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 6 of 48.
Phanikumar said:
7 years ago
Speed of the train relative to man = 125/10 m/sec (meters per second).
But you need to find the speed in KMPh.
Ex; 1kmph = 5/18 m/sec,
so that 1m/sec = 18/5 kmph.
Because of that 25/2 is multiplied with 18/5, then we got the speed in KMPh.
Similarly (25/2)*(18/5)=45kmph.
But the both man and train is moving in same direction.
So, X-5=45.
x=50kmph.
But you need to find the speed in KMPh.
Ex; 1kmph = 5/18 m/sec,
so that 1m/sec = 18/5 kmph.
Because of that 25/2 is multiplied with 18/5, then we got the speed in KMPh.
Similarly (25/2)*(18/5)=45kmph.
But the both man and train is moving in same direction.
So, X-5=45.
x=50kmph.
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.
Shweta said:
9 years ago
A solution to this problem is so simple.
Speed of train = s1;
Speed of men = s2;
S2 = 5km/hr.
The length of train =125m.
Time =10 sec.
So, convert the time and sec i.e. m/s in km/hr.
125/10 * 18/5; //convert the m/s to km/hr formula is a * 18/5.
Formula is: s1- s2 = (L1/T).
We got the equation.
S1 - 5 = (125/10) * (18/5).
S1- 5 = 45km/hr;.
S1 = 45 + 5.
S1 = 50km/hr.
Speed of train = s1;
Speed of men = s2;
S2 = 5km/hr.
The length of train =125m.
Time =10 sec.
So, convert the time and sec i.e. m/s in km/hr.
125/10 * 18/5; //convert the m/s to km/hr formula is a * 18/5.
Formula is: s1- s2 = (L1/T).
We got the equation.
S1 - 5 = (125/10) * (18/5).
S1- 5 = 45km/hr;.
S1 = 45 + 5.
S1 = 50km/hr.
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.
MANAV said:
10 years ago
Everything is ok and easy explanation.
But one thing is questionable.
If the train is crossing a man who is running in the same direction with a speed of 5km/hr, than than how the distance travelled in 10 secs is equal to the length of the train. Don't you think dear friends the distance travelled by the man should also be added to the length of the train.
But one thing is questionable.
If the train is crossing a man who is running in the same direction with a speed of 5km/hr, than than how the distance travelled in 10 secs is equal to the length of the train. Don't you think dear friends the distance travelled by the man should also be added to the length of the train.
Rahul said:
2 decades ago
@ vijayagopal
here both man and train are moving in the same direction and the train crosses the man in 10 seconds.
speed = distance(or length)/time.
As both man and train are moving, relative motion comes into picture.
therefore:
relative speed= (length of the train)/(time taken by the train to cross the man who is in motion)
relative speed= 125/10
here both man and train are moving in the same direction and the train crosses the man in 10 seconds.
speed = distance(or length)/time.
As both man and train are moving, relative motion comes into picture.
therefore:
relative speed= (length of the train)/(time taken by the train to cross the man who is in motion)
relative speed= 125/10
Anitha Alavala said:
1 decade ago
See the answer for 18/5 is how?
1 km=1000m & 1hr=3600sec by this we get 1km/1hr = 5/18 m/sec.But
the answer must be in km/hr....But 5/18 is in m/sec. we have to convert it to km/hr.
1 km = 1000m => m = 1 km/1000 hr/3600)
= (1km/1000)*(3600/1hr)
= 3600km/1000hr => 18/5 km/hr
friends...its a perfect answer...have a look.
1 km=1000m & 1hr=3600sec by this we get 1km/1hr = 5/18 m/sec.But
the answer must be in km/hr....But 5/18 is in m/sec. we have to convert it to km/hr.
1 km = 1000m => m = 1 km/1000 hr/3600)
= (1km/1000)*(3600/1hr)
= 3600km/1000hr => 18/5 km/hr
friends...its a perfect answer...have a look.
Shani sharma said:
2 years ago
If we consider man as stationary then the velocity of train is;
Vt=125/10
=12.5 m/s.
Here, the velocity of man is given which is Vm= 5 km/hr.
We have to convert the velocity of the train m/s into km/s.
Then we get,
Vt= 12.5 × 3600/1000 km/hr.
= 45km/hr.
Now, the velocity of the train wrt man.
Vtm= Vt +Vm.
= 45 +5.
= 50 km/hr.
Vt=125/10
=12.5 m/s.
Here, the velocity of man is given which is Vm= 5 km/hr.
We have to convert the velocity of the train m/s into km/s.
Then we get,
Vt= 12.5 × 3600/1000 km/hr.
= 45km/hr.
Now, the velocity of the train wrt man.
Vtm= Vt +Vm.
= 45 +5.
= 50 km/hr.
(91)
Shanker said:
1 decade ago
We know that 1 km = 1000 meters,
1 hr = 60 minutes, 1 minute = 60 seconds.
And 1 hr = 60*60 seconds (i.e.360 seconds).
Then 1 km/hr = 1000 meters/360 seconds.
This becomes 5 meters/18 seconds((5/18) m/s).
So 1km/hr=5/18 m/s //line1.
But in this question it is given in m/s.
Then from the above line 1 we can write 1 m/s=18/5 km/hr.
1 hr = 60 minutes, 1 minute = 60 seconds.
And 1 hr = 60*60 seconds (i.e.360 seconds).
Then 1 km/hr = 1000 meters/360 seconds.
This becomes 5 meters/18 seconds((5/18) m/s).
So 1km/hr=5/18 m/s //line1.
But in this question it is given in m/s.
Then from the above line 1 we can write 1 m/s=18/5 km/hr.
Sanal said:
1 decade ago
A different way to calculate this:
Length of the train: 125 m.
Distance traveled by man in 10 sec: 5*(5/18) m/s = 25/18 m for 1 sec.
i.e 4(25/18)*10 = 125/9 m for 10 sec.
Total distance traveled by train: 125 + 125/9 = 1250/9 m in ten seconds.
Speed for traveling 1250/9 m in 1 sec = (1250/9)/10 = 13.888.
In km = 13.888*(18/5) = 50 km/hr.
Length of the train: 125 m.
Distance traveled by man in 10 sec: 5*(5/18) m/s = 25/18 m for 1 sec.
i.e 4(25/18)*10 = 125/9 m for 10 sec.
Total distance traveled by train: 125 + 125/9 = 1250/9 m in ten seconds.
Speed for traveling 1250/9 m in 1 sec = (1250/9)/10 = 13.888.
In km = 13.888*(18/5) = 50 km/hr.
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