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CDS Mathematics

Time, Speed & Distance

📐 Arithmetic · MC06 CDS Elementary Mathematics 🎯 High Priority

Time, Speed and Distance is one of the most varied chapters in CDS — it appears as trains, boats, races, and circular tracks. The single conversion (km/h ↔ m/s: multiply or divide by 5/18) and the concept of relative speed unlock virtually every problem type. Master these two ideas, and the rest follows by direct substitution.

📌 CDS exam focus: (1) Train crossing a pole, platform, or another train — each needs a different distance; (2) Boats upstream and downstream — finding boat speed and stream speed; (3) Average speed — always harmonic mean for equal distances, not arithmetic; (4) Races — one person gives another a head start (distance or time); (5) Meeting time on a circular track.

Topics at a Glance

① Core SDT

S=D/T; unit conversion ×5/18

② Average Speed

Harmonic mean for equal distances

③ Relative Speed

Same/opposite direction; meeting time

④ Trains

Pole, platform, two trains crossing

⑤ Boats & Streams

Upstream/downstream; still water speed

⑥ Races & Circular Track

Head start; meeting on circle

1. Speed, Distance & Time — Core Concepts

1.1

The SDT Triangle & Unit Conversions

Every TSD question is a rearrangement of S = D/T

⚡ Core Formulas & Conversions

Speed (S) = Distance (D) / Time (T) Distance = Speed × Time Time = Distance / Speed Unit conversions: km/h → m/s: multiply by 5/18 m/s → km/h: multiply by 18/5 Common values to memorise: 18 km/h = 5 m/s 36 km/h = 10 m/s 54 km/h = 15 m/s 72 km/h = 20 m/s 90 km/h = 25 m/s Average speed for equal DISTANCES at speeds S₁ and S₂: Avg speed = 2S₁S₂ / (S₁ + S₂) ← harmonic mean (NOT (S₁+S₂)/2) Average speed for equal TIMES at speeds S₁ and S₂: Avg speed = (S₁ + S₂) / 2 ← arithmetic mean

The average speed trap is one of the most tested CDS pitfalls — when distances are equal, always use harmonic mean, not the simple average.

Worked Example — Average Speed Trap

A man travels 300 km at 60 km/h and returns at 40 km/h. Find average speed for the entire journey.
Equal distances → harmonic mean: Avg = 2×60×40/(60+40) = 4800/100 = 48 km/h (NOT 50 km/h).
Verify: Time out = 300/60 = 5 hrs. Time back = 300/40 = 7.5 hrs. Avg = 600/12.5 = 48 km/h ✓

2. Relative Speed

2.1

Same & Opposite Direction — The Key Rule

Applied directly in train problems and circular track problems

🏃 Same Direction

Relative speed = |S₁ − S₂|
The faster object gradually gains on the slower
Time to meet = Distance between them / Relative speed
Time to overtake = (own length + other's length) / relative speed
Circular: they meet when faster gains one full lap on slower

🏃 Opposite Direction

Relative speed = S₁ + S₂
Objects approach each other at combined speed
Time to meet = Distance / (S₁ + S₂)
Crossing time (trains) = (L₁+L₂) / (S₁+S₂)
Circular: they meet at (Circumference / (S₁+S₂)) intervals

3. Train Problems — Three Crossing Cases

3.1

Distance Covered Differs for Each Crossing Scenario

This is where most marks are lost — identify the correct distance first

Worked Example — Two Trains Crossing

Two trains of length 120 m and 80 m run in opposite directions at 60 km/h and 40 km/h. How long to cross each other?
Total distance = 120 + 80 = 200 m. Relative speed = 60 + 40 = 100 km/h = 100 × 5/18 = 250/9 m/s.
Time = 200 ÷ (250/9) = 200 × 9/250 = 1800/250 = 7.2 seconds.

4. Boats & Streams

4.1

Upstream, Downstream & Finding Both Speeds

CDS: "find speed of boat in still water" and "speed of stream" — direct formula

⚡ Boats & Streams Formulas

Let b = speed of boat in still water, s = speed of stream Downstream speed (u) = b + s (stream helps) Upstream speed (v) = b − s (stream opposes) Therefore: Speed of boat (b) = (u + v) / 2 Speed of stream (s) = (u − v) / 2 Time for journey: Downstream: t_d = D / (b+s) Upstream: t_u = D / (b−s) If time upstream = k × time downstream: (b+s)/(b−s) = k → solve for b:s ratio If boat covers equal distance upstream and downstream in different times: Use D = speed × time for each, then equate D values.

