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 11 of 48.
Mozammel hossain said:
7 years ago
There have man speed but not train speed.
So we known that;
speed = distanc/time.
= 125/10,
= 25/2.
Now convert into km/h.
25/2 * 18/5.
= 45km/h.
Lets train speed is x km/h.
So, the relative speed is x-5 km/h.
x- 5 = 45,
X = 50km/h.
So we known that;
speed = distanc/time.
= 125/10,
= 25/2.
Now convert into km/h.
25/2 * 18/5.
= 45km/h.
Lets train speed is x km/h.
So, the relative speed is x-5 km/h.
x- 5 = 45,
X = 50km/h.
Nithish said:
7 years ago
Thanks all for the explanations.
Muhammad Zulafqar said:
7 years ago
The simplest answer is;
10sec/1000Meter =0.01.
Then multiply with 5KM i.e 0.01 * 5000meter = 50KM.
10sec/1000Meter =0.01.
Then multiply with 5KM i.e 0.01 * 5000meter = 50KM.
Vinod said:
7 years ago
Thanks all for explaining the answer.
Anjali said:
7 years ago
Thanks all for explaining the answer.
Abuthahir said:
7 years ago
X-5 = 45.
Here x = 45+5,
Then x = 50.
Here x = 45+5,
Then x = 50.
Narendra said:
7 years ago
In same direction, T. S-M. S = 125/10 * 18/5 = 45.
M.S = 5KMph.
T.S - 5 = 45.
T.S = 50sec.
M.S = 5KMph.
T.S - 5 = 45.
T.S = 50sec.
Lavanya said:
7 years ago
Why do we to subtract 45-5? please clarify that.
Hari kumar said:
7 years ago
50 is the right answer.
Chandan Azad said:
7 years ago
T(time required) = Lt(length of train) + Lo(length of object)/ Ut(speed of Train)+,-Uo(speed of object) .
("-" when train and object goes in same direction.
"+" when train and object goes in opposite direction.)
10=(125+0)/(Ut-5*5/18).
Ut=125/9.
Ut=(125/9)*(18/5).
Ut=50km/hr.
("-" when train and object goes in same direction.
"+" when train and object goes in opposite direction.)
10=(125+0)/(Ut-5*5/18).
Ut=125/9.
Ut=(125/9)*(18/5).
Ut=50km/hr.
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