Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 2)
2.
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
Answer: Option
Explanation:
| Speed of the train relative to man = |  |
125 |  m/sec |
| 10 |
| = | |
25 | m/sec. |
| 2 |
| = | |
25 | x | 18 | km/hr |
| 2 | 5 |
= 45 km/hr.
Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr.
x - 5 = 45 
x = 50 km/hr.
Discussion:
478 comments Page 33 of 48.
Mubarak said:
9 years ago
Thank you guys for your best explanations really easily understandable, for the non maths learners. You people are doing good, good things never stop. Please keep on posting.
Pramod.M said:
9 years ago
Length(Distance) of the train 125m, time is 10 sec. The speed of man is 5 km/hr.
Speed = D/T i.e. 125/10 = 25/2m/s.
Therefore speed in km = 25/2 * 18/5 we got 45km/hr.
Moving in the same direction we must subtract i.e. 45 - 5 = 40 km/hr.
If the man moves against the train we must add 5km/hr to 45 km/hr.
Speed = D/T i.e. 125/10 = 25/2m/s.
Therefore speed in km = 25/2 * 18/5 we got 45km/hr.
Moving in the same direction we must subtract i.e. 45 - 5 = 40 km/hr.
If the man moves against the train we must add 5km/hr to 45 km/hr.
Jaydev ramnath said:
9 years ago
Why it is 18/5? Please explain.
Er.deepak said:
9 years ago
Man speed is 5km/hr.
So, km = 1000metrs.
Per hour = 60mintes = 3600second.
3600/1000 = 36/10 = 18/5.
Trains cross the man in 10 seconds.
Length of the train is =125m.
Relative man trains speed = 125/10 = 25/2.
So 25/2 x 18/5 = 45km/hr.
We let train speed is = x km/hr.
Relative to man turn speed = x - 5 = 45.
x = 45 + 5 = 50km/hr.
So, km = 1000metrs.
Per hour = 60mintes = 3600second.
3600/1000 = 36/10 = 18/5.
Trains cross the man in 10 seconds.
Length of the train is =125m.
Relative man trains speed = 125/10 = 25/2.
So 25/2 x 18/5 = 45km/hr.
We let train speed is = x km/hr.
Relative to man turn speed = x - 5 = 45.
x = 45 + 5 = 50km/hr.
Sathiyaraj said:
9 years ago
Since train and man run in the same direction,
So, in 10 seconds, train crosees 125 mts (train length) + 125/9 mts,
(5 km/hr => 5000/3600, 10 seconds => 125/9),
So train speed , 36(125 + 125/9) (10 seconds x 36 => 3600 seconds => 1 hr),
Answer : 50, 000 mts = 3600 secs => 50 km/hr.
So, in 10 seconds, train crosees 125 mts (train length) + 125/9 mts,
(5 km/hr => 5000/3600, 10 seconds => 125/9),
So train speed , 36(125 + 125/9) (10 seconds x 36 => 3600 seconds => 1 hr),
Answer : 50, 000 mts = 3600 secs => 50 km/hr.
David said:
9 years ago
@Jaydev Ramnath
To convert km/hr into m/s we should multiply with 18/5.
To convert km/hr into m/s we should multiply with 18/5.
Peter said:
9 years ago
Train Is 125 M.
Man Speed 5 KM/H.
Train Passed Man in 10 Seconds.
HR/KM = 3600 seconds/1000 MR = 36/10 divided both on 2 = 18/5.
Train Passed in 10 seconds with his tall 125 M so 125/10 divided on 5 = 25/2.
(18/5 * 25/2) = 45 KM/H speed + man speed 5 KM/H = 50 :).
Man Speed 5 KM/H.
Train Passed Man in 10 Seconds.
HR/KM = 3600 seconds/1000 MR = 36/10 divided both on 2 = 18/5.
Train Passed in 10 seconds with his tall 125 M so 125/10 divided on 5 = 25/2.
(18/5 * 25/2) = 45 KM/H speed + man speed 5 KM/H = 50 :).
Nandhini said:
9 years ago
The given speed is 5 km/hr. How it comes 18/5?
PREM KUMAR said:
9 years ago
What are the formulas used to find the time?
Eldhose said:
9 years ago
Time(t) = Distance(s)/Speed.
S = ut + .5at^2
v = u + at
2as = (v^2 - u^2)
Where, a = acceleration, v = final vel, u = ini.vel, s = dist, t = time.
In these train equation, v and a are taken as zero.
S = ut + .5at^2
v = u + at
2as = (v^2 - u^2)
Where, a = acceleration, v = final vel, u = ini.vel, s = dist, t = time.
In these train equation, v and a are taken as zero.
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