Time, Speed & Distance

high importance~2 Q in Tier 120 formulas⚡ 15 shortcuts5 subtopics

Overview Learn (5)Formulas (20)Examples (5)Practice (63)

Trains, relative speed, boats & streams, average-speed traps and races — two dependable slots per Tier 1 shift and more in Tier 2. Everything hangs on D = S×T, the 5/18 conversion, and the add/subtract rule for relative speed.

📖

Learn — 5 subtopics

One page per subtopic: detailed notes, every question type, formulas, tricks and practice sets.

⚡

Formula sheet

Every formula on one printable page, grouped by subtopic.

✍️

Solved examples

5 exam-level questions worked step by step.

🎯

Practice & test

63 questions — untimed practice or a timed test with analysis.

Track record in the exam

avg 1.0 Q / shift2024: 1–2 Q2025: 1 Q

Questions per shift in recent SSC CGL papers.

Test difficulty mix (63 questions)

20 easy33 medium10 hard

Question patterns exams keep repeating

Taken from previous-year papers. If a pattern is marked "very common", expect to see it in your exam.

Train crossing a pole or a man

very common

Spot it:

Length of train given, time to pass a point asked (or vice versa).

How to solve: $t = \frac{L}{S}$ with S in m/s — convert with $\frac{5}{18}$ first. A pole, man or signal is a point: only the train's own length passes it.

Example: A 150 m long train running at 90 km/h crosses a pole in:

$S = 90 \times \frac{5}{18} = 25$ m/s → $\frac{150}{25} = 6$ seconds.

Learn this in “Trains Crossing Poles, Platforms & Trains” →

Train crossing a platform, bridge or another train

very common

Spot it:

Two lengths involved; time for the train to completely cross.

How to solve: $t = \frac{L + P}{S}$ (platform) or $\frac{L_1 + L_2}{S_1 \pm S_2}$ (train). Pole + platform times together solve for both $L$ and $S$ .

Example: A train running at 54 km/h crosses a pole in 12 s and a platform in 30 s. The platform's length is:

$S = 15$ m/s, $L = 180$ m → platform $= 15 \times 30 - 180 = 270$ m.

Learn this in “Trains Crossing Poles, Platforms & Trains” →

Boats & streams

very common

Spot it:

Downstream/upstream times, boat or stream speed asked; sometimes two double trips.

How to solve: $u = b + s$ , $v = b - s$ ; recover with $b = \frac{u+v}{2}$ , $s = \frac{u-v}{2}$ . Two trip equations are linear in $\frac1u$ and $\frac1v$ .

Example: A boat goes downstream at 15 km/h and upstream at 10 km/h. The speed of the stream is:

$s = \frac{15 - 10}{2} = 2.5$ km/h (boat: 12.5 km/h).

Learn this in “Boats & Streams” →

Relative speed — meeting or overtaking

common

Spot it:

Two vehicles move towards or along each other; find meeting time or crossing time.

How to solve: Opposite: add the speeds; same direction: subtract. Distance = the gap (or the sum of both lengths for trains).

Example: Two towns 100 km apart send buses towards each other at 20 km/h and 30 km/h. They meet after:

$\frac{100}{20 + 30} = 2$ hours.

Learn this in “Relative Speed” →

Average speed for legs of a journey

common

Spot it:

Out-and-back or multi-leg trips; the trap is averaging the speeds.

How to solve: Total distance ÷ total time; equal legs ⇒ $\frac{2ab}{a+b}$ ; three equal legs ⇒ $3 \div (\frac1a+\frac1b+\frac1c)$ .

Example: A car goes out at 40 km/h and returns the same road at 60 km/h. The average speed is:

$\frac{2 \times 40 \times 60}{40 + 60} = 48$ km/h — not 50.

Learn this in “Speed, Distance, Time & Unit Conversion” →

Speed change with early/late arrival

common

Spot it:

Walking slower/faster makes the person late/early by known minutes.

How to solve: Same distance both ways: $\frac{d}{s_1} - \frac{d}{s_2} =$ delay in hours; a speed fraction $f$ multiplies the time by $\frac1f$ .

Example: Walking at $\frac34$ of his usual speed a man is 20 minutes late. His usual time is:

New time $= \frac43 t$ → late by $\frac{t}{3} = 20$ → $t = 60$ minutes.

Learn this in “Speed, Distance, Time & Unit Conversion” →

Races and starts

common

Spot it:

'A beats B by x m / t s', or start handicaps; find speeds or ratios.

How to solve: Distances covered in equal time are in the speed ratio: beats by $x$ m ⇒ $D : (D-x)$ ; chain margins multiplicatively.

Example: In a 200 m race A beats B by 20 m. The ratio of their speeds is:

$S_A : S_B = 200 : 180 = 10 : 9$ .

Learn this in “Races & Handicaps” →

Unit conversion embedded in other questions

common

Spot it:

Any train/m/s question quietly needs km/h ⇄ m/s.

How to solve: $\frac{5}{18}$ and $\frac{18}{5}$ — know the 36/54/72/90 table by heart.

Example: A 120 m train crosses a pole in 8 seconds. Its speed in km/h is:

$\frac{120}{8} = 15$ m/s $= 54$ km/h.

Learn this in “Speed, Distance, Time & Unit Conversion” →

Your next step

New here? Start with subtopic 1 in Learn. Revision mode? Jump straight to the test and let it tell you what to fix.