Episode 8 — Aptitude and Reasoning / 8.12 — Problems on Trains

8.12.b Tips, Tricks, and Shortcuts -- Problems on Trains

Tip 1: Always Convert to m/s First

The number one mistake in train problems is a unit mismatch. Train lengths are in metres, so convert speed to m/s immediately.

    STEP 1 of every train problem:  Convert km/h to m/s

    Speed (m/s) = Speed (km/h) x 5/18

    Quick table:
    36 km/h  = 10 m/s         72 km/h  = 20 m/s
    45 km/h  = 12.5 m/s       90 km/h  = 25 m/s
    54 km/h  = 15 m/s        108 km/h  = 30 m/s
    60 km/h  = 50/3 m/s      120 km/h  = 100/3 m/s

Tip 2: The "What Crosses What" Decision Framework

Before writing any equation, answer these two questions:

    Question 1: What is the TOTAL DISTANCE to cover?

    +----------------------------------------------------+
    | Scenario               | Distance                  |
    |------------------------|---------------------------|
    | Train passes pole      | L_train                   |
    | Train passes platform  | L_train + L_platform      |
    | Train passes train     | L_train1 + L_train2       |
    | Train passes man on    |                           |
    |   another train        | L_passing_train ONLY      |
    +----------------------------------------------------+

    Question 2: What is the RELATIVE SPEED?

    +----------------------------------------------------+
    | Scenario               | Relative Speed            |
    |------------------------|---------------------------|
    | Object is stationary   | Speed of train            |
    | Same direction         | Difference of speeds      |
    | Opposite direction     | Sum of speeds             |
    +----------------------------------------------------+

    Then:  Time = Distance / Relative Speed

Tip 3: The Two-Equation Method

When a train crosses a pole in T1 seconds and a platform of length P in T2 seconds:

    Speed = P / (T2 - T1)        [Shortcut -- no need for two equations]

    Then:  Length of train = Speed x T1

    WHY THIS WORKS:
    The EXTRA time (T2 - T1) is spent covering the platform length P.
    So P / (T2 - T1) gives the speed directly.

Example:

    A train crosses a pole in 12 seconds and a 240 m platform in 24 seconds.

    Speed = 240 / (24 - 12) = 240 / 12 = 20 m/s
    Length = 20 x 12 = 240 m

Tip 4: Same Train, Two Platforms -- Direct Formula

    If a train crosses Platform 1 (length P1) in T1 seconds
    and Platform 2 (length P2) in T2 seconds:

    Speed = (P2 - P1) / (T2 - T1)        [if P2 > P1]

    Length = Speed x T1 - P1

Tip 5: Two Trains Crossing Each Other -- Speed Unknowns

If two trains cross each other in opposite directions in T_opp seconds and in the same direction in T_same seconds:

    S1 + S2 = (L1 + L2) / T_opp
    S1 - S2 = (L1 + L2) / T_same

    Adding:    2.S1 = (L1+L2)(1/T_opp + 1/T_same)
    Subtracting: 2.S2 = (L1+L2)(1/T_opp - 1/T_same)

Tip 6: "Man on Train" vs "Man on Platform" Trap

This is the most common trap in train problems.

    "A train crosses a man standing on a platform"
    --> Man is stationary. Distance = L_train. Speed = S_train.

    "A train crosses a man sitting in another train (opposite direction)"
    --> Man is a point. Distance = L_passing_train ONLY.
        Speed = S1 + S2 (opposite).

    "Two trains cross each other"
    --> Both have length. Distance = L1 + L2.
        Speed = S1 + S2 (opposite) or |S1 - S2| (same).

The key distinction:

  • "crosses a man" --> only the train's length matters
  • "crosses another train" --> both lengths matter

Tip 7: Ratio Shortcut for Same Train Crossing Different Objects

    If a train crosses a pole in T1 seconds and a platform in T2 seconds:

    L_train / (L_train + L_platform) = T1 / T2

    This gives: L_platform / L_train = (T2 - T1) / T1
                L_platform = L_train x (T2 - T1) / T1

Example:

    Train length = 200 m, crosses pole in 10 s, platform in 25 s.
    Platform length = 200 x (25-10)/10 = 200 x 1.5 = 300 m

Tip 8: Speed of Train When Crossing a Walking Man

    Man walking SAME direction as train:
    Speed of train = L / T + Speed of man     (in same units)

    Man walking OPPOSITE to train:
    Speed of train = L / T - Speed of man

    NOTE: L/T gives the relative speed. Add/subtract the man's speed
    to get the train's actual speed.

