Problems on Trains - Aptitude Tricks & Shortcuts & Formulas

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Quantitative aptitude questions on trains are a special case of time and distance problems. The only difference is that the length of the train and length of the object that the train crosses has to be considered to solve them. This is because the distance has to be covered by the entire length of train.

 

Basic Concepts to Solve Questions on Trains

In case the length of the object, either stationary (for e.g. a bridge or platform) or moving (another train) is not negligible,
`"Distance covered" = "length of object" + "length of train"`

This concept forms the basis of solving all problems on trains.

 

Handy Formulas for Solving Questions on Trains

  • Time taken to cross a moving object in the direction of train
    Given Length of object as `L_o`,
    Length of train as `L_t`
    Speed of moving object as `S_o` and
    Speed of train as `S_t`.
    `"Time taken to cross a moving object in the direction of train " =` `("Length of Train " + " Length of Object") / ("Speed of train " – " Speed of Object")`
    `t = (L_t + L_o) / (S_t – S_o)`
  • Special Cases of Basic Formula
    • If the length of the object is negligible, then `L_o = 0`
    • If the object is stationary, then `S_o = 0`
    • If the object moves in the opposite direction of train then relative speed is `"speed of train " + " speed of object"`.
      Since the directions are opposite we consider speed of object `- S_o`. Thus denominator of equation becomes `S_t + S_o`
    Now let's see these special cases in application
    1. Time taken by the train to cross a stationary object like a tree, electric pole, a standing man
      `t = L_t/S_t`
    2. Time taken by the train to cross a stationary object for e.g. a bridge or a platform of length `L_o`
      `t = (L_t + L_o)/S_t`
    3. Time taken by the train to cross a moving object like a walking man of negligible length
      When object moves in the same direction of train, `t = L_t/(S_t – S_o)`
      When object moves in the opposite direction of train, `t = L_t/(S_t + S_o)`
    4. Time taken by the train to cross a moving object like another moving train of length `L_o`
      When object moves in the same direction of train, `t = (L_t+L_o)/(S_t – S_o)`
      When object moves in the opposite direction of train, `t = (L_t+L_o)/(S_t + S_o)`
    5. Let two trains of lengths `L_1` and `L_2` takes `t_1 s` time to cross each other if they travel in opposite direction and `t_2 s` is they travel in same direction. Let `S_f` and `S_s` be the speed of the faster and slower train repectively, then
      `(S_f + S_s) = (L_1 + L_2) / t_1`
      `(S_f - S_s) = (L_1 + L_2) / t_2`
      Speed of faster train, `S_f = ((L_1 + L_2)/2) (1/t_1 + 1/t_2)`
      Speed of slower train, `S_s = ((L_1 + L_2)/2) (1/t_1 - 1/t_2)`

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