# Problems on Trains - Aptitude Tricks & Shortcuts & Formulas

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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