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 7 of 48.
Kavya reddy said:
1 decade ago
Why we are using 18/5 instead of 5/8 is as follows,
1km = 1000m.
1hr = 3600s.
So 1km/hr = (1000/3600)m/s.
Hence 1km/hr = (5/8)m/s.
Then 5/8 becomes reverse as 8/5 and it will becomes as,
(8/5)km/hr = m/s.
Again you get a doubt that why it converts into m/s because check our problem has 125 m and also they gave 10 seconds.
So (125/10)m/s.
1km = 1000m.
1hr = 3600s.
So 1km/hr = (1000/3600)m/s.
Hence 1km/hr = (5/8)m/s.
Then 5/8 becomes reverse as 8/5 and it will becomes as,
(8/5)km/hr = m/s.
Again you get a doubt that why it converts into m/s because check our problem has 125 m and also they gave 10 seconds.
So (125/10)m/s.
Jaimik chaudhary said:
1 decade ago
Distance travel by man in 10s = (5*1000*10)/(3600) = 13.89 m.
Now this much more distance the train has to travel due to mobile reference which is man.
Total distance travel by train = (125+13.89)m = 138.89 m.
Speed = (138.89*3600)/(10000) = 50 km/hr.
Comparison:
If I split the 138.89 into 125+13.89.
= 125/10+13.89/10 = 45+5 = 50 km/hr.
Now this much more distance the train has to travel due to mobile reference which is man.
Total distance travel by train = (125+13.89)m = 138.89 m.
Speed = (138.89*3600)/(10000) = 50 km/hr.
Comparison:
If I split the 138.89 into 125+13.89.
= 125/10+13.89/10 = 45+5 = 50 km/hr.
Ashik Multani said:
2 years ago
Given
Length of train = 125m
Speed of man = 5km/hr = 5 * 5/18 = 25/18
Time = 10sec.
Speed = distance/time
Man running in the same direction so;
Let theh speed of the train is x.
speed = x - 25/18;
then by formula;
x - 25/18 = 125/10
x = 125/10 + 25 /18
x = 250/18 sec/m
Then;
convert into km/hr.
speed = 250/18 * 18/5.
speed = 50 => ans.
Length of train = 125m
Speed of man = 5km/hr = 5 * 5/18 = 25/18
Time = 10sec.
Speed = distance/time
Man running in the same direction so;
Let theh speed of the train is x.
speed = x - 25/18;
then by formula;
x - 25/18 = 125/10
x = 125/10 + 25 /18
x = 250/18 sec/m
Then;
convert into km/hr.
speed = 250/18 * 18/5.
speed = 50 => ans.
(45)
Salahuddin said:
1 decade ago
To avoid confusion
Simple convert the trains speed for 1 hr - metres/seconds, 125m/10 sec. So 125 * 6 * 60 / 10 * 6 * 60 = 45000 mtrs/ 3600 . So to convert it, divide by 1000, its 45 kms/hr.
As the man running in the same direction at a speed of 5kms/hr add 45 + 5, if the person moves in opposite direction then subtract. Just logic guys
Simple convert the trains speed for 1 hr - metres/seconds, 125m/10 sec. So 125 * 6 * 60 / 10 * 6 * 60 = 45000 mtrs/ 3600 . So to convert it, divide by 1000, its 45 kms/hr.
As the man running in the same direction at a speed of 5kms/hr add 45 + 5, if the person moves in opposite direction then subtract. Just logic guys
Kunal Pardeshi said:
6 years ago
Hi. please solve my query.
Since the man is also moving, why the distance travelled by train in 10 seconds is considered equivalent to the length of the train when in actuality the distance covered will be more than that as the man is not stationary. Please, somebody explain this clearly. I want to know the concept and logic behind this.
Since the man is also moving, why the distance travelled by train in 10 seconds is considered equivalent to the length of the train when in actuality the distance covered will be more than that as the man is not stationary. Please, somebody explain this clearly. I want to know the concept and logic behind this.
Narinder singh said:
6 years ago
D = S X T.
(125 + 0) = (X-25/18) X 10 --> (i)
{Whereas 125 = length of the train, 0 = length of man, X is taken as the speed of the train, 25/18 is 5X5/18 to convert m/s into km/h, "-" Minus is used because they are running in the same direction, 10 is time to cross man}
Now solve the equation (i) and we will get the answer.
(125 + 0) = (X-25/18) X 10 --> (i)
{Whereas 125 = length of the train, 0 = length of man, X is taken as the speed of the train, 25/18 is 5X5/18 to convert m/s into km/h, "-" Minus is used because they are running in the same direction, 10 is time to cross man}
Now solve the equation (i) and we will get the answer.
Er.deepak said:
9 years ago
Man speed is 5km/hr.
So, km = 1000metrs.
Per hour = 60mintes = 3600second.
3600/1000 = 36/10 = 18/5.
Trains cross the man in 10 seconds.
Length of the train is =125m.
Relative man trains speed = 125/10 = 25/2.
So 25/2 x 18/5 = 45km/hr.
We let train speed is = x km/hr.
Relative to man turn speed = x - 5 = 45.
x = 45 + 5 = 50km/hr.
So, km = 1000metrs.
Per hour = 60mintes = 3600second.
3600/1000 = 36/10 = 18/5.
Trains cross the man in 10 seconds.
Length of the train is =125m.
Relative man trains speed = 125/10 = 25/2.
So 25/2 x 18/5 = 45km/hr.
We let train speed is = x km/hr.
Relative to man turn speed = x - 5 = 45.
x = 45 + 5 = 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.
KIRUBHA said:
1 decade ago
FRNDS is it correct or not tell me plz....
MAN'S SPEED = 5 KM/HR
Train length =125 m.
let assume train speed=x.
then we get relative speed (rs) of train and man is become
rs=x-5.
To conver km/hr into m/s.
= (x-5) / (5/18)
= 5x-25 / 18
speed = distance/time
125/10 = 5x-25 / 18
2250 = 50x-250
x = 2000 / 50
x = 40
Is it correct?
MAN'S SPEED = 5 KM/HR
Train length =125 m.
let assume train speed=x.
then we get relative speed (rs) of train and man is become
rs=x-5.
To conver km/hr into m/s.
= (x-5) / (5/18)
= 5x-25 / 18
speed = distance/time
125/10 = 5x-25 / 18
2250 = 50x-250
x = 2000 / 50
x = 40
Is it correct?
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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