Problems On Trains 2
Problems on Trains 2 Aptitude tests will test your ability to perform tasks and react to situations at work. This includes problem-solving, prioritisation and numerical skills, amongst other things.Train problems form an integral part of the time and speed questions which are frequently asked in the quantitative aptitude section of various Government exams. These questions are different from the basic speed, distance and time questions and require a different approach to be answered.
Important Formulas For Problems on Trains 2
To solve any numerical ability question a candidate needs to memorise the related formulas to be able to answer the questions easily and efficiently.
Given below are the important train-based questions formulas which shall help candidates answer the questions based on this topic:
- km/hr to m/s conversion:a km/hr =a x5/18m/s.
- m/s to km/hr conversion:a m/s =a x18/5km/hr.
- Formulas for finding Speed, Time and Distance
- Time taken by a train of length l metres to pass a pole or standing man or a signal post is equal to the time taken by the train to cover l metres.
- Time taken by a train of length l metres to pass a stationery object of length b metres is the time taken by the train to cover (l + b) metres.
- Suppose two trains or two objects bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relative speed is = (u – v) m/s.
- Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s.
- If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then:The time taken by the trains to cross each other =(a + b)sec.(u + v)
- If two trains of length a metres and b metres are moving in the same direction at u m/s and v m/s, then:The time taken by the faster train to cross the slower train =(a + b)sec.(u – v)
- If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then:(A’s speed) : (B’s speed) = (b : a)