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 14 of 48.
Vinothini said:
1 decade ago
Hai friends I want to give small tips.
If there train passes in same direction just subtract it.
If train passes in different or opposite direction direction just add it. If you need to convert km/pr use this formula (a*5%18) for speed and m/sec the formula is (a*18%5).
If there train passes in same direction just subtract it.
If train passes in different or opposite direction direction just add it. If you need to convert km/pr use this formula (a*5%18) for speed and m/sec the formula is (a*18%5).
Bibin said:
1 decade ago
Why you made it 18/5, earlier it was 5/18, please explain me.
Jayan said:
1 decade ago
Here x=45 is speed of per 1km. question is 5km/hr .
So
x-5=45
x=45+5
x=50
Thats all.
So
x-5=45
x=45+5
x=50
Thats all.
Anvesh said:
1 decade ago
l=125m;
s=5*5/18;
t=10s;
l=t*s;
s=t/l;
s=125/10;
s=12.5*18/5;
the man was also going same direction so,
s=45+5;
s=50;
s=5*5/18;
t=10s;
l=t*s;
s=t/l;
s=125/10;
s=12.5*18/5;
the man was also going same direction so,
s=45+5;
s=50;
Charles @ Redhills said:
1 decade ago
Let the speed of the train is 'x'
Length=125m
Speed=10s
--> as per "Length of the train = speed X time"
--> 125m = 10s.x
--> x=(125m/10s) X (360/360)=(360 X 125m)/3600s
--> =45000m/3600s = 45km/1hr
As the above value is relative with the speed of man whose speed of running is 5km/1hr also to be added for the exact speed of the train.
There fore..
-->exact speed of the train is = (45km/1hr)+(5km/1hr)=50km/1hr
I believe its simple and right answer.
Length=125m
Speed=10s
--> as per "Length of the train = speed X time"
--> 125m = 10s.x
--> x=(125m/10s) X (360/360)=(360 X 125m)/3600s
--> =45000m/3600s = 45km/1hr
As the above value is relative with the speed of man whose speed of running is 5km/1hr also to be added for the exact speed of the train.
There fore..
-->exact speed of the train is = (45km/1hr)+(5km/1hr)=50km/1hr
I believe its simple and right answer.
Jagadeesh said:
1 decade ago
Let relative velocity of train with respect to man be vtm,actual velocity of train Be vt and velocity of man be vm.
We know that vtm=vt-vm,
But from problem vtm is =length of train/time taken
=(125*18/5)/10
=45 km/hr.
From above actual speed of train is,vt=vtm+vm
=45+5=50 km/hr.
We know that vtm=vt-vm,
But from problem vtm is =length of train/time taken
=(125*18/5)/10
=45 km/hr.
From above actual speed of train is,vt=vtm+vm
=45+5=50 km/hr.
Preethi said:
1 decade ago
A train of length 110mtr travelling at a speed of 80km/hr. Find the time taken by that train to overtake a train having length 130mtr travelling at a speed of 62km/hr?
Malik said:
1 decade ago
@Preethi.
Relative speed u-v = 80-62= 18km/hr=(18* 5/18)= 5 m/s.
Total length a+b = 110+130= 240 m.
Time to overtake = (a+b)/(u-v)[see formula 9 from formula section]
= 240/5
= 48 sec.
Am I correct?
Relative speed u-v = 80-62= 18km/hr=(18* 5/18)= 5 m/s.
Total length a+b = 110+130= 240 m.
Time to overtake = (a+b)/(u-v)[see formula 9 from formula section]
= 240/5
= 48 sec.
Am I correct?
Diksha said:
1 decade ago
How we come to know that the speed 5km\hr is the speed of the man because it is not given in the question?
Aruntracer said:
1 decade ago
@Diksha.
A train 125 m long passes a man, running at 5 km/hr in the same direction means the , in between man and running is for confusing you it should be taken as 125 m long train passes a man running at 5 km/hr.
A train 125 m long passes a man, running at 5 km/hr in the same direction means the , in between man and running is for confusing you it should be taken as 125 m long train passes a man running at 5 km/hr.
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