Trains

Some points to remember:
1. Time taken by A train of length L metres to overtake a pole or a standing man or a signal post is equal to the time taken by the express train to cover L Metres.
2. Time taken by A train of length L metres to overtake a stationary object of length b metres is the time taken by the express train to cover (L + b) metres.
3. If two trains or two bodies are running in the same direction at u m/s and v m/s , where u > v, then their relative speed = (u – v) m/s.
4. Suppose two express trains or two bodies are running in opposite directions at u m/s and v m/s , then their relative speed = ( u + v) m/s.
5. If two express trains of length a metres and b metres are running in opposite directions at u m/s and v m/s, then time taken by the express trains to pass each other = (a+b) / (u+ v) sec.
6. If two express trains of length a metres and b metres are running in the same direction at u m/s and v m/s , then the time taken by the faster express train to pass the slower express train = ( a+b) / (u + v ) sec.
7. If two express train ( or bodies) start at the same time from points A and B towards each other and after passing they take a and b sec in reaching B and A respectively, then ( A speed) : ( B speed) = ( ?b : ?a ).

Example1: A. express train moves at (3/4)th its original speed. Due to this, it is 20 min late. Find the original time for the journey.
Method1:
Think about 2 diff. situations, 1st with accident and another without accident. As distance in both the cases is constant
So,
V₁/V2=T₂/T₁
=> V₁/[(3/4) * V₁]=(T₁+20)/T₁
=> 4/3=(T₁+20)/T₁
=> T₁=60

Method 2:
Velocity decreases by 25% (3/4 of original speed => decrement by 1/4) so time will increase by 33.3% (4/3 of original time
=> increase by 1/3
now, 33.3%=20 min
=> 100%=60 min