SSC GD Time and Distance Questions with Answers: Quick Answer
This page solves 2 official SSC GD time and distance questions with full step-by-step working. Both questions come from one official paper: the SSC GD 2025 Elementary Mathematics section paper held on 7 February 2025, Shift 1. The answers are 0.5 minutes for the two-trains question and 160 seconds for the circular-race question. That count of 2 is the honest total for this paper: every one of its 20 questions was checked, and only these 2 are genuine speed, time and distance questions. The other 18 questions in the same paper cover topics such as averages, ratios, percentages, interest and mensuration, and they are not repeated here under a time-and-distance label.
The rest of this page is a complete method guide for the topic, so that the 2 solved questions are useful rather than just readable. It covers the speed formula, converting between km/h and m/s, relative speed, train problems and circular tracks, each with a short worked method illustration. Those illustrations use plain practice numbers and are clearly labelled as method illustrations. They are not previous-year questions, and nothing on this page claims otherwise. As of 2 November 2026, the 2 questions below are the only official time-and-distance questions from this paper on this site.
SSC GD Time and Distance Formula Box
Keep these five ideas in front of you while you work through the questions. Almost every SSC GD time-and-distance question is one of them in disguise.
- Speed = distance ÷ time. Rearranged: distance = speed × time, and time = distance ÷ speed. If any two of the three are known, the third follows.
- km/h to m/s: multiply by 5/18. m/s to km/h: multiply by 18/5. A speed in km/h and a distance in metres cannot be divided directly. Convert first, every time.
- Relative speed, same direction: subtract. Relative speed, opposite directions: add. The faster body gains on the slower one at the difference of their speeds when both move the same way.
- Train crossing: distance covered = length of the train + length of the object crossed. A pole or a person counts as length zero, so the distance is just the train's own length. A platform, a bridge or another train adds its full length.
- Circular track, meeting at the starting point: take the LCM of the individual lap times. Each runner must have completed a whole number of laps, so the first time they are together again at the start is the smallest time that is a whole number of laps for both.
Shortcut habit: convert units before you calculate, and check the unit the question asks for before you pick an option. In the two-trains question below, the working naturally gives seconds, but the question asks for minutes. That single conversion is where marks are most often lost in this topic.
Speed, Distance and Time: The Core Method
Speed tells you how much distance is covered in one unit of time. A speed of 60 km/h means 60 kilometres in one hour. A speed of 25 m/s means 25 metres in one second. The formula speed = distance ÷ time is a definition, not a trick, and the other two forms follow from it by ordinary rearrangement. If a car covers 180 km in 3 hours, its speed is 180 ÷ 3 = 60 km/h. If the same car travels at 60 km/h for 2.5 hours, it covers 60 × 2.5 = 150 km. If it must cover 150 km at 50 km/h, it needs 150 ÷ 50 = 3 hours. Write the units beside every number as you work, and a wrong rearrangement usually shows itself at once, because the units will not make sense.
Method illustration (not a previous-year question): A cyclist covers 36 km in 2 hours. Her speed is 36 ÷ 2 = 18 km/h. At that speed, covering 45 km would take 45 ÷ 18 = 2.5 hours. The method is identical whether the numbers are large or small, and whether the vehicle is a train, a cyclist or a runner. What changes between questions is only which quantity is unknown and which unit conversion is needed.
One common error is to divide a distance in metres by a speed in km/h and treat the result as seconds. It is not. Metres divided by kilometres per hour gives a number whose unit is a mixture, and no answer option built from it will be right. Convert the speed to m/s first if the distance is in metres, or convert the distance to kilometres first if the speed stays in km/h. Either route works, provided both quantities end up in matching units before the division happens.
Converting km/h and m/s
One kilometre is 1,000 metres and one hour is 3,600 seconds, so 1 km/h equals 1,000 ÷ 3,600 = 5/18 m/s. That fraction is the whole conversion. To turn km/h into m/s, multiply by 5/18. To turn m/s into km/h, multiply by 18/5, which is the same fraction flipped over. Memorise both directions, because questions move in both directions and the conversion is needed in train problems, circular-track problems and average-speed problems alike.
Method illustration (not a previous-year question): A speed of 72 km/h becomes 72 × 5/18 = 20 m/s. Working it out: 72 ÷ 18 = 4, and 4 × 5 = 20. In the other direction, 15 m/s becomes 15 × 18/5 = 54 km/h, because 15 ÷ 5 = 3 and 3 × 18 = 54. Notice how dividing by 18 or 5 first keeps the arithmetic small. In the examination, always look for that cancellation before reaching for long multiplication.
