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Numerical Ability · AFCAT

NA07 — Time, Speed & Distance

🚀 Numerical Ability – NA07 AFCAT Level ★ High Priority
📌 AFCAT Focus 2022–2026: TSD gives 2–3 questions per paper — the highest in NA. Tested sub-topics: train crossing (platform/pole/another train), boats & streams (upstream/downstream), and relative speed. Master the unit conversion (×5/18 for km/h→m/s) and the three train-crossing cases — they account for over 60% of AFCAT TSD questions.

1. Speed, Distance & Time — Basics

Fig. 1.1 — The SDT Triangle: Derive Any Variable Instantly
D S × T Speed = D ÷ T km/h or m/s Distance = S × T km · m · miles Time = D ÷ S hours · min · sec 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
TRAINS

2. Train Problems — Three Cases

Fig. 2.1 — Train Crossing: What Distance Does the Train Cover?
TRAIN CROSSING — Distance for Each Case CASE 1 Pole / Person Distance = Length of Train only A pole is a point — train covers its own length to fully pass Time = L_train ÷ Speed CASE 2 Platform / Bridge Distance = Train length + Platform length Train must completely clear both its body and the platform Time = (L_train + L_platform) ÷ Speed CASE 3 Two Trains Distance = L₁ + L₂  (sum of both train lengths) Opposite direction → Relative speed = S₁ + S₂ Same direction → Relative speed = | S₁ − S₂ |
BOATS & STREAMS

3. Boats & Streams

Fig. 3.1 — Upstream & Downstream: Speed Relationships
← Stream flows this way → DOWNSTREAM ▶ Speed = b + s Boat & stream go same direction ⛵ ▶ ◀ UPSTREAM Speed = b − s Stream pushes against the boat ◀ ⛵ Boat speed (still water) b = (D + U) / 2 D = downstream, U = upstream Speed of stream s = (D − U) / 2 D = downstream, U = upstream

🏁 Relative Speed

  • Same direction: Relative speed = |S₁ − S₂|. Use when overtaking, gap closing
  • Opposite direction: Relative speed = S₁ + S₂. Use for head-on meeting
  • Time to meet: = Distance / Relative speed
  • Circular track: Same dir: meet when faster covers 1 extra lap; Opp dir: meet when combined covers 1 full lap

🏁 Race Problems

  • "A beats B by x metres": When A finishes the full distance, B is still x m short of the finish line
  • "A beats B by t seconds": B takes t seconds longer than A to complete the race
  • "A gives B a start of x metres": B begins from x m ahead of the starting line; A starts from 0
  • "A gives B a start of t seconds": B starts t seconds before A; both finish race of same distance
  • Dead heat: Both finish together — same time, same position
  • Speed ratio = Distance ratio (when time is the same for both)
  • Winning margin: In a race of L m, if A beats B by x m: when A covers L, B covers (L−x) m
  • Circular track: Same dir: meet when relative gain = 1 lap (L/|S₁−S₂|). Opposite dir: meet every L/(S₁+S₂)
✎ Worked Example — Train Crossing Platform
A train 240 m long passes a platform 160 m long in 20 seconds. Find the speed of the train.
Distance = Train + Platform = 240 + 160 = 400 m
Speed = Distance / Time = 400 / 20 = 20 m/s
In km/h: 20 × 18/5 = 72 km/h
✔ Speed = 72 km/h
✎ Worked Example — Boats & Streams
A boat goes 30 km upstream in 3 hours and 30 km downstream in 2 hours. Find the speed of the stream.
Upstream speed U = 30/3 = 10 km/h
Downstream speed D = 30/2 = 15 km/h
Speed of stream = (D − U)/2 = (15 − 10)/2 = 2.5 km/h
✔ Stream speed = 2.5 km/h
✎ Worked Example — Race Problem
In a 100 m race, A beats B by 10 m and C by 15 m. By how much does B beat C in the same race?
When A runs 100 m → B runs 90 m, C runs 85 m.
So when B runs 90 m, C runs 85 m.
When B runs 100 m, C runs: 85 × 100/90 = 94.44 m
B beats C by: 100 − 94.44 = 5.56 m ≈ 5⁵⁄₉ m
✔ B beats C by 5⁵⁄₉ metres

📐 Formula Sheet — NA07

Core SDT
S = D/T   D = S×T   T = D/S
km/h → m/s: ×5/18
m/s → km/h: ×18/5
Avg speed (equal dist): 2s₁s₂/(s₁+s₂)
Trains
Past pole: dist = L train
Past platform: dist = L train + L platform
Two trains: dist = L₁ + L₂
Opp dir: speed = S₁+S₂ | Same: |S₁−S₂|
Boats & Streams
Downstream: D = b + s
Upstream: U = b − s
Boat speed: b = (D+U)/2
Stream speed: s = (D−U)/2
Relative Speed
Same direction: |S₁ − S₂|
Opposite direction: S₁ + S₂
Circular (same dir): meet every L/|S₁−S₂| sec
Circular (opp dir): meet every L/(S₁+S₂) sec
Race Shortcuts
"A beats B by x m": B is x m short
"A gives B start of x": B starts x m ahead
Speed ratio = Distance ratio (same time)
Time ratio = Distance/Speed
Quick Numbers
36 km/h = 10 m/s
54 km/h = 15 m/s
72 km/h = 20 m/s
90 km/h = 25 m/s

📝 Topic-Wise PYQs — NA07

Q1. A train 200 m long passes a pole in 10 seconds. What is the speed in km/h? AFCAT PYQ
(a) 60 km/h(b) 72 km/h(c) 80 km/h(d) 54 km/h
✔ Answer: (b) 72 km/h
Speed = 200/10 = 20 m/s = 20×18/5 = 72 km/h. Passing a pole → distance = train length only.
Q2. Two trains 120 m and 80 m long run in opposite directions at 50 and 40 km/h. Time to cross? AFCAT PYQ
(a) 6 sec(b) 8 sec(c) 9 sec(d) 10 sec
✔ Answer: (b) 8 sec
Relative speed = 50+40 = 90 km/h = 90×5/18 = 25 m/s. Distance = 120+80 = 200 m. Time = 200/25 = 8 sec.
Q3. Downstream speed = 18 km/h, upstream = 10 km/h. Find speed of boat in still water. AFCAT PYQ
(a) 14 km/h(b) 12 km/h(c) 4 km/h(d) 8 km/h
✔ Answer: (a) 14 km/h
Boat speed = (D+U)/2 = (18+10)/2 = 14 km/h. Stream = (18−10)/2 = 4 km/h.
Q4. A man covers equal distances at 30 km/h and 60 km/h. Average speed? ⚡ Tricky
(a) 45 km/h(b) 40 km/h(c) 42 km/h(d) 50 km/h
✔ Answer: (b) 40 km/h
Equal distance → harmonic mean: 2×30×60/(30+60) = 3600/90 = 40 km/h. (45 = arithmetic mean — wrong for equal distances!)
Q5. In a race of 1 km, A beats B by 100 m. B beats C by 100 m. By how much does A beat C? ⚡ Tricky
(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. So when B runs 900 m, C runs 900×900/1000 = 810 m. A beats C by 1000−810 = 190 m.

🧠 Quick Memory Chart — NA07

⚡ Core SDT
  • S = D/T | D = S×T | T = D/S
  • km/h→m/s: ×5/18
  • m/s→km/h: ×18/5
  • Avg speed (equal D): 2s₁s₂/(s₁+s₂)
🚂 Trains
  • Pole: L train
  • Platform: L train + L platform
  • Two trains: L₁+L₂
  • Opp: S₁+S₂ | Same: |S₁−S₂|
⛵ Boats
  • Downstream: b + s
  • Upstream: b − s
  • b = (D+U)/2
  • s = (D−U)/2
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