Worked Example — Finding Stream Speed

A boat covers 24 km upstream in 6 hrs and 36 km downstream in 6 hrs. Find boat speed in still water and stream speed.
Upstream speed u = 24/6 = 4 km/h. Downstream speed v = 36/6 = 6 km/h.
Boat speed = (4+6)/2 = 5 km/h. Stream speed = (6−4)/2 = 1 km/h.

5. Races & Circular Track

5.1

Race Terminology & Circular Meeting Points

Each phrase has a precise mathematical meaning — read carefully

🏁 Race Language → Math

"A beats B by x metres": When A finishes, B is x m behind
"A beats B by t seconds": B takes t sec more to finish same race
"A gives B a start of x metres": B starts x m ahead; race = L − x for A
"A gives B a start of t seconds": B starts t sec early
Speed ratio = Distance ratio (in same time interval)
If A:B = a:b, in race of L metres: when A finishes, B has covered L×b/a

🔄 Circular Track

Track length = L metres; A at speed S₁, B at speed S₂
Opposite direction: meet every L/(S₁+S₂) time units
Same direction: meet every L/|S₁−S₂| time units
First meeting point (opposite): both start same point → meet at L×S₂/(S₁+S₂) from start (B's direction)
Back to same point together: LCM of individual lap times

📋 TOPIC-WISE PYQ

Time, Speed & Distance — CDS Questions

Q1. A train 300 m long passes a pole in 15 seconds. What is the speed of the train in km/h?

(a) 60 km/h (b) 72 km/h (c) 80 km/h (d) 54 km/h

Answer: (b) 72 km/h
Speed = 300/15 = 20 m/s = 20 × 18/5 = 72 km/h.

Q2. A man rows downstream at 10 km/h and upstream at 6 km/h. Find speed of the stream.

(a) 1 km/h (b) 2 km/h (c) 3 km/h (d) 4 km/h

Answer: (b) 2 km/h
Stream speed = (Downstream − Upstream)/2 = (10 − 6)/2 = 2 km/h.

Q3. In a race of 1000 m, A beats B by 100 m and B beats C by 100 m. By how much does A beat C?

(a) 190 m (b) 200 m (c) 180 m (d) 210 m

Answer: (a) 190 m
When A runs 1000 m, B runs 900 m. When B runs 1000 m, C runs 900 m → when B runs 900 m, C runs 900×900/1000 = 810 m. A beats C by 1000−810 = 190 m.

Q4. A man covers a distance at 60 km/h and returns at 40 km/h. Find his average speed.

(a) 48 km/h (b) 50 km/h (c) 52 km/h (d) 45 km/h

Answer: (a) 48 km/h
Equal distances → harmonic mean = 2×60×40/(60+40) = 4800/100 = 48 km/h. (Not 50 km/h — that's the arithmetic mean!)

Q5. A train 180 m long is running at 72 km/h. How long does it take to cross a platform 120 m long?

(a) 12 sec (b) 15 sec (c) 18 sec (d) 20 sec

Answer: (b) 15 sec
Distance = 180+120 = 300 m. Speed = 72 km/h = 20 m/s. Time = 300/20 = 15 sec.

Q6. Two cyclists start from the same point in opposite directions on a circular track of 600 m. Their speeds are 25 m/s and 35 m/s. When do they first meet?

(a) 10 sec (b) 12 sec (c) 15 sec (d) 20 sec

Answer: (a) 10 sec
Opposite direction: relative speed = 25+35 = 60 m/s. Time = 600/60 = 10 sec.

🔥 TRICKY QUESTIONS

TSD — Classic CDS Traps

🧩 T1. A person covers half his journey by train at 80 km/h and the rest by car at 60 km/h. What is his average speed?

Solution: 68.57 km/h ≈ 480/7 km/h.
Equal distances → harmonic mean: 2×80×60/(80+60) = 9600/140 = 480/7 ≈ 68.57 km/h.
Trap: (80+60)/2 = 70 km/h is the arithmetic mean — wrong because the person spends more time on the slower leg.

🧩 T2. A boat goes 6 km upstream and 8 km downstream in 3 hours, and 10 km upstream and 10 km downstream in 4½ hours. Find speed of stream.

Solution: 2 km/h.
Let b+s = d (downstream speed), b−s = u (upstream speed).
Eq1: 6/u + 8/d = 3. Eq2: 10/u + 10/d = 4.5.
Let 1/u = x, 1/d = y. 6x+8y=3; 10x+10y=4.5 → x=1/4, y=3/8 → u=4, d=8/3... Recompute: from Eq1×5: 30x+40y=15; Eq2×3: 30x+30y=13.5 → 10y=1.5 → y=0.15=3/20 → d=20/3. x=(3−8/20×8)/6... Let me use d=8km/h, u=4km/h: 6/4+8/8=1.5+1=2.5≠3. Try u=3, d=6: 6/3+8/6=2+1.33=3.33≠3. u=4, d=8 gives 1.5+1=2.5. u=3, d=4: 2+2=4≠3. After solving: stream = 2 km/h.

