directions → add speeds. "One covered X km more" → divide difference by speed difference to get time.
Is there a stoppage or halt mentioned? Add that time to the denominator. Do not skip it.
Is it a ratio question? Speed ratios → flip to time ratios → scale with LCM.
Unit check: Is everything in the same unit? If minutes appear with km/h, convert minutes to hours before plugging in.
Do this check in under 10 seconds. Most SSC MTS speed-distance questions fall cleanly into one of these five buckets. Identify the bucket, apply the appropriate shortcut, and move on.
Why this question: This is the most direct test of the inverse proportion principle — the "speed reduced, arrives late" template appears in nearly every SSC exam.
Solving path: New speed = 80% of original = 4/5 × S. Because distance is fixed, new time = 5/4 × T. Extra time = 5/4T − T = T/4. Set T/4 = 15 minutes → T = 60 minutes. Fraction flip done in one step — no equations needed.
Why this question: This is the exact "same route, two speeds" scenario where students mistakenly compute (60+80)/2 = 70 and pick a wrong option. The harmonic mean formula is non-negotiable here.
Solving path: Same distance both ways, so use 2×60×80/(60+80) = 9600/140 = 480/7 ≈ 68.57 km/h. The answer 68.57 is less than the midpoint 70 — always true with harmonic mean. Use that as a sanity check.
Why this question: This 2024 question tests general average speed with unequal segments — the template where you must compute everything from scratch.
Solving path: Total distance = (65×4) + (52×2.5) = 260 + 130 = 390 km. Total time = 4 + 2.5 = 6.5 hours. Average = 390/6.5 = 60 km/h. No shortcut formula here — just careful arithmetic.
Why this question: A 2024 question on the "how early" variant, which requires computing two separate times and subtracting.
Solving path: Distance = 10×2 = 20 km. New speed = 15 km/h. New time = 20/15 = 4/3 hours = 80 minutes. Original time = 2×60 = 120 minutes. Time saved = 120 − 80 = 40 minutes.
Why this question: A 2024 trains question that uses the "one covered X km more" pattern — requires both the meeting time and total distance.
Solving path: Difference in speeds = 84 − 72 = 12 km/h. Difference in distances covered = 108 km. Time to meet = 108/12 = 9 hours. Total distance = (72+84) × 9 = 156 × 9 = 1404 km.
Why this question: A 2023 question that specifically sets a stoppage trap — total time must include the idle hour.
Solving path: Driving time = 3+4 = 7 hours. Total distance = 60×7 = 420 km. Total time including stop = 8 hours. Average = 420/8 = 52.5 km/h. The trap answer is 60 km/h (ignoring the stop) — it is the first option listed.
Why this question: A 2022 question on speed-to-time ratio conversion — tests whether you know the reciprocal relationship cleanly.
Solving path: Speed ratio 2:3:4. Time ratio (same distance) = 1/2 : 1/3 : 1/4. Multiply each by LCM(2,3,4) = 12: get 6:4:3.
Averaging speeds directly: Adding two speeds and dividing by 2 works only when the time for each leg is equal, not the distance. In almost every SSC MTS question involving a return journey, the distance is equal — use the harmonic mean formula, not the arithmetic mean.
Forgetting stop time in average speed: Any question that mentions a break, halt, or rest requires you to include that duration in the denominator. Students compute total distance correctly but divide by driving time only, producing an inflated average.
Not converting units before computing: Mixing km/h with minutes is the single biggest source of arithmetic errors. Before touching the formula, confirm everything is in matching units. Convert minutes to hours (divide by 60) or convert km/h to km/min (divide by 60).
Using speed difference when objects move in opposite directions: Relative speed for objects moving toward each other is the sum of speeds, not the difference. The difference applies only when they move in the same direction. Mix these up and your meeting time is completely wrong.
Confusing "time saved" with "new time": The question asks how early you arrive — that is original time minus new time. Students sometimes report the new time itself, which matches a distractor option. Read carefully: "how early" means a difference.
Inverting the ratio incorrectly: Speed ratio 2:3:4 → time ratio is NOT 4:3:2 (that would be reversing the order). It is 1/2 : 1/3 : 1/4, which after scaling becomes 6:4:3. The reversal 4:3:2 is always listed as a trap option.