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 34 of 48.
Pabitra mohan das said:
9 years ago
Why it is divided by 18/5?
Hasan Mahmud said:
9 years ago
Good job.
Laxman said:
9 years ago
Nice & excellent explanations, Thank you all.
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.
Vignesh.Rv said:
9 years ago
Anyone tell me why its divided by 18/5?
Ramya said:
9 years ago
Please give the shortcut to solve the problem.
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.
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.
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.
Vinutha said:
8 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?
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