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 27 of 48.
Suman said:
1 decade ago
Here given that (x-5) km/hr. What is 5? Please explain.
Sohail said:
1 decade ago
Why did you put 18/5 instead 5/18?
Where 5 denotes the meter and 18 denotes the hours (1000/3600) this should be as (25/2 x 5/18)?
I am a little unclear.
Where 5 denotes the meter and 18 denotes the hours (1000/3600) this should be as (25/2 x 5/18)?
I am a little unclear.
Velu said:
1 decade ago
Here they are asking the train speed. But why we are finding the relative speed.
Deepak said:
1 decade ago
How we where we divide and multiply by 5 and 18?
Abid said:
1 decade ago
How to calculate what Km/Hr is train running if its length is 130 m and cross it own start point in 15 seconds?
PPRakshit said:
1 decade ago
Speed of the train relative to man = 125/10.
How they wrote it?
How they wrote it?
MammuChowdary said:
1 decade ago
Moving in same direction : SUBTRACT.
Moving in opposite direction : ADD.
Moving in opposite direction : ADD.
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?
Jaimik chaudhary said:
1 decade ago
Distance travel by man in 10s = (5*1000*10)/(3600) = 13.89 m.
Now this much more distance the train has to travel due to mobile reference which is man.
Total distance travel by train = (125+13.89)m = 138.89 m.
Speed = (138.89*3600)/(10000) = 50 km/hr.
Comparison:
If I split the 138.89 into 125+13.89.
= 125/10+13.89/10 = 45+5 = 50 km/hr.
Now this much more distance the train has to travel due to mobile reference which is man.
Total distance travel by train = (125+13.89)m = 138.89 m.
Speed = (138.89*3600)/(10000) = 50 km/hr.
Comparison:
If I split the 138.89 into 125+13.89.
= 125/10+13.89/10 = 45+5 = 50 km/hr.
Sanal said:
1 decade ago
A different way to calculate this:
Length of the train: 125 m.
Distance traveled by man in 10 sec: 5*(5/18) m/s = 25/18 m for 1 sec.
i.e 4(25/18)*10 = 125/9 m for 10 sec.
Total distance traveled by train: 125 + 125/9 = 1250/9 m in ten seconds.
Speed for traveling 1250/9 m in 1 sec = (1250/9)/10 = 13.888.
In km = 13.888*(18/5) = 50 km/hr.
Length of the train: 125 m.
Distance traveled by man in 10 sec: 5*(5/18) m/s = 25/18 m for 1 sec.
i.e 4(25/18)*10 = 125/9 m for 10 sec.
Total distance traveled by train: 125 + 125/9 = 1250/9 m in ten seconds.
Speed for traveling 1250/9 m in 1 sec = (1250/9)/10 = 13.888.
In km = 13.888*(18/5) = 50 km/hr.
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