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 10 of 48.
Bikramjit said:
1 decade ago
1 km/hr -------------> 5/18 m/sec.
5/18 m/sec ------------> 1 km/hr.
1 m/sec -------------> 1/(5/18)km/hr = 18/5 km/hr.
25/2 m/sec -----------> (18/5)*(25/2) = 45 km/hr.
If train speed x km/hr, then (x-5)km/hr.
x-5 = 45.
x = 45+5.
x = 50 km/hr(answer).
5/18 m/sec ------------> 1 km/hr.
1 m/sec -------------> 1/(5/18)km/hr = 18/5 km/hr.
25/2 m/sec -----------> (18/5)*(25/2) = 45 km/hr.
If train speed x km/hr, then (x-5)km/hr.
x-5 = 45.
x = 45+5.
x = 50 km/hr(answer).
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 :).
Vishwa Telagadi said:
6 years ago
Man's speed=5 km/hr = 5 * 5/18 = (25/18)m/s.
The distance covered by man in 10 sec = (25/18)*10=(250/18)m,
Therefore distance covered by train for crossing the man=125+250/18=1250/9m,
Therefore speed=(1250/9)/10=125/9 m/s,
=(125/9)*3600/1000,
=(125/9)*(18/5)=50 km/hr.
The distance covered by man in 10 sec = (25/18)*10=(250/18)m,
Therefore distance covered by train for crossing the man=125+250/18=1250/9m,
Therefore speed=(1250/9)/10=125/9 m/s,
=(125/9)*3600/1000,
=(125/9)*(18/5)=50 km/hr.
Yuvraj said:
8 years ago
Speed=Distance(length of train)/Time;
s=125/10;
But s is (speed of train and man together i.e man's speed is 5km/hr so convert 5*(5/18)& train and man going in same direction that's why -X (let the speed of train)
(X*(5/18))-(5*(5/18))=125/10,
x-5=45,
x=50.
s=125/10;
But s is (speed of train and man together i.e man's speed is 5km/hr so convert 5*(5/18)& train and man going in same direction that's why -X (let the speed of train)
(X*(5/18))-(5*(5/18))=125/10,
x-5=45,
x=50.
Raja said:
1 decade ago
18/5 is to convert m/s to km/hr.
5/18 is to convert km/hr to m/sec.
1 km = 1000 m
1 hr = 3600 s.
M = 1 km/1000
s =1 hr/3600.
M/s = (3600/1000) km/hr.
= (5/18) km/hr.
------------------------------------------------.
Km/hr = (1000/3600) m/s.
= (5/18) m/s.
5/18 is to convert km/hr to m/sec.
1 km = 1000 m
1 hr = 3600 s.
M = 1 km/1000
s =1 hr/3600.
M/s = (3600/1000) km/hr.
= (5/18) km/hr.
------------------------------------------------.
Km/hr = (1000/3600) m/s.
= (5/18) m/s.
Mangaldeep Ghosh said:
6 years ago
Why ((a+b)/(u-v)) = t does not work?
Where a=length of the train (125 meters).
b= length of object( 0 meter),
u = speed of train ( let, x m/s),
t = time taken to cross the object,
v = speed of object (5 km/hr).
Where a=length of the train (125 meters).
b= length of object( 0 meter),
u = speed of train ( let, x m/s),
t = time taken to cross the object,
v = speed of object (5 km/hr).
Ashis said:
8 years ago
5 km/h convert into m/s =5*1000/3600 or 5*5/18 =1.38,
Distance traveled by man in 10 sec = 1.38*10=13.8,
Total distance traveling by train =125m +13.8m =138.8,
Final speed of train in 10 sec = 13.88m/s=49.968km/h.
Some error because of 1.38 is approximate.
Distance traveled by man in 10 sec = 1.38*10=13.8,
Total distance traveling by train =125m +13.8m =138.8,
Final speed of train in 10 sec = 13.88m/s=49.968km/h.
Some error because of 1.38 is approximate.
Praveen said:
1 decade ago
Friends its a simple logic that if a men travel along the same direction of the train then the train speed is subtracted if it travel in opposite direction then train speed is added.
Hope you understand take this logic in your mind and solve any problem.
Hope you understand take this logic in your mind and solve any problem.
Samcena said:
1 decade ago
I made a relation like this (125m/v1-5*5/18) =10 sec. But the answer is not coming ? I converted 5 km/hr to m/s. Then I put it in above equation.
I put all the terms in m and sec (not in km and hour). My answer is coming to be 30. How is it? Help me out
I put all the terms in m and sec (not in km and hour). My answer is coming to be 30. How is it? Help me out
Sonu kumar from bihar said:
7 years ago
@Suraj Kumar
It is X-5=45,
So, X=45+5=50.
@Bouvthika.
45km/hr is the relative speed of train w.r.t. the man running speed, but we let the actual speed of the train is X km/hr, so actual speed of train- man running speed = relative speed of the train.
It is X-5=45,
So, X=45+5=50.
@Bouvthika.
45km/hr is the relative speed of train w.r.t. the man running speed, but we let the actual speed of the train is X km/hr, so actual speed of train- man running speed = relative speed of the train.
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