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 20 of 48.
Ajay said:
8 years ago
As per my calculation, 50 km/hr is the right answer.
Sushanth ratan said:
9 years ago
@Jothi.
-5 if it goes on the other side it becomes plus (+).
-5 if it goes on the other side it becomes plus (+).
Vinutha said:
9 years ago
Can anyone explain, how we will get to know that, the given length and time is for relative speed, and why can't it be related to speed of train?
JEEVAN said:
9 years ago
Shortcut
125/10 = trains speed -mans speed.
Lowest form = trains speed - 5.
25/2 into km/h = 25/2 * 18/5.
= 5/1 * 9/1.
= 45km/h.
45km/h = trains speed - mans speed.
45km/h = trains speed - 5.
THEREFORE 5 + 45 = TRAINS SPEED (transposing method).
= 50km/h.
125/10 = trains speed -mans speed.
Lowest form = trains speed - 5.
25/2 into km/h = 25/2 * 18/5.
= 5/1 * 9/1.
= 45km/h.
45km/h = trains speed - mans speed.
45km/h = trains speed - 5.
THEREFORE 5 + 45 = TRAINS SPEED (transposing method).
= 50km/h.
CHINNU said:
9 years ago
Here, one thing I got to know is that, you all have missed a point and that was making a sense and hence considerable.
We can assume a stationary man or a pole of negligible length but this man was moving with a speed of 5km/hr and hence covered almost (10 * (5 * 5/18)) meters in 10 seconds and so the length should be considered. Thus the answer is 55Km/hr instead of 50Km/hr.
We can assume a stationary man or a pole of negligible length but this man was moving with a speed of 5km/hr and hence covered almost (10 * (5 * 5/18)) meters in 10 seconds and so the length should be considered. Thus the answer is 55Km/hr instead of 50Km/hr.
Ashish Dagar said:
9 years ago
Here, one thing I got to know is that, you all have missed a point and that was making a sense and hence considerable.
We can assume a stationary man or a pole of negligible length but this man was moving with a speed of 5km/hr and hence covered almost (10*(5*5/18)) meters in 10 seconds and so the length should be considered. Thus the answer is 55Km/hr instead of 50Km/hr.
We can assume a stationary man or a pole of negligible length but this man was moving with a speed of 5km/hr and hence covered almost (10*(5*5/18)) meters in 10 seconds and so the length should be considered. Thus the answer is 55Km/hr instead of 50Km/hr.
Ramya said:
9 years ago
Please give the shortcut to solve the problem.
Vignesh.Rv said:
9 years ago
Anyone tell me why its divided by 18/5?
SRIKANTH said:
9 years ago
Another method friends but bit lengthy.
D = S X T.
125 = (A - 25/18)10,
25 = (18A - 25)/9,
18A - 25 = 225,
18A = 250,
A = 250/18 X 18/5 = 50 KMPH.
D = S X T.
125 = (A - 25/18)10,
25 = (18A - 25)/9,
18A - 25 = 225,
18A = 250,
A = 250/18 X 18/5 = 50 KMPH.
Laxman said:
9 years ago
Nice & excellent explanations, Thank you all.
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