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 44 of 48.
Nithish said:
7 years ago
Thanks all for the explanations.
Mozammel hossain said:
6 years ago
There have man speed but not train speed.
So we known that;
speed = distanc/time.
= 125/10,
= 25/2.
Now convert into km/h.
25/2 * 18/5.
= 45km/h.
Lets train speed is x km/h.
So, the relative speed is x-5 km/h.
x- 5 = 45,
X = 50km/h.
So we known that;
speed = distanc/time.
= 125/10,
= 25/2.
Now convert into km/h.
25/2 * 18/5.
= 45km/h.
Lets train speed is x km/h.
So, the relative speed is x-5 km/h.
x- 5 = 45,
X = 50km/h.
Kunal Pardeshi said:
6 years ago
Hi. please solve my query.
Since the man is also moving, why the distance travelled by train in 10 seconds is considered equivalent to the length of the train when in actuality the distance covered will be more than that as the man is not stationary. Please, somebody explain this clearly. I want to know the concept and logic behind this.
Since the man is also moving, why the distance travelled by train in 10 seconds is considered equivalent to the length of the train when in actuality the distance covered will be more than that as the man is not stationary. Please, somebody explain this clearly. I want to know the concept and logic behind this.
P. Sudha said:
6 years ago
How 18/5 possible? Please explain to me.
K. Abhishek said:
6 years ago
@All.
Here, Train took 10 seconds to cross the person who is running at the speed of 5km/hr, then during this 10s the person might have travelled a certain distance, right? so train travelled (125m + distance covered by the man in 10s) in 10s then why the relative speed of train = 125/10?
How, will anyone explain it for me?
Here, Train took 10 seconds to cross the person who is running at the speed of 5km/hr, then during this 10s the person might have travelled a certain distance, right? so train travelled (125m + distance covered by the man in 10s) in 10s then why the relative speed of train = 125/10?
How, will anyone explain it for me?
Yugapooja said:
6 years ago
Thanks for the explanation @Rahul.
ABHISHEK said:
6 years ago
The speed of train with respect to ground=125/10=25/2 (m/sec).
Now to covert into km/hr:(25/2)*(3600/1000)=45 Km/hr.
Now there is a formula:
Velocity of train w.r.t. man(45 km/hr)=velocity of train w.r.t. ground(let it be x) - velocity of man w.r.t. ground(5km/hr).
So,
45 = x-5.
i.e. x = 50 Km/hr.
Now to covert into km/hr:(25/2)*(3600/1000)=45 Km/hr.
Now there is a formula:
Velocity of train w.r.t. man(45 km/hr)=velocity of train w.r.t. ground(let it be x) - velocity of man w.r.t. ground(5km/hr).
So,
45 = x-5.
i.e. x = 50 Km/hr.
Sai ganesh said:
6 years ago
as per my opinion The question is given in m/sec so the answer should be converted into 1 hr = 3600 sec and 1 km = 1000 m.
(3600/1000) m/sec as per given question.
125m and 10 sec,
(125/10)m/sec=45m/sec.
Here the man is moving in the same direction so we should subtract.
(x-5)=45.
x = 50km/hr.
(3600/1000) m/sec as per given question.
125m and 10 sec,
(125/10)m/sec=45m/sec.
Here the man is moving in the same direction so we should subtract.
(x-5)=45.
x = 50km/hr.
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.
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