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:
475 comments Page 14 of 48.
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.
Nishant said:
9 years ago
Guys, we easily get 45 km/hr then we subtract from 45-5 then we get an answer 40 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.
Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr.
x - 5 = 45.
x = 50 km/hr.
Murugan MJ said:
1 decade ago
@Hemavathy. Since man is moving in same direction with train the relative speed is s-5 so,
s-5 = 45.
s = 50 km/hr.
Suppose if man move in opposite direction of train then,
s+5 = 45.
s = 40 km/hr.
s-5 = 45.
s = 50 km/hr.
Suppose if man move in opposite direction of train then,
s+5 = 45.
s = 40 km/hr.
Aayushi said:
1 decade ago
In 10 sec man will cover 125/9 m. In the relative motion distance should be added in either cases (opposite direction or same direction). So relative speed should be (125+(125/9))/10. Please explain.
Jaimin Patel said:
2 years ago
Let X km/hr be the speed of the train.
Length = 125m = 0.125km.
Relative Speed= (X-5)km/hr (as train and man moving in the same direction hence it is subtracted).
0125/(X-5) = 10/3600.
X = 50km/hr.
Length = 125m = 0.125km.
Relative Speed= (X-5)km/hr (as train and man moving in the same direction hence it is subtracted).
0125/(X-5) = 10/3600.
X = 50km/hr.
(26)
Chinnupradhan said:
2 years ago
S = L/T.
S = 125/10
S = 25/2.
then,
M/sec to km/hr convert.
1km/hr = [1000m/3600sec] = 5/18m/sec,
m/sec = 18/5km/hr.
i.e
(25/2 * 18/5)km/hr,
45km/hr.
Therefore first train x-5 = 45.
X = 50 km/hr.
S = 125/10
S = 25/2.
then,
M/sec to km/hr convert.
1km/hr = [1000m/3600sec] = 5/18m/sec,
m/sec = 18/5km/hr.
i.e
(25/2 * 18/5)km/hr,
45km/hr.
Therefore first train x-5 = 45.
X = 50 km/hr.
(30)
Rahul said:
2 decades ago
@gowtham
It would be only speed if the man(REFERENCE POINT) was stationary and only the train was moving. Here as both trian and REFERENCE POINT(man) are moving relative motion comes into picture.
It would be only speed if the man(REFERENCE POINT) was stationary and only the train was moving. Here as both trian and REFERENCE POINT(man) are moving relative motion comes into picture.
SHYAM said:
1 decade ago
Can anyone please explain me why it is divided by 18/5 in place of 5/18?
Reason is 1 km/hr = 1000 m/3600 sec = 10/36 = 5/18.
So why it is used 18/5 in place of 5/18?
Please help me understand?
Reason is 1 km/hr = 1000 m/3600 sec = 10/36 = 5/18.
So why it is used 18/5 in place of 5/18?
Please help me understand?
Kushal M V said:
4 years ago
They took 18/5 because the length of train is in meter and time given which passes the man is given in seconds. To convert m/s to km/ hr they multiplied with 18/5. And the answers also in km/hr.
(2)
Kushal M V said:
4 years ago
They took 18/5 because the length of train is in meter and time given which passes the man is given in seconds. To convert m/s to km/ hr they multiplied with 18/5. And the answers also in km/hr.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers