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 20 of 48.
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.
Hasan Mahmud said:
9 years ago
Good job.
Rishabh Suthar said:
9 years ago
Same direction then we do subtract and opposite direction then we do addition. Why?
Pabitra mohan das said:
9 years ago
Why it is divided by 18/5?
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