A second conversion also appears regularly: minutes and seconds. One minute is 60 seconds, so seconds divided by 60 give minutes, and minutes multiplied by 60 give seconds. The two-trains question below asks for minutes while the natural working gives 30 seconds, so the last step is 30 ÷ 60 = 0.5 minutes. Skipping that step, or dividing by 100 instead of 60, turns a correct method into a wrong option. Finish every question by asking: does my answer carry the unit the question asked for?
Relative Speed: Same Direction and Opposite Directions
When two bodies move, what matters for meeting or crossing is not either speed alone but how fast the gap between them is closing. That closing speed is the relative speed. If two trains move in the same direction, the faster one gains on the slower one at the difference of their speeds. If they move towards each other, the gap closes at the sum of their speeds, because both are contributing to closing it.
Method illustration (not a previous-year question): Two cyclists ride in the same direction at 20 km/h and 14 km/h. The faster gains at 20 − 14 = 6 km/h. If the slower cyclist starts 12 km ahead, the faster one catches up after 12 ÷ 6 = 2 hours. If instead they ride towards each other from 12 km apart, the gap closes at 20 + 14 = 34 km/h, and they meet after 12 ÷ 34 hours, which is about 21 minutes. The distance is the same in both cases; only the relative speed changes, and with it the time.
Relative speed also explains why a train crossing problem and a meeting problem feel similar: both ask how long a gap takes to close at a given relative speed. In a crossing problem the gap is the combined length that must pass; in a meeting problem the gap is the starting distance between the two bodies. Once the relative speed and the distance to be covered are both identified, the time is simply distance divided by relative speed, exactly as in the core formula.
Train Problems: Lengths Add Up
A train is not a point. When a train crosses a pole, the whole train must pass the pole, so the distance covered is the train's own length. When it crosses a platform, a bridge or a tunnel, the whole train must clear the far end, so the distance covered is the train's length plus the platform's, bridge's or tunnel's length. When one train crosses another, both trains must fully pass each other, so the distance covered is the sum of both lengths. Forgetting to add a length is the most common train-problem error, ahead even of unit conversion.
Method illustration (not a previous-year question): A train 200 m long travelling at 54 km/h crosses a platform 100 m long. The distance to cover is 200 + 100 = 300 m. The speed in m/s is 54 × 5/18 = 15 m/s. The time taken is 300 ÷ 15 = 20 seconds. If the same train crossed a pole instead, the distance would be just 200 m and the time would be 200 ÷ 15 = 13.33 seconds. Same train, same speed, different object crossed, different answer: the object decides the distance.
For two trains moving in the same direction, combine both ideas. The distance is the sum of the two lengths, because the faster train must pass the slower train completely, from its front reaching the slower train's rear to its rear clearing the slower train's front. The speed is the relative speed, the difference of the two speeds, because both trains are moving forward. The two-trains question below is exactly this combination, with the extra step of converting seconds into minutes at the end.
Circular Tracks: Lap Times and Meeting Points
On a circular track, each runner returns to the starting point after one lap time, which is the track length divided by that runner's speed. Two runners who start together at the same point will next be together at the starting point only when both have completed a whole number of laps. The first such time is the least common multiple (LCM) of their individual lap times. This is different from the first time they meet anywhere on the track. They may meet on the far side of the track much sooner, because the faster runner laps round and catches the slower one from behind. The question's wording decides which answer is wanted: "at the starting point" means the LCM method, while "meet for the first time" without that phrase usually means the relative-speed method on the track length.
Method illustration (not a previous-year question): Two runners on a 400 m track run at 8 m/s and 10 m/s. Their lap times are 400 ÷ 8 = 50 seconds and 400 ÷ 10 = 40 seconds. The LCM of 50 and 40 is 200 seconds, so they are together at the starting point again after 200 seconds. They meet for the first time anywhere on the track after 400 ÷ (10 − 8) = 200 seconds as well in this particular case, because the speed difference is small, but that coincidence does not hold in general, so do not treat the two questions as the same. Read the wording, then choose the method.
To find an LCM of two lap times, use prime factors or list multiples. For lap times of 32 seconds and 20 seconds, the multiples of 32 are 32, 64, 96, 128, 160, and the multiples of 20 are 20, 40, 60, 80, 100, 120, 140, 160. The first multiple shared by both lists is 160. By prime factors: 32 = 2⁵ and 20 = 2² × 5, so the LCM takes the highest power of each prime, 2⁵ × 5 = 160. Either route reaches the same answer, and the prime-factor route stays reliable when the numbers are larger.
Average Speed: Total Distance Over Total Time
Average speed is total distance divided by total time. It is not, in general, the average of the speeds. If a journey has two parts at different speeds, the part that takes longer carries more weight in the average, because average speed is weighted by time, not by distance. The simple average of the two speeds is correct only when the two parts take equal times. When the two parts cover equal distances, a different result, the harmonic mean, applies instead.
Method illustration (not a previous-year question): A bus travels 60 km at 30 km/h and then 60 km at 60 km/h. The first part takes 60 ÷ 30 = 2 hours and the second takes 60 ÷ 60 = 1 hour. Total distance is 120 km and total time is 3 hours, so the average speed is 120 ÷ 3 = 40 km/h. The plain average of the speeds would be (30 + 60) ÷ 2 = 45 km/h, which is wrong, because the bus spent twice as long at the slower speed. Whenever a question gives speeds for parts of a journey, work out each part's time first, then divide total distance by total time.
Solved Question 1: Two Trains in the Same Direction
Question: Two trains having lengths of 150 m and 350 m are running at speeds of 60 km/h and 120 km/h, respectively, in the same direction. The time taken (in minutes) by the faster train, coming from behind, to completely cross the other train is:
(A) 2
(B) 1.2
(C) 3.5
(D) 0.5
Answer: (D) 0.5 minutes.
Previously asked in SSC GD 2025, 7 February 2025, Shift 1 (Elementary Mathematics section paper, Question 5).
Shortcut: Relative speed = 120 − 60 = 60 km/h. Distance to cover = 150 + 350 = 500 m. Convert 60 km/h to m/s: 60 × 5/18 = 50/3 m/s. Time = 500 ÷ (50/3) = 30 seconds = 0.5 minutes.
Full working:
Step 1 — Find the relative speed. Both trains move in the same direction, and the faster train comes from behind, so it gains on the slower train at the difference of their speeds: 120 − 60 = 60 km/h. If the speeds had been added instead, that would be the relative speed for trains moving towards each other, which is not the situation described here.
Step 2 — Find the distance to be covered. For the faster train to completely cross the slower train, its front must travel past the slower train's front and its rear must clear the slower train's rear. The distance covered is therefore the sum of both lengths: 150 + 350 = 500 m. Using only one train's length here would understate the distance and give a time that is too short.
Step 3 — Convert the speed to matching units. The distance is in metres, so the speed must be in metres per second. Multiply by 5/18: 60 × 5/18 = 300/18 = 50/3 m/s, which is about 16.67 m/s. Dividing 500 m by 60 km/h directly, without this conversion, is the error this question is designed to catch.
Step 4 — Divide distance by relative speed. Time = 500 ÷ (50/3) = 500 × 3/50 = 30 seconds. The faster train needs 30 seconds to gain 500 m at a relative speed of 50/3 m/s.
Step 5 — Convert to the unit asked for. The question asks for minutes, so divide by 60: 30 ÷ 60 = 0.5 minutes. The answer is option (D). A final check confirms the size of the answer: at roughly 17 m/s relative speed, 500 m takes roughly 30 seconds, and half a minute is 30 seconds, so 0.5 minutes is consistent.
Solved Question 2: Circular Race and the Starting Point
Question: In a circular race of 800 m length, Ram and Mohan start at speeds of 25 m/sec and 40 m/sec, respectively, at the same time from the same point. After how much time will they meet for the first time at the starting point, when running in the same direction?
(A) 136 seconds
(B) 120 seconds
(C) 152 seconds
(D) 160 seconds
Answer: (D) 160 seconds.
Previously asked in SSC GD 2025, 7 February 2025, Shift 1 (Elementary Mathematics section paper, Question 8).
Shortcut: Ram's lap time = 800 ÷ 25 = 32 seconds. Mohan's lap time = 800 ÷ 40 = 20 seconds. First meeting at the starting point = LCM of 32 and 20 = 160 seconds.
Full working:
Step 1 — Find each runner's lap time. The track is 800 m long. Ram runs at 25 m/s, so one lap takes 800 ÷ 25 = 32 seconds. Mohan runs at 40 m/s, so one lap takes 800 ÷ 40 = 20 seconds. Both speeds are already in m/s and the track length is already in metres, so no unit conversion is needed in this question.
