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 45 of 48.
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).
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.
Arun said:
6 years ago
How we say 45 km/hr is a relative speed?
Omsingh more said:
6 years ago
They should mention "relative speed" in the question. Why it is not so? It confuses the examiner!
Suresh said:
6 years ago
125m
5km/hr
10sec
Speed=distance/time.
Therefore S=km/hr(it's convert to m/sec)
S=1000m/3600sec
S=5/18.
Therefore S=D/T
D=125m
T=10sec
S=125/10
S=12.5, S=5/18.
Therefore 12.5=5/18=S(cross multiplication)
S = 12.5 * 18/5.
S = 225/5,
S = 45km/hr.
Remaining train running speed is added.
Therefore S = 45 + 5.
= 50km/hr.
5km/hr
10sec
Speed=distance/time.
Therefore S=km/hr(it's convert to m/sec)
S=1000m/3600sec
S=5/18.
Therefore S=D/T
D=125m
T=10sec
S=125/10
S=12.5, S=5/18.
Therefore 12.5=5/18=S(cross multiplication)
S = 12.5 * 18/5.
S = 225/5,
S = 45km/hr.
Remaining train running speed is added.
Therefore S = 45 + 5.
= 50km/hr.
Narinder singh said:
6 years ago
D = S X T.
(125 + 0) = (X-25/18) X 10 --> (i)
{Whereas 125 = length of the train, 0 = length of man, X is taken as the speed of the train, 25/18 is 5X5/18 to convert m/s into km/h, "-" Minus is used because they are running in the same direction, 10 is time to cross man}
Now solve the equation (i) and we will get the answer.
(125 + 0) = (X-25/18) X 10 --> (i)
{Whereas 125 = length of the train, 0 = length of man, X is taken as the speed of the train, 25/18 is 5X5/18 to convert m/s into km/h, "-" Minus is used because they are running in the same direction, 10 is time to cross man}
Now solve the equation (i) and we will get the answer.
Jatin said:
6 years ago
Why 5 - X not taken?
Sonali said:
6 years ago
If only one of the speed is given then how will we consider that which one is to be subtracted or not?
Siddharth said:
6 years ago
We 1st Find the speed of train relative to man Vr.
Man and train moved in the same direction = (V-U).
V = Speed of the train.
U = Speed of man.
Vr = (V-U).
Man and train moved in the same direction = (V-U).
V = Speed of the train.
U = Speed of man.
Vr = (V-U).
Pradeep said:
6 years ago
It's 50 m/s unit.
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