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Problems On Trains

 

 

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.

 
Important formulae and tricks and shortcuts of Problems 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

 

How to solve Problems on trains?

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 by step Method to solve time and distance question:

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.

 

Problems on train Shortcut tricks:

Lets discuss all formulas and tricks to solve problems,

Tricks 1: To convert km/hr to m/sec and m/sec to km/hr.

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?

Solution:

By formula:

x km/hr = x × 5÷18 m/sec, where x is 18.

18 km/hr = 18 × 5÷18 m/sec

= 5 m/sec answer

By derivation:

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

Tricks 2: A train running on S speed having L length and it is crossing a standing man then total time taken by train to cross the man is,

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.

Tricks 3: Relative Speed of train in Same Direction

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.

Tricks 4: Relative Speed of train in Opposite Direction

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.

Tricks 5: Total Distance

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.

Tricks 6: Time taken by two trains to cross each other (opposite direction)

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.

Tricks 7: Time taken by two trains to cross each other (same direction)

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.

Tricks 8: Crossing trains each other

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: