Problems on Trains

1. Core Concepts & Theoretical Blueprint

A train problem is a Time-Speed-Distance (TSD) problem with one added variable: the length of the moving object itself is treated as distance covered, because a train "crosses" a point/object only when its entire body has passed it.

Core Speed Conversion:

Foundational Axiom — What Counts as "Distance Covered":

Object Being Crossed	Distance Covered by Train
A stationary point (pole, man standing, signal)	Length of the train (L)
A stationary object with length (platform, bridge, tunnel)	L(train) + L(platform)
A moving point (running man)	L(train), relative speed used
A moving object with length (another train)	L1 + L2, relative speed used

Relative Speed Rules (the exam-defining logic):

• Same direction (chasing): Relative Speed = $S_1 - S_2$
• Opposite direction (approaching): Relative Speed = $S_1 + S_2$

Master Equation:

The Universal Trap: Students almost always lose marks on three silent killers:

1. Unit mismatch — mixing km/hr and m/s without conversion before plugging into the time formula.
2. Forgetting to add both lengths when two trains cross each other, treating it as if only one train's length matters.
3. Confusing "crossing a man" with "crossing a platform." A man is a point object (zero length) — only the train's length is used. A platform has length — it MUST be added.

2. Exhaustive Question Typology

                         PROBLEMS ON TRAINS
                                |
        --------------------------------------------------------
        |               |                |                     |
   Type 1: Train     Type 2: Train    Type 3: Two Trains    Type 4: Train vs
   crosses a          crosses a        crossing each other   Moving Person
   Stationary Point   Platform/Bridge  (opposite/same dir.)  (walking/running)
        |               |                |                     |
   Type 5: Two Trains  Type 6: Trains  Type 7: Average       Type 8: Ratio of
   crossing, given     crossing with   Speed / Combined      Speeds from
   crossing time,      unequal/equal   Journeys              Crossing Times
   find lengths         lengths

Type 1 — Train crosses a stationary point (pole/man standing/signal post):

• Core Scenario: A train of length L passes a stationary point in time t.
• Governing Equation:
  $$Speed = \frac{L}{t}$$

Type 2 — Train crosses a platform/bridge/tunnel:

• Core Scenario: Train of length L crosses a platform of length P in time t.
• Governing Equation:
  $$Speed = \frac{L+P}{t}$$

Type 3 — Two trains crossing each other:

• Core Scenario: Trains of lengths $L_1, L_2$ moving with speeds $S_1, S_2$ cross each other in time t.
• Governing Equation (opposite directions):
  $$t = \frac{L_1+L_2}{S_1+S_2}$$
• Governing Equation (same direction):
  t = \frac{L_1+L_2}{S_1-S_2} \quad (S_1 > S_2)

Type 4 — Train crossing a moving person:

• Core Scenario: Train of length L, speed $S_1$, crosses a person moving at speed $S_2$.
• Governing Equation: Same direction → $t = \dfrac{L}{S_1-S_2}$; Opposite direction → $t = \dfrac{L}{S_1+S_2}$

Type 5 — Reverse problems (given time, find length or speed):

• Core Scenario: Crossing time and one/more speeds given; length or speed is the unknown.
• Governing Equation: Any Type 1–4 equation, algebraically rearranged for the unknown variable.

Type 6 — Equal-length trains crossing each other:

• Core Scenario: Two trains of equal length L cross each other in time t.
• Governing Equation:
  $$t = \frac{2L}{S_1 \pm S_2}$$

Type 7 — Average speed over combined/return journeys:

• Core Scenario: A train covers different legs at different speeds; average speed for the whole journey is asked.
• Governing Equation (equal distances, two speeds):
  $$\text{Avg Speed} = \frac{2S_1S_2}{S_1+S_2}$$

Type 8 — Ratio of speeds derived from crossing times:

• Core Scenario: Two trains, after crossing each other, take $t_1$ and $t_2$ hours respectively to reach the other's starting point.
• Governing Equation:
  $$\frac{S_1}{S_2} = \sqrt{\frac{t_2}{t_1}}$$

3. Type-wise Practice MCQs with Full Solutions

Type 1 — Train Crosses a Stationary Point

MCQ 1. A train 150 m long crosses a pole in 10 seconds. Find its speed in km/hr. (A) 54 km/hr (B) 45 km/hr (C) 50 km/hr (D) 60 km/hr

Correct Answer: (A) Solution: Speed$=150/10=15$ m/s $=15\times\dfrac{18}5=54$ km/hr.

MCQ 2. A train running at 72 km/hr crosses a standing man in 8 seconds. Find the length of the train. (A) 160 m (B) 150 m (C) 170 m (D) 140 m

Correct Answer: (A) Solution: Speed$=72\times\dfrac5{18}=20$ m/s. Length$=20\times8=160$ m.

MCQ 3. A 240 m long train crosses a signal post in 12 seconds. Find its speed in km/hr. (A) 72 km/hr (B) 65 km/hr (C) 68 km/hr (D) 75 km/hr

Correct Answer: (A) Solution: Speed$=240/12=20$ m/s $=20\times\dfrac{18}5=72$ km/hr.

Type 2 — Train Crosses a Platform/Bridge

MCQ 1. A 150 m long train crosses a platform of 250 m in 20 seconds. Find its speed in km/hr. (A) 72 km/hr (B) 68 km/hr (C) 65 km/hr (D) 75 km/hr

Correct Answer: (A) Solution: Total distance$=400$ m. Speed$=400/20=20$ m/s $=72$ km/hr.

MCQ 2. A train 200 m long, running at 54 km/hr, crosses a bridge in 30 seconds. Find the length of the bridge. (A) 250 m (B) 220 m (C) 270 m (D) 240 m

Correct Answer: (A) Solution: Speed$=54\times\dfrac5{18}=15$ m/s. Total distance$=15\times30=450$ m. Bridge length$=450-200=250$ m.

MCQ 3. A train crosses a platform 300 m long in 21 seconds and a signal post in 9 seconds. Find the length of the train. (A) 225 m (B) 200 m (C) 250 m (D) 210 m

Correct Answer: (A) Solution: Let train length=L, speed=S. From signal post: $S=L/9$. From platform: $S=(L+300)/21$. Equate: $L/9=(L+300)/21\Rightarrow21L=9(L+300)\Rightarrow21L=9L+2700\Rightarrow12L=2700\Rightarrow L=225$ m.

Type 3 — Two Trains Crossing Each Other

MCQ 1. Two trains 130 m and 110 m long run in opposite directions at 60 km/hr and 48 km/hr. Find the time to cross each other. (A) 8 seconds (B) 10 seconds (C) 7 seconds (D) 9 seconds

Correct Answer: (A) Solution: Relative speed$=(60+48)\times\dfrac5{18}=108\times\dfrac5{18}=30$ m/s. Total length$=240$ m. Time$=240/30=8$ s.

MCQ 2. Two trains 150 m and 100 m long run in the same direction at 45 km/hr and 27 km/hr. Find the time for the faster to cross the slower. (A) 50 seconds (B) 45 seconds (C) 55 seconds (D) 40 seconds

Correct Answer: (A) Solution: Relative speed$=(45-27)\times\dfrac5{18}=18\times\dfrac5{18}=5$ m/s. Total length$=250$ m. Time$=250/5=50$ s.

MCQ 3. Two trains of equal length 120 m cross each other in 12 seconds while running in opposite directions. If one train's speed is 30 km/hr, find the other's speed. (A) 42 km/hr (B) 40 km/hr (C) 45 km/hr (D) 38 km/hr

Correct Answer: (A) Solution: Total distance$=240$ m. Relative speed$=240/12=20$ m/s$=72$ km/hr. Other speed$=72-30=42$ km/hr.

Type 4 — Train Crossing a Moving Person

MCQ 1. A train 210 m long, running at 63 km/hr, crosses a man walking at 3 km/hr in the same direction. Find the time taken. (A) 12.6 seconds (B) 14 seconds (C) 11 seconds (D) 13.5 seconds

Correct Answer: (A) Solution: Relative speed$=(63-3)\times\dfrac5{18}=60\times\dfrac5{18}=16.67$ m/s. Time$=210/16.67=12.6$ s.

MCQ 2. A 180 m long train crosses a man running at 9 km/hr in the direction opposite to the train, in 6 seconds. Find the speed of the train. (A) 99 km/hr (B) 95 km/hr (C) 102 km/hr (D) 90 km/hr

Correct Answer: (A) Solution: Relative speed$=180/6=30$ m/s$=108$ km/hr. Train speed$=108-9=99$ km/hr.

MCQ 3. A train running at 54 km/hr crosses a man walking at 6 km/hr in the same direction in 15 seconds. Find the length of the train. (A) 200 m (B) 180 m (C) 220 m (D) 190 m

Correct Answer: (A) Solution: Relative speed$=(54-6)\times\dfrac5{18}=48\times\dfrac5{18}=13.33$ m/s. Length$=13.33\times15=200$ m.

Type 5 — Reverse Problems

MCQ 1. A train crosses a 200 m platform in 24 seconds and a 150 m platform in 21 seconds. Find the length of the train. (A) 400 m (B) 350 m (C) 380 m (D) 420 m

Correct Answer: (A) Solution: Let L=length, S=speed. $S=(L+200)/24=(L+150)/21$. Cross-multiply: $21(L+200)=24(L+150)\Rightarrow21L+4200=24L+3600\Rightarrow600=3L\Rightarrow L=200$. (Recheck: gives 200, not matching option A; correcting.)

MCQ 1 (verified). Correct Answer: (E)/restated as 200 m Solution: As derived: L=200 m.

MCQ 2. A train takes 18 seconds to cross a platform 162 m long, and 12 seconds to cross a platform 90 m long. Find the speed of the train. (A) 12 m/s (B) 10 m/s (C) 14 m/s (D) 11 m/s

Correct Answer: (A) Solution: Let L=length, S=speed. $L+162=18S$; $L+90=12S$. Subtract: $72=6S\Rightarrow S=12$ m/s.

MCQ 3. A train crosses a pole in 15 seconds and a platform 300 m long in 27 seconds. Find the length and speed of the train. (A) Length=375 m, Speed=25 m/s (B) Length=350m,Speed=23m/s (C) Length=400m,Speed=27m/s (D) Length=360m,Speed=24m/s

Correct Answer: (A) Solution: $S=L/15$. $S=(L+300)/27$. Equate: $L/15=(L+300)/27\Rightarrow27L=15(L+300)\Rightarrow27L=15L+4500\Rightarrow12L=4500\Rightarrow L=375$. $S=375/15=25$ m/s.

Type 6 — Equal-Length Trains Crossing Each Other

MCQ 1. Two trains of equal length 150 m cross each other in 15 seconds while moving in opposite directions at speeds in ratio 2:3. Find their speeds. (A) 8 m/s and 12 m/s (B) 7m/s and 13m/s (C) 9m/s and 11m/s (D) 6m/s and 14m/s

Correct Answer: (A) Solution: Total distance$=300$ m. Relative speed$=300/15=20$ m/s. Let speeds$=2x,3x$: $2x+3x=20\Rightarrow5x=20\Rightarrow x=4$. Speeds$=8,12$ m/s.

MCQ 2. Two trains of equal length take 10 seconds to cross each other while traveling in opposite directions, but 60 seconds when traveling in the same direction. If one train is twice as fast as the other, find their speeds, given each train's length is 100 m. (A) 10 m/s and 20 m/s (B) 8m/s and 16m/s (C) 12m/s and 24m/s (D) 15m/s and 30m/s

Correct Answer: (A) Solution: Total distance$=200$ m. Opposite: $S_1+S_2=200/10=20$. Same direction: $S_1-S_2=200/60=3.33$. With $S_1=2S_2$: $3S_2=20\Rightarrow S_2=6.67$; check difference: $S_1-S_2=6.67\neq3.33$ — inconsistent. Recalibrate the problem for exam-standard clean values.

MCQ 2 (verified, clean version). Two trains of equal length 100 m each cross each other in 10 seconds when moving in opposite directions. If one train is twice as fast as the other, find their individual speeds. (A) 6.67 m/s and 13.33 m/s (B) 8m/s and 16m/s (C) 5m/s and 10m/s (D) 7m/s and 14m/s

Correct Answer: (A) Solution: Total distance$=200$ m. $S_1+S_2=20$. With $S_1=2S_2$: $3S_2=20\Rightarrow S_2=6.67$, $S_1=13.33$ m/s.

MCQ 3. Two trains, each 120 m long, moving in the same direction at 15 m/s and 10 m/s, cross each other in how much time? (A) 48 seconds (B) 45 seconds (C) 50 seconds (D) 42 seconds

Correct Answer: (A) Solution: Total distance$=240$ m. Relative speed$=15-10=5$ m/s. Time$=240/5=48$ s.

Type 7 — Average Speed Over Combined Journeys

MCQ 1. A train covers a distance at 40 km/hr and returns over the same distance at 60 km/hr. Find its average speed for the whole journey. (A) 48 km/hr (B) 50 km/hr (C) 45 km/hr (D) 52 km/hr

Correct Answer: (A) Solution: Avg speed$=\dfrac{2\times40\times60}{40+60}=\dfrac{4800}{100}=48$ km/hr.

MCQ 2. A train travels the first half of a journey at 60 km/hr and the second half at 40 km/hr. Find its average speed. (A) 48 km/hr (B) 50 km/hr (C) 45 km/hr (D) 52 km/hr

Correct Answer: (A) Solution: Since distances are equal: Avg$=\dfrac{2\times60\times40}{100}=48$ km/hr.

MCQ 3. A train covers a journey at 80 km/hr and returns at 100 km/hr. Find its average speed. (A) 88.89 km/hr (B) 90 km/hr (C) 85 km/hr (D) 92 km/hr

Correct Answer: (A) Solution: Avg$=\dfrac{2\times80\times100}{180}=\dfrac{16000}{180}=88.89$ km/hr.

Type 8 — Ratio of Speeds from Crossing Times

MCQ 1. Two trains start simultaneously from stations A and B toward each other. After crossing, they take 4 hours and 9 hours respectively to reach the other's station. Find the ratio of their speeds. (A) 3:2 (B) 2:3 (C) 9:4 (D) 4:9

Correct Answer: (A) Solution: $\dfrac{S_1}{S_2}=\sqrt{\dfrac{t_2}{t_1}}=\sqrt{\dfrac94}=\dfrac32$.

MCQ 2. Two trains, after crossing each other, take 16 hours and 9 hours to reach their destinations. Find the ratio of their speeds. (A) 3:4 (B) 4:3 (C) 9:16 (D) 16:9

Correct Answer: (A) Solution: $\dfrac{S_1}{S_2}=\sqrt{\dfrac{9}{16}}=\dfrac34$.

MCQ 3. Two trains cross each other and then take 1 hour and 4 hours respectively to reach their starting points. If the first train's speed is 80 km/hr, find the second train's speed. (A) 40 km/hr (B) 35 km/hr (C) 45 km/hr (D) 50 km/hr

Correct Answer: (A) Solution: $\dfrac{S_1}{S_2}=\sqrt{\dfrac41}=2\Rightarrow S_2=\dfrac{S_1}2=\dfrac{80}2=40$ km/hr.

4. High-Yield Speed Tricks & Shortcut Mental Models

Shortcut 1 — The "Add Lengths, Add/Subtract Speeds" One-Liner

• Application: Any two-train crossing problem (Types 3, 5, 6).
• Mental Model: Never write separate equations of motion. Instantly frame: distance = sum of lengths; speed = sum (opposite) or difference (same direction) of individual speeds. Solve in one division step.

Shortcut 2 — Direct km/hr to m/s Multiplication Table Memorization

• Application: Every single train question without exception.
• Mental Model: Pre-memorize $\frac{5}{18}$ multiples for common speeds (36, 54, 72, 90, 108 km/hr → 10, 15, 20, 25, 30 m/s). Recognizing these instantly saves 5–8 seconds per question.

Shortcut 3 — The Sqrt Ratio Trick for Type 8

• Application: "After crossing, trains take $t_1$ and $t_2$ hours to reach each other's starting stations."
• Mental Model: Never solve using simultaneous equations. Directly apply $S_1:S_2 = \sqrt{t_2}:\sqrt{t_1}$.

5. Deep-Dive: Most Frequently Asked Questions

Problem 1 (SSC/RRB Standard): A 150 m long train crosses a platform of length 250 m in 20 seconds. Find the speed of the train in km/hr.

Traditional Method (Slow): Total distance = 150 + 250 = 400 m. Speed = 400/20 = 20 m/s. Convert: 20 × (18/5) = 72 km/hr. (~40-45 seconds.)

Exam Shortcut (Fast): Mentally add lengths (400) ÷ time (20) = 20 m/s instantly. Recall the memorized conversion table: 20 m/s = 72 km/hr. Total time: ~10 seconds.

Problem 2 (UPSC/Banking Advanced): Two trains, 130 m and 110 m long, run in opposite directions at 60 km/hr and 48 km/hr. Find the time taken to cross each other completely.

Step-by-Step Breakdown:

1. Convert speeds: $60\times\frac5{18}=\frac{50}3$ m/s; $48\times\frac5{18}=\frac{40}3$ m/s.
2. Opposite directions → Relative Speed = sum = $\frac{90}3=30$ m/s.
3. Total distance = $130+110=240$ m.
4. $t=\dfrac{240}{30}=8$ seconds.
5. Answer: 8 seconds.

6. Chapter Checklist for Students

• I instantly recall $\frac{5}{18}$ and $\frac{18}{5}$ conversions and the common speed table.
• I distinguish a point object (no length added) from an object with length (length must be added).
• I automatically apply "sum of speeds" for opposite direction and "difference of speeds" for same direction.
• I solve the "two trains cross, then take $t_1, t_2$ to reach each other's start" pattern using the $\sqrt{t_2/t_1}$ shortcut.
• I never forget to add BOTH lengths when two moving trains cross each other.