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 16 of 48.
Gayu said:
1 decade ago
Why its divided by 15/8? need some brief explanation.
Akansha said:
1 decade ago
18/5 is used to change meter per second to kilometer per hour.
And I have no idea why it says -5 as it should be +5.
And I have no idea why it says -5 as it should be +5.
Veena said:
1 decade ago
Why train speed has to be subtracted with man speed?
Navaneedhar said:
1 decade ago
Please explain the subtract value of (x-5)?
Arjun said:
1 decade ago
Speed of man*time = 5*10 = 50.
Iyyanar said:
1 decade ago
Please give a clear cut definition for relative velocity.
Aditya said:
1 decade ago
The train and man are moving in the same direction so. If we want to take out the speed of any one of them, then we have to take one object as stationery.
Roopa said:
1 decade ago
Hi friends,
Here length of train = speed*time.
Likewise speed of train = length/time.
-> 125/10 = (25/2)m/sec.
Now given 5km/hr.
So speed is (x-5)km/hr or (x-5)(5/18).
Here we goes -> (25/2)m/sec = (x-5)(5/18).
-> (25/2)*(18/5) = (x-5).
-> 45 = (x-5).
-> x = 45+5.
-> x = 50 km/hr.
Here length of train = speed*time.
Likewise speed of train = length/time.
-> 125/10 = (25/2)m/sec.
Now given 5km/hr.
So speed is (x-5)km/hr or (x-5)(5/18).
Here we goes -> (25/2)m/sec = (x-5)(5/18).
-> (25/2)*(18/5) = (x-5).
-> 45 = (x-5).
-> x = 45+5.
-> x = 50 km/hr.
Shanker said:
1 decade ago
We know that 1 km = 1000 meters,
1 hr = 60 minutes, 1 minute = 60 seconds.
And 1 hr = 60*60 seconds (i.e.360 seconds).
Then 1 km/hr = 1000 meters/360 seconds.
This becomes 5 meters/18 seconds((5/18) m/s).
So 1km/hr=5/18 m/s //line1.
But in this question it is given in m/s.
Then from the above line 1 we can write 1 m/s=18/5 km/hr.
1 hr = 60 minutes, 1 minute = 60 seconds.
And 1 hr = 60*60 seconds (i.e.360 seconds).
Then 1 km/hr = 1000 meters/360 seconds.
This becomes 5 meters/18 seconds((5/18) m/s).
So 1km/hr=5/18 m/s //line1.
But in this question it is given in m/s.
Then from the above line 1 we can write 1 m/s=18/5 km/hr.
SHAFI said:
1 decade ago
Simple trick to handle:
S= D/T Where S-speed, D-distance, T-time
Now:
Train data.
D = 125m, T = 10sec.
find speed S = 125m/10sec = 25/2mps.
Note object running in same direction of train will be positive and in opposite direction will be negative.
SPEED OF MEN 5 km/hr+(25/2x5/18).
ANSWER 50 KM/HR.
S= D/T Where S-speed, D-distance, T-time
Now:
Train data.
D = 125m, T = 10sec.
find speed S = 125m/10sec = 25/2mps.
Note object running in same direction of train will be positive and in opposite direction will be negative.
SPEED OF MEN 5 km/hr+(25/2x5/18).
ANSWER 50 KM/HR.
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