Step 2 — Understand what "at the starting point" requires. Both runners must be at the starting point at the same moment. That happens only after each has completed a whole number of laps. Ram must have run for a multiple of 32 seconds and Mohan for a multiple of 20 seconds, and the two times must be equal. The first such time is the smallest number that is a multiple of both 32 and 20: their least common multiple.
Step 3 — Find the LCM of 32 and 20. By prime factors, 32 = 2 × 2 × 2 × 2 × 2 and 20 = 2 × 2 × 5. The LCM takes the highest power of each prime factor: 2⁵ × 5 = 32 × 5 = 160. By listing multiples, the multiples of 32 run 32, 64, 96, 128, 160, and 160 is the first of them that is also a multiple of 20. Either way, the LCM is 160 seconds.
Step 4 — Check the answer against whole laps. In 160 seconds, Ram completes 160 ÷ 32 = 5 laps and Mohan completes 160 ÷ 20 = 8 laps. Both are whole numbers, so both are back at the starting point together, and no earlier time works: at 128 seconds Ram has done 4 laps but Mohan has done 6.4 laps, and at 120 seconds Mohan has done 6 laps but Ram has done only 3.75 laps. The answer is option (D), 160 seconds.
Step 5 — Do not confuse this with meeting anywhere on the track. The faster runner gains on the slower one at 40 − 25 = 15 m/s, so they first meet somewhere on the track after 800 ÷ 15 = 53.33 seconds. That is a different question from the one asked. The words "at the starting point" are what make the LCM method correct here, and reading them is worth as much as the arithmetic.
Practise the Full Paper and the Maths Mock
The 2 questions above come from a single 20-question section paper. Attempting the whole paper under time pressure shows how time-and-distance questions sit among the other Elementary Mathematics topics and how quickly the method must be recalled when the clock is running.
- Attempt the full paper: SSC GD 2025 Question Paper 7 February Shift 1 — the complete 80-question paper from the same shift, with the official answer key as labelled on this site.
- Practise maths under exam conditions: SSC GD Maths Mini Mock Test 01 — a timed maths mock for the same examination.
When you attempt either test, solve every time-and-distance question by writing down the relative speed and the distance to be covered before you calculate. If you catch yourself dividing metres by km/h, stop and convert first. That one habit removes the largest single source of error in this topic.
SSC GD Time and Distance Questions FAQs
How many official time and distance questions are solved on this page?
Two. Both come from the SSC GD 2025 Elementary Mathematics section paper of 7 February 2025, Shift 1, and every one of that paper's 20 questions was checked to reach that count. The rest of the page is a method guide with clearly labelled illustrations, not additional previous-year questions, so the count is never overstated.
What is the most important formula for SSC GD time and distance questions?
Speed = distance ÷ time, together with the conversion between km/h and m/s. Most questions add one further idea to that core: relative speed for moving bodies, added lengths for trains, or the LCM of lap times for circular tracks. If the units are converted first and the distance to be covered is identified correctly, the core formula finishes the question.
Why is the two-trains answer in minutes when the working gives seconds?
Because the question asks for minutes. The working gives 30 seconds, and 30 ÷ 60 = 0.5 minutes. Always read the unit in the question stem before choosing an option. A correct method with the wrong final unit is still a wrong answer, and option sets in this topic are often built around that exact slip.
Do Ram and Mohan meet before 160 seconds in the circular-race question?
Yes, they meet on the track after about 53 seconds, when the faster runner catches the slower one from behind. But the question asks when they meet for the first time at the starting point, which needs both runners to have completed whole laps. That happens after 160 seconds, when Ram has done 5 laps and Mohan has done 8 laps. The wording decides the method.
Should I learn train problems and circular-track problems separately?
Learn them as one idea with two settings. Both ask how long a distance takes to close at a relative speed, or when two lap counts line up. Train problems add lengths to get the distance; circular-track problems use lap times and their LCM. Practising them together, as this page does, makes the shared structure visible and stops the two question types from feeling like unrelated topics.
Sources
- This site's own test file: SSC GD 2025 Maths Question Paper 7 February Shift 1, marked official with the official answer key, 20 questions, exam date 7 February 2025, Shift 1. Both solved questions, their options and their answers are taken from this file, and the remaining 18 stems were read to confirm that only 2 are time-and-distance questions.
- Staff Selection Commission portal, ssc.gov.in — examination notices and answer-key notices for the SSC Constable (GD) Examination 2025.
Last checked: 2 November 2026. This page will be updated when further official SSC GD section papers are added to this site and their time-and-distance questions can be verified from the site's own files.