Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 5)
5.
A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
Answer: Option
Explanation:
| Speed = |  |
54 x | 5 |  m/sec = 15 m/sec. |
| 18 |
Length of the train = (15 x 20)m = 300 m.
Let the length of the platform be x metres.
| Then, | x + 300 | = 15 |
| 36 |
x + 300 = 540
x = 240 m.
Discussion:
276 comments Page 2 of 28.
Sara said:
2 decades ago
We can take 15x20 as the train length only if the man is standing at the beginning of the platform. Am I correct.
Anj said:
2 decades ago
The answer is perfectly right and logical.
Rahul said:
2 decades ago
Length of the train is always calculated considering the train passes through a REFERENCE POINT. Here platform cannot be a REFERENCE POINT as it has some length. Thus man is taken as REFERENCE POINT to calculate the length of the train. Here train takes 20 seconds to cross REFERENCE POINT (man). Thus the train speed is multiplied by 20 seconds (15*20) to calculate the length of the train.
Amit Kr. Soni said:
2 decades ago
Finally I solve question in my own language.
Time to cross the man = 20 sec
Speed of train = 54 km/hr, = 54*5/18, 15 m/sec .......... (1)
Now length of train = 20*15 (fm distns = spd * time) = 300 mtr
Now let length (ditns) of pltfrm = x mtr.
Totl lenth (distns) = 300 + x
Spd = 15 m/sec ........from eqan(1)
According to question,
300 + x/15 = 36, [b'coz trn complet d totl distns in 36 sec]
So x = 540 - 300 = 240 Answer
Time to cross the man = 20 sec
Speed of train = 54 km/hr, = 54*5/18, 15 m/sec .......... (1)
Now length of train = 20*15 (fm distns = spd * time) = 300 mtr
Now let length (ditns) of pltfrm = x mtr.
Totl lenth (distns) = 300 + x
Spd = 15 m/sec ........from eqan(1)
According to question,
300 + x/15 = 36, [b'coz trn complet d totl distns in 36 sec]
So x = 540 - 300 = 240 Answer
Frank said:
1 decade ago
Whenever u get two timings subtract the highest from the lowest.
Eg: 36 - 20 = 16
16*15 (after converting it to m/sec) = 240.
Eg: 36 - 20 = 16
16*15 (after converting it to m/sec) = 240.
Nagashree said:
1 decade ago
Here man is standing in the platform, and we have to calculate the length of platform.
Hence we hav to multiply by 20 & not by36. Man in platform is reference point to us.
Hence we hav to multiply by 20 & not by36. Man in platform is reference point to us.
Remy said:
1 decade ago
How come (300+x) /15?its not said the man stands at the beginning of platform. If hes somwhere in the middle how come we add the length of the train 300 m ?
Gaurang said:
1 decade ago
Amit thanks for answer correctly.
Magesh said:
1 decade ago
Guys listen.
Take a point as man standing on the platform. So the moving train pass the point at the time of 20 sec.
So, Length of the train = (15 x 20)m = 300 m.
Take a point as man standing on the platform. So the moving train pass the point at the time of 20 sec.
So, Length of the train = (15 x 20)m = 300 m.
Karthik said:
1 decade ago
Shortcut:
Minus the both time of the train to cross the platform as well as man (reference point).
T = 36 sec, t = 20 sec.
T - t = 16 sec.
With the normal formula Length = Speed * Time
(i.e) 15 m/s * 16 sec = 240 metres.
Minus the both time of the train to cross the platform as well as man (reference point).
T = 36 sec, t = 20 sec.
T - t = 16 sec.
With the normal formula Length = Speed * Time
(i.e) 15 m/s * 16 sec = 240 metres.
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