Important formulae and tricks and shortcuts of Problems On Trains:
We are providing you Important Concepts and Tricks on Problems On Trains which are usually asked in almost all competitive exams like IBPS, Govt. Exams and CAT and in quantitative aptitude of MNCs. Use these below given tricks to solve questions within minimum time. These tricks will be very helpful for your upcoming Competitive Exam.
If you are looking for tricks to solve problems on train. here let me present how to approach the problems in train question and get the correct answer. This way you can solve problems very quickly.
Step 1: Find out all the values given in the question and note down.
Step 2: Draw a basic sketch of distance and train direction and write values on that.
Step 3: Find if train is crossing a station/bridge or it is crossing a standing man/pole.
Step 4: Covert all the values in single unit (either km/h or m/sec)
Step 5: If it is pole/standing man then count only train length as whole distance other wise add the bridge length + train length for total length.
Step 6: Try to find all the possible values from given values.
Step 7: Put all the finding in equation and get the desired answer.
Lets discuss all formulas and tricks to solve problems,
We know, km/hr is bigger value and m/sec is smaller so what we do to convert bigger value into smaller?
We divide by a value which has greater denominator than numerator. so to convert km/hr into m/sec we multiply by 5÷18, where 18 (denominator) is greater than 5 (numerator).
x km/hr = x × 5÷18 m/sec
Similarly, To convert m/sec to km/hr, we do the reverse process.
x × 5÷18 m/sec = x km/hr
Example: How to convert 18 km/hr into m/sec?
x km/hr = x × 5÷18 m/sec, where x is 18.
18 km/hr = 18 × 5÷18 m/sec
= 5 m/sec answer
first enlist the value of all the unit.
1 km = 1000 meter.
1 hr = 60 min = 60 x 60 seconds ⇒ 3600 seconds
Now, 18 km/hr = 18 × 1000 ÷ 3600 m/sec ⇒ 18 × 10 ÷ 36 m/sec
= 10 ÷ 2 ⇒ 5 m/sec answer
T = L/S
T = Time taken
L =Length of train
S= Speed of train
Note: Time taken by train ( L length) to cross the signal/standing man is equal to the time taken by the train to cover L distance.
If two trains/Objects are moving in the same direction at v1 m/s and v2 m/s respectively where v1>v2, then their relative speed (v1−v2) m/s.
If two trains/Objects are moving in the Opposite direction at v1 m/s and v2 m/s respectively, then their relative speed (v1+v2) m/s.
If train(length L) is crossing a standing man/pole then total distance traveled will be L.
If a train(length L1) is crossing a bridge/station(length L2) then total distance traveled will be L1 + L2.
If one train (length L1) is crossing another train (length L2) in same direction then total distance traveled will be L1 + L2.
If one train (length L1) is crossing another train (length L2) in opposite direction then total distance traveled will be L1 + L2.
If two trains of length a metres and b metres are moving in opposite directions at u m / s and v m/s, then time taken by the trains to cross each other = (a + b)/(u+v) sec.
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)/(u-v) sec.
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) =(√b1 : √a1 )
For solved problems on above formulas please visit below sections: