Time, Speed and Distance — SSC CGL Quantitative Aptitude

Every TSD question is $d=s\times t$ wearing a disguise — a train, a boat, a thief. What SSC actually tests is whether you convert units instantly, whether you know an average speed is never the plain average, and whether you remember that a train must travel its own length past whatever it crosses. Four tools cover the whole chapter.

1. What SSC actually asks

Tier 1: 1–2 Q · Tier 2: 2–3 Q. The disguises: plain $d=st$ with a twist (late/early arrivals, fraction-of-speed walks), average speed round trips, trains crossing poles/platforms/each other, boats up- and downstream, and pursuit ("a thief is spotted…").

2. Tool 1 — the 5/18 conversion

$1\text{ km/h}=\frac{5}{18}\text{ m/s}\text{km/h}\stackrel{\times 5/18}{\to }\text{m/s}\stackrel{\times 18/5}{\to }\text{km/h}$

Train lengths come in metres, speeds in km/h — every train question begins with this conversion. Useful anchors: 36 km/h = 10 m/s, 54 = 15, 72 = 20, 90 = 25, 108 = 30.

3. Tool 2 — average speed (the harmonic trap)

Equal distances at speeds $x$ and $y$:

$s_{avg}=\frac{2xy}{x+y}(\text{NOT }\frac{x+y}{2})$

Go at 60, return at 40 → $\frac{2\cdot 60\cdot 40}{100}=48$, not 50. The average tilts toward the slower speed because you spend longer at it. Plain averaging of speeds is correct only when the times are equal — SSC lists the arithmetic mean as an option every single time.

4. Tool 3 — relative speed (trains and chases)

• Opposite directions: speeds add. Same direction: speeds subtract.
• A train crossing a pole/man covers its own length; crossing a platform/bridge covers own length + platform length; crossing another train covers the sum of both lengths at the relative speed.
• Pursuit: gap ÷ relative speed = catch-up time. A 200 m head start against a 2 km/h speed advantage closes in $0.2/2=0.1$ h = 6 minutes.

5. Tool 4 — ratios for late/early problems

Speed and time are inversely proportional for fixed distance. Walking at $\frac{3}{4}$ speed → time becomes $\frac{4}{3}$ → the extra $\frac{1}{3}$ of usual time equals the stated delay:

At $\frac{3}{4}$ speed, 10 min late → $\frac{1}{3}t=10$ → usual time 30 min. No equations.

Two-speed late/early: if speed $s_1$ makes you $t_1$ late and $s_2$ makes you $t_2$ early, then $\frac{d}{s_1}-\frac{d}{s_2}=t_1+t_2$ — one line to the distance.

6. Boats and streams

Boat speed $u$ (still water), stream $v$: downstream $u+v$, upstream $u-v$, and inverting:

$u=\frac{\text{down}+\text{up}}{2},v=\frac{\text{down}-\text{up}}{2}$

Downstream 10 km/h, upstream 6 → boat 8, stream 2. Round trips: compute each leg's time separately — never average the two speeds (harmonic trap again).

7. Solved PYQ-style examples

Q1. 54 km/h in m/s? Solution. $54\times \frac{5}{18}=$ 15 m/s.

Q2. A 150 m train at 54 km/h crosses a pole in… Solution. $150÷15=$ 10 s (a pole has no length — the train covers only itself).

Q3. A 240 m train at 72 km/h crosses a 360 m platform in… Solution. $(240+360)÷20=$ 30 s.

Q4. Trains 120 m and 180 m, at 42 and 66 km/h, opposite directions — time to cross? Solution. Relative $=108$ km/h $=30$ m/s; distance $=300$ m → 10 s.

Q5. At 40 km/h a man is 10 min late; at 50 km/h he is 5 min early. Distance? Solution. $\frac{d}{40}-\frac{d}{50}=\frac{15}{60}\Rightarrow \frac{d}{200}=\frac{1}{4}\Rightarrow d=$ 50 km.

8. Exam protocol

1. Convert units before anything else; keep the 36/54/72/90/108 anchors memorised.
2. Average speed of a round trip: harmonic ($2xy/(x+y)$) — the arithmetic mean option is always planted.
3. Train questions: ask "what total distance must the train cover?" — own length, plus platform, plus other train.
4. Late/early questions: go straight to time ratios (inverse of speed ratios); equations are the slow path.
5. Boats: resolve to $u+v$ and $u-v$ immediately; each leg gets its own time.