Thursday, August 20, 2015

Maths Problems on Trains- Part 5th

Important Types questions on Trains Continues:-
Concept 3rd :- Relative speed Concept:-

Q13. A train running at the speed of 25 km/hr takes 18 secs. To pass a platform. Next, it takes 12 seconds to pass a man walking at 5 km/hr in the opposite direction. Find the length of the train and that of the platform.

Sol. Analyse the given data in the question:
Speed of the train = 25 km/hr.
Speed of the man = 5 km/hr.
Let the length of the train = l1
Length of the platform = l2
Distance travelled by train in passing the moving man = length of the train (l1)

Distance travelled by train in passing the platform = length of the {train (l1) + platform (l2)}
                                                                                    = (l1 + l2) meters.
Time taken by the train to cross the man, moving in opposite direction = 12 secs.
Relative speed while both are moving in opposite direction = 25 + 5
                                                                              = 30 km/hr.
                                                                                   = 30 × (5/18) = (25/3) m/sec.
Time taken by the train to pass the moving man = Distance/ relative speed
                                                                            12 = l1/(25/3)
                                                                             l1 = 12 × (25/3) =100 m.
length of the train = l1 = 100 metres.
Time taken by the train to cross the platform = 18 secs.
Speed of the train = 25 km/hr = 25 × (5/18) = (125/18) m/sec.
Speed of the train = Distance/time
              (125/18) = (l1 + l2)/18
                 (l1 + l2) = 125.
                100 + l2 = 125.
                          l2 = 125 – 100 = 25 m
Length of the platform = 25 m

Hence the required length of the train = 100 m and length of the platform = 25 m.

Concept 5th :- Combination of two different types of Questions

Q14. A train passes a pole in 20 seconds and passes a platform 100 m long in 30 seconds. Find the length and its speed.

Sol. Method 1st :- Analyse the given data in the question:
Let the speed of the train = x m/sec.
Let the length of the train = L m
Time taken to cross the pole = 20 secs.
Speed = distance/time
x = L/20
L = 20x
Length of the platform = 100 m
Distance travelled by the train to cross the platform = (L + 100) m
Time taken to cross the platform = 30 seconds.
Again speed = distance/time
x = (20x + 100)/30
30x = 20x +100
10x = 100
x = 10 m/sec
Now L = 20x = 20 × 10
L = 200m
Hence the length of the train = 200 m and speed of the train = 10 m/sec.

Method 2nd :- Concept :- In these type of questions the train covers its length in time t1 seconds and covers its length including the length of the platform in t2 seconds. This means it covers the length of the platform in (t2 – t1) seconds. Thus we can calculate the speed of the train direct.

Sol. Time taken by the train to cover its length (cross a pole) = 20 secs.
Time taken by the train to its length including platform (cross a platform) = 30 secs.
Hence the actual time taken by the train to cover the length of platform = 30 – 20 = 10 secs.
Length of the platform = distance travelled = 100 m
Speed of the train = distance / time.
Speed = 100/10 = 10 m/sec.
Length of the train = distance travelled in crossing the pole = speed × time
Length of the train = 20 × 10 = 200 m.

Hence the length of the train = 200 m and speed of the train = 10 m/sec.

(Note you may say that this is so long but dear this is the only way to explain the whole concept of the question to you now its upto you that how fast and short handedly you can solve these type of questions)

Hope you have enjoyed the base concept of this chapter till now. More Conceptual enjoyment with the Various Questions related to this chapter which are frequently asked in the examinations will be presented to your for self study on this blog soon, just keep visiting this blog regularly and writing me your related queries through comment or forum.
More types of questions will be followed soon on this blog........................

Best of Luck for CGL Exams. To be continued soon........................

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