Module 3 · SSC CGL Quantitative Aptitude
Speed, Distance & Time
Relative speed, trains, boats & streams.
Relative Speed · Trains · Boats & Streams
Learning Objectives
- Understand the fundamental concepts of Speed, Distance & Time
- Apply key formulas and techniques to solve problems
- Practice with exam-level questions to build speed and accuracy
Key Concepts
What is Speed, Distance & Time?
Speed, Distance, and Time form the backbone of one of the most high-scoring topics in SSC CGL Quantitative Aptitude. Every year, 4–6 questions appear from this chapter in Tier 1 and Tier 2. The basic relationship is Distance = Speed × Time. Problems span trains, boats & streams, circular tracks, races, and clocks — making this a versatile topic that rewards conceptual clarity and shortcut mastery.
SSC CGL Weightage
In the last 5 years, Speed-Distance-Time contributes 5–6 questions in Tier 1 and 4–5 in Tier 2. Average speed, train-platform, boats & streams, and relative speed are the most frequently tested sub-topics.
1. Basic Formula & Unit Conversion
Distance = Speed × Time. This can be rearranged as Speed = Distance / Time and Time = Distance / Speed. Always ensure units are consistent before plugging values.
Unit Conversion: 1 km/h = (1000 m) / (3600 s) = 5/18 m/s. Conversely, 1 m/s = 18/5 km/h.
Tip 1 — Quick Unit Conversion
To convert km/h to m/s, multiply by 5/18. To convert m/s to km/h, multiply by 18/5. Memorise: 36 km/h = 10 m/s, 54 km/h = 15 m/s, 72 km/h = 20 m/s, 90 km/h = 25 m/s, 108 km/h = 30 m/s. These repeating values save time.
Tip 2 — DST Triangle
Cover the quantity you need: D over S×T. If you need speed, cover S → D/T. If you need time, cover T → D/S. This visual trick prevents formula errors under exam pressure.
Example 1 — Unit Conversion
A train runs at 25 m/s. What is its speed in km/h?
Solution: Speed = 25 × (18/5) = 5 × 18 = 90 km/h. Quick check: 25 m/s = 90 km/h.
2. Average Speed
Average speed is not the arithmetic mean of speeds unless times are equal. It is Total Distance ÷ Total Time.
Case 1 — Equal Distances: A body travels from A to B at speed a and returns at speed b. Average speed = 2ab / (a + b).
Case 2 — Equal Times: A body travels at speed a for time t and at speed b for another t. Average speed = (a + b) / 2.
Tip 3 — Average Speed Shortcut
When distances are equal (common in SSC CGL), never average the speeds directly. Use 2ab/(a+b). Example: 40 km/h and 60 km/h give average = 2×40×60/(100) = 48 km/h, not 50.
Tip 4 — When Times Are Equal
If the body moves at speed a for t hours and speed b for the next t hours, average speed is simply (a+b)/2. This is rarely tested directly but appears in multi-leg journey problems.
Example 2 — Average Speed (Equal Distances)
A person covers half the distance at 30 km/h and the other half at 50 km/h. Find the average speed for the entire journey.
Solution: Here distances are equal (each is half). Average speed = 2ab/(a+b) = 2×30×50/(30+50) = 3000/80 = 37.5 km/h.
Example 3 — Average Speed (Equal Times)
A car travels at 40 km/h for the first 2 hours and at 60 km/h for the next 2 hours. Find the average speed.
Solution: Total distance = 40×2 + 60×2 = 80 + 120 = 200 km. Total time = 4 hours. Average speed = 200/4 = 50 km/h. Alternatively, (40+60)/2 = 50 km/h.
3. Relative Speed
When two objects move, their relative speed determines how fast the distance between them changes.
Opposite Direction: Relative speed = S₁ + S₂. Time to meet = Initial distance / (S₁ + S₂).
Same Direction: Relative speed = |S₁ − S₂|. Time to catch up / overtake = Initial separation / |S₁ − S₂|.
Tip 5 — Relative Speed Caution
Always convert km/h to m/s before multiplying with time in seconds, or keep everything in km and hours. Mixing units is the #1 SSC CGL trap in this chapter.
Example 4 — Opposite Direction
Two friends start from points A and B 450 m apart and walk towards each other at 3 m/s and 2 m/s respectively. How long will they take to meet?
Solution: Relative speed = 3 + 2 = 5 m/s. Time = 450/5 = 90 seconds = 1.5 minutes.
Example 5 — Same Direction (Chase)
A thief runs at 8 km/h. A policeman spots him 200 m away and chases at 10 km/h. How far does the policeman run before catching the thief?
Solution: Relative speed = 10 − 8 = 2 km/h = 2 × 5/18 = 5/9 m/s. Time to catch = 200 ÷ (5/9) = 360 s. Distance run by policeman = 10 × (5/18) × 360 = 10 × 100 = 1000 m.
4. Train Problems
Train problems hinge on one key insight: when a train passes an object, the distance covered equals the train's length plus the object's length (if the object has length).
- Passing a pole / person: Distance = Length of train. Time = Length / Speed.
- Passing a platform / bridge / tunnel: Distance = Length of train + Length of platform.
- Passing another train (opposite direction): Distance = Sum of lengths. Relative speed = Sum of speeds.
- Passing another train (same direction): Distance = Sum of lengths. Relative speed = Difference of speeds.
Tip 6 — Train Platform Shortcut
For train + platform problems, SSC CGL often asks speed in km/h. Solve in m/s first, then multiply by 18/5. Remember: Total distance = L_train + L_platform.
Tip 7 — Two Trains Crossing
When two trains cross each other (opposite direction), time = (L₁+L₂)/(S₁+S₂). For same direction, time = (L₁+L₂)/|S₁−S₂|. Keep lengths in metres and speeds in m/s.
Example 6 — Train + Platform
A 240 m long train crosses a 360 m long platform in 40 seconds. Find the speed of the train in km/h.
Solution: Total distance = 240 + 360 = 600 m. Time = 40 s. Speed = 600/40 = 15 m/s = 15 × 18/5 = 54 km/h.
Example 7 — Train Passing a Pole
A 180 m long train passes a pole in 9 seconds. What is its speed in km/h?
Solution: Distance = 180 m (only train length). Time = 9 s. Speed = 180/9 = 20 m/s = 20 × 18/5 = 72 km/h.
Example 8 — Two Trains Crossing
Two trains of length 200 m and 300 m run towards each other at 54 km/h and 72 km/h. How long will they take to cross each other?
Solution: Relative speed = 54 + 72 = 126 km/h = 126 × 5/18 = 35 m/s. Total length = 200 + 300 = 500 m. Time = 500/35 = 100/7 ≈ 14.3 seconds.
5. Boats & Streams
Still water speed of a boat is its speed when there is no current. Stream speed adds or subtracts depending on direction.
Downstream (D): Speed along the current = Boat speed + Stream speed.
Upstream (U): Speed against the current = Boat speed − Stream speed.
Boat speed (in still water): (D + U) / 2.
Stream speed: (D − U) / 2.
Tip 8 — Boats Formula Memory
Downstream: D = B + S (both help). Upstream: U = B − S (boat minus stream). Boat speed = (D+U)/2 (average). Stream speed = (D−U)/2 (half the difference).
Example 9 — Downstream & Upstream Time
A boat's speed in still water is 12 km/h and stream speed is 4 km/h. How long will it take to go 48 km downstream and return?
Solution: Downstream speed = 12 + 4 = 16 km/h. Time downstream = 48/16 = 3 h. Upstream speed = 12 − 4 = 8 km/h. Time upstream = 48/8 = 6 h. Total = 9 h.
Example 10 — Find Stream Speed
A boat covers 36 km downstream in 6 hours and 36 km upstream in 9 hours. Find the speed of the stream.
Solution: Downstream speed = 36/6 = 6 km/h. Upstream speed = 36/9 = 4 km/h. Stream speed = (6 − 4)/2 = 1 km/h.
6. Circular Track Problems
On a circular track of length L, two runners start from the same point.
Meeting for the first time (opposite direction): Time = L / (S₁ + S₂).
Meeting for the first time (same direction): Time = L / |S₁ − S₂|.
Meeting at the starting point again: LCM of individual lap times = LCM(L/S₁, L/S₂).
Tip 9 — Circular Track LCM
To find when runners meet again at the starting point, compute time each takes for one lap (L/S), then take LCM of those times. This is a common SSC CGL Tier 2 question.
Example 11 — Circular Track
Two runners start from the same point on a 400 m circular track at 8 m/s and 12 m/s. When will they meet for the first time if they run in opposite directions? Same direction?
Solution: Opposite: relative speed = 8 + 12 = 20 m/s. Time = 400/20 = 20 s. Same direction: relative speed = 12 − 8 = 4 m/s. Time = 400/4 = 100 s.
7. Race Problems
In races, one runner may give another a head start or beat them by a distance. The key ratio is distance : time = speed.
Tip 10 — Race Problems
When A beats B by d metres in a race of L metres, in the time A covers L, B covers (L − d). Use this proportion to find speeds or times.
Example 12 — Race Problem
In a 1000 m race, A beats B by 200 m. Find the ratio of their speeds.
Solution: When A runs 1000 m, B runs 800 m in the same time. Speed ratio = Distance ratio = 1000 : 800 = 5 : 4.
8. Clock Problems as Speed Problems
The minute hand and hour hand move at constant speeds. Minute hand speed = 6° per minute. Hour hand speed = 0.5° per minute. Relative speed = 5.5° per minute.
Tip 11 — Clock Hand Speeds
Minute hand: 360° in 60 min = 6°/min. Hour hand: 360° in 720 min = 0.5°/min. Relative speed = 5.5°/min. Hands meet every 360/5.5 = 720/11 = 65 5/11 minutes.
Example 13 — Clock Problem
At what time between 2 and 3 o’clock will the hands of a clock be together?
Solution: At 2:00, hour hand is 60° ahead. Relative speed = 5.5°/min. Time to cover 60° = 60/5.5 = 120/11 = 10 10/11 min. So time = 2:10:54.5 approximately.
9. SSC CGL Shortcuts & Tricks Compilation
Tip 12 — Average Speed (Equal Distances)
Average speed when distances are equal: 2ab/(a+b). This is the most heavily tested SSC CGL formula in this chapter.
Tip 13 — Average Speed (Equal Times)
Average speed when times are equal: (a+b)/2. Use when a body travels at speed a for t hours and speed b for the next t hours.
Tip 14 — Relative Speed Table
km/h to m/s: multiply by 5/18. Common values: 36→10, 54→15, 72→20, 90→25, 108→30, 126→35, 144→40. Memorise for instant recall.
Tip 15 — Train & Platform
If a train passes a platform in t seconds, speed = (L_train + L_platform)/t. To get km/h, multiply result by 18/5.
Tip 16 — Time to Overtake
If two bodies start simultaneously from the same point in the same direction, time to meet again on a circular track = L/|S₁−S₂|. Opposite direction = L/(S₁+S₂).
Tip 17 — Boat Speed Shortcut
If a boat takes t₁ hours downstream and t₂ hours upstream for the same distance D, boat speed = D(t₁+t₂)/(2 t₁ t₂). Stream speed = D(t₂−t₁)/(2 t₁ t₂).
Tip 18 — SSC CGL Exam Strategy
In the exam, always write the formula first, convert units to matching pairs (km with h or m with s), then plug numbers. This structured approach prevents 90% of careless errors.
10. Solved Examples — Advanced Level
Example 14 — Train + Man (Same Direction)
A 180 m long train travelling at 72 km/h overtakes a man walking at 6 km/h in the same direction. How long will the train take to pass the man?
Solution: Relative speed = 72 − 6 = 66 km/h = 66 × 5/18 = 55/3 m/s. Time = 180 ÷ (55/3) = 180 × 3/55 = 540/55 = 108/11 ≈ 9.82 seconds.
Example 15 — Two Trains Passing (Same Direction)
Two trains of length 160 m and 240 m run on parallel tracks at 60 km/h and 42 km/h in the same direction. How long will they take to cross each other?
Solution: Relative speed = 60 − 42 = 18 km/h = 18 × 5/18 = 5 m/s. Total length = 160 + 240 = 400 m. Time = 400/5 = 80 seconds.
11. Advanced Practice Problems
Advanced 1 — Average Speed Variation
A person travels from A to B at 30 km/h and returns at 20 km/h. Next day, he travels A to B at 40 km/h and returns at 30 km/h. Compare the average speeds.
Solution: Day 1 average = 2×30×20/50 = 24 km/h. Day 2 average = 2×40×30/70 ≈ 34.3 km/h.
Advanced 2 — Train Crossing a Tunnel
A 300 m long train enters a 500 m long tunnel at 72 km/h. How long from the moment the engine enters till the last coach exits?
Solution: Distance = 300 + 500 = 800 m. Speed = 72 × 5/18 = 20 m/s. Time = 800/20 = 40 seconds.
Advanced 3 — Boat Intermediate
A boat takes 9 hours downstream from A to B and 12 hours upstream. Stream speed is 2 km/h. Find AB.
Solution: Let boat speed = b. (b+2)/(b−2) = 12/9 = 4/3. 3b+6 = 4b−8 → b = 14 km/h. AB = (14+2)×9 = 144 km.
Advanced 4 — Circular Track Three Runners
Three runners start from the same point on a 600 m circular track at 6 m/s, 8 m/s, and 10 m/s. When will all three meet again at the starting point?
Solution: Lap times: 600/6=100 s, 600/8=75 s, 600/10=60 s. LCM(100,75,60)=300 s = 5 min.
Advanced 5 — Race with Head Start
In a 200 m race, A gives B a 20 m head start and beats B by 5 seconds. A's speed is 10 m/s. Find B's speed.
Solution: A’s time = 200/10 = 20 s. B covers 180 m in 25 s. B’s speed = 180/25 = 7.2 m/s.
Advanced 6 — Relative Speed with Head Start
Train leaves Delhi at 6:00 AM at 60 km/h. Another train leaves at 7:30 AM at 75 km/h same direction. At what distance from Delhi will the second overtake the first?
Solution: Head start = 1.5×60 = 90 km. Relative speed = 15 km/h. Time = 90/15 = 6 h. Distance = 75×6 = 450 km.
Advanced 7 — Boat with Total Time
A man rows 24 km downstream and back in 10 hours. Stream speed is 2 km/h. Find his rowing speed in still water.
Solution: Let boat speed = b. 24/(b+2) + 24/(b−2) = 10. 24(2b)/(b²−4) = 10 → 48b = 10b²−40 → 5b²−24b−20=0 → (5b+4)(b−5)=0 → b=5 km/h.
Advanced 8 — Clock Angle
Find the angle between the hour and minute hands at 3:40.
Solution: Minute hand at 40 min = 240° from 12. Hour hand at 3:40 = 3×30 + 40×0.5 = 90+20 = 110°. Difference = 130°. Smaller angle = 130°.