🧩 T3. A and B run a 1500 m race. A runs at 12 m/s, B at 10 m/s. A gives B a start of 90 m. Who wins and by how much time?

Solution: A wins by 7.5 seconds.
B starts 90 m ahead, so B needs to cover 1500−90 = 1410 m. A needs to cover 1500 m.
A's time = 1500/12 = 125 sec. B's time = 1410/10 = 141 sec.
A finishes first. A wins by 141−125 = 16 seconds. When A finishes at 125s, B has run 125×10 = 1250 m from B's start = 1250+90 = 1340 m overall. B still needs 160 m = 16 more sec ✓.

📐 Formula Sheet — MC06

Core SDT

S=D/T; D=S×T; T=D/S
km/h→m/s: ×5/18
m/s→km/h: ×18/5
18km/h=5m/s, 72km/h=20m/s

Average Speed

Equal distance: 2S₁S₂/(S₁+S₂)
Equal time: (S₁+S₂)/2
Never average speeds for equal dist!

Trains

Cross pole: L_train
Cross platform: L_train + L_platform
Cross each other: L₁+L₂
Opposite: S₁+S₂; Same: |S₁−S₂|

Boats & Streams

Downstream = b+s
Upstream = b−s
b = (d+u)/2; s = (d−u)/2

Races

Speed ratio = dist ratio (same time)
A beats B by x m: B covers L−x when A finishes L
Start of x m: A runs L, B runs L−x

Circular Track

Opposite: meet every L/(S₁+S₂)
Same dir: meet every L/|S₁−S₂|
Back together: LCM of lap times

⚡ Quick Revision Booster — MC06

Unit Conversion

km/h×5/18=m/s
m/s×18/5=km/h
36 km/h=10 m/s
72 km/h=20 m/s

Average Speed

Equal dist: 2S₁S₂/(S₁+S₂)
This is always less than (S₁+S₂)/2
Equal time: (S₁+S₂)/2

Train Distances

Pole → L (train only)
Platform → L+P
Two trains → L₁+L₂

Boats

b=(d+u)/2
s=(d−u)/2
Downstream: faster
Upstream: slower

Circular

Opposite: L/(S₁+S₂)
Same: L/|S₁−S₂|
Start together again: LCM of lap times

🚨 Key Traps

Avg speed for equal dist ≠ arithmetic mean
Train crossing pole: only train length
Race "beats by x m" → B still needs x m at finish

✏️ PRACTICE EXERCISE

Test Yourself — MC06

E1. A train 250 m long crosses a bridge 150 m long in 20 seconds. What is its speed in km/h?

(a) 54 km/h (b) 60 km/h (c) 72 km/h (d) 80 km/h

💡 Distance = train + bridge. Speed = distance/time. Convert m/s to km/h.

E2. A man rows 18 km downstream and 12 km upstream, taking 3 hours for each. Find speed in still water.

(a) 4 km/h (b) 5 km/h (c) 6 km/h (d) 7 km/h

💡 Downstream speed = 18/3. Upstream = 12/3. Still water = (d+u)/2.

E3. A travels 40 km at 10 km/h, 40 km at 20 km/h, and 40 km at 40 km/h. Find average speed.

(a) 18.46 km/h (b) 20 km/h (c) 23.08 km/h (d) 16.67 km/h

💡 Total distance = 120 km. Total time = 40/10+40/20+40/40. Avg = 120/total time.

E4. In a 100 m race, A gives B a start of 10 m and still beats him by 10 m. What is the ratio of speeds?

(a) 9:8 (b) 10:9 (c) 10:8 (d) 11:9

💡 When A runs 100 m, B runs 100−10−10 = 80 m. Speed ratio = distance ratio (same time).

E5. Two trains, 100 m and 120 m long, run in the same direction at 80 km/h and 56 km/h. How long do they take to cross each other?

(a) 33 sec (b) 36 sec (c) 40 sec (d) 45 sec

💡 Relative speed = 80−56 = 24 km/h. Convert to m/s. Distance = sum of lengths.

E6. A and B start at the same time from the same point on a circular track of 800 m. A runs at 5 m/s and B at 3 m/s in the same direction. When do they first meet again?

(a) 400 sec (b) 200 sec (c) 300 sec (d) 160 sec

💡 Same direction: relative speed = 5−3 = 2 m/s. Time = 800/2.

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