Tip 9: Finding Train Length from Crossing Times

If a train crosses two objects of known lengths in known times:

    L + P1     L + P2
    ------  =  ------    (same speed, so equate S)
      T1         T2

    Cross multiply: T2(L + P1) = T1(L + P2)
    T2.L + T2.P1 = T1.L + T1.P2
    L(T2 - T1) = T1.P2 - T2.P1
    L = (T1.P2 - T2.P1) / (T2 - T1)

Tip 10: Visualization Technique

Draw the scenario BEFORE writing equations. Use these ASCII templates:

    Pole problem:
    +---L---+
    |       |---->
    +-------+  |
               POLE

    Platform problem:
    +---L---+
    |       |====P====|
    +-------+         |

    Two trains (opposite):
    +---L1---+        +---L2---+
    |        |---->   |<-------|
    +--------+        +--------+

    Two trains (same direction):
    +---L1---+   +---L2---+
    |        |-->|        |-->
    +--------+   +--------+
    (faster)      (slower)

Tip 11: The "Passenger on the Train" Perspective

When a passenger ON a moving train watches something:

    Watching a pole:        Time = 0 (pole is a point, passes instantly)
                            Actually, the pole appears to zip past.

    Watching a platform:    Time = L_platform / S_train
                            (Only platform length matters)

    Watching another train (opposite): Time = L_other / (S1 + S2)
                            (Only OTHER train's length matters)

    Watching another train (same dir): Time = L_other / |S1 - S2|
                            (Only OTHER train's length matters)

Tip 12: Common Problem Patterns and Quick Approaches

Pattern 1: "Find the speed of the train"

    Usually given: Length + Time to cross something
    Approach: S = Total distance / Time

Pattern 2: "Find the length of the train"

    Usually given: Speed + Time to cross a pole
    Approach: L = S x T

Pattern 3: "Find the length of the platform"

    Usually given: Train length + Speed + Time to cross platform
    Approach: L_p = S x T - L_train

Pattern 4: "Two trains crossing, find time"

    Given: Both lengths + Both speeds + Direction
    Approach: T = (L1 + L2) / Relative speed

Pattern 5: "Find when trains meet between two stations"

    Given: Distance between stations + Both speeds
    Approach: T = D / (S1 + S2)  [opposite direction]

Tip 13: Mental Math for Common Train Speeds

Railway problems often use these standard speeds:

    Express train:   72 km/h = 20 m/s
    Passenger train: 54 km/h = 15 m/s
    Freight train:   36 km/h = 10 m/s
    Man walking:     5 km/h  = 25/18 m/s (approx 1.39 m/s)
    Man running:     10 km/h = 25/9 m/s (approx 2.78 m/s)

Tip 14: Double-Check With Reasonableness

    A train of 200 m at 72 km/h (20 m/s) crossing a pole:
    T = 200/20 = 10 seconds   (reasonable)

    If your answer is 200 seconds for the same problem,
    something is wrong (likely a unit error).

    Rule of thumb:
    - Crossing a pole: 5-30 seconds typically
    - Crossing a platform: 15-60 seconds typically
    - Crossing another train (opposite): 5-15 seconds
    - Crossing another train (same dir): 30-120 seconds

Tip 15: Working Backwards from Time

If the time is given and you need speed or length:

    From T = (L1 + L2) / S_rel:

    S_rel = (L1 + L2) / T     --> to find speed
    L1 = S_rel x T - L2       --> to find length of one train

Summary of Key Shortcuts

    1. ALWAYS convert to m/s first
    2. Pole: T = L/S
    3. Platform: T = (L+P)/S
    4. Two equations: Speed = P/(T2-T1)
    5. Same direction: use DIFFERENCE of speeds
    6. Opposite direction: use SUM of speeds
    7. "Man on train" = point observer = only other object's length
    8. "Two trains cross" = both lengths added
    9. Draw before you solve

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