Aptitude - Problems on Trains - Discussion

S = L/T.
S = 125/10
S = 25/2.

then,
M/sec to km/hr convert.
1km/hr = [1000m/3600sec] = 5/18m/sec,
m/sec = 18/5km/hr.

i.e
(25/2 * 18/5)km/hr,
45km/hr.

Therefore first train x-5 = 45.
X = 50 km/hr.

Why was the formula reversed from 5/18 to 18/5? Please explain me.

m/sec = [1/1000 km] / [1/3600 sec]
= [1/1000] *[3600/1],
= [3600/1000].
= [18/5]km/hr.

Very nice, thanks everyone for explaining the answer.

Given:

Length of the train = 125 m.
Speed of the man = 5 km/hr.
Time taken to pass the man = 10 seconds.

First, we need to convert the speed of the man from kilometres per hour to meters per second:
5 km hr=5×1000 m3600 s=50003600≈1.39 m s.
5 km hr=5×3600 s1000 m​=36005000​≈1.39 m s.

Next, we calculate the speed of the train using the formula:
Speed=DistanceTime.
Speed=TimeDistance​.

The distance in this case is equal to the length of the train (125 m), and it passes the man in 10 seconds:
Speed of train relative to man =125 m10 s=12.5 m s.
Speed of train relative to man =10 s125 m​=12.5 m s.

Since this is the relative speed (train's speed minus man's speed), we can express it as:
Speed of train − Speed of man = 12.5 m s.
Speed of train − Speed of man =12.5 m s.

Substituting in the man's speed:
Speed of train−1.39 m s=12.5 m s
Speed of train−1.39 m s=12.5 m s.

Now, solving for the speed of the train:
Speed of train=12.5+1.39=13.89 m s
Speed of train=12.5+1.39=13.89 m s

Finally, converting this back to kilometres per hour:
13.89 m s=13.89×36001000=50 km hr
13.89 m s=13.89×10003600​=50 km hr.

Therefore, the speed of the train is 50 km/hr.

How it 18÷5 will become? Please explain.

Here is the solution for this question:

The length of a train is given as 125m (here it is mentioned as m/s), and the speed is given as 5km/hr (here it is mentioned as km/hr).

While converting km/hr into m/s, we are multiplying the speed by 5/18; therefore while converting m/s into km/hr, we have to take the reciprocal of 5/18, that is 18/5. The speed of the train should be calculated in km/hr; therefore, we take 18/5.

To convert m/s to km/hr we use direct formula,
i.e 18/5.
Here,1km/hr = 5/18 m/sec.

To convert a given data from m/sec to km/hr we use 18/5.

Consider we need to convert m/sec to km/hr..
If we need to convert something from m to km, 1km=1000m. Thus 1m= (1/1000)km
So we multiply the numerator by 1/1000.

Similarly, (denominator part) for a sec to hr, 1hr=3600sec. Thus 1sec=(1/3600)hr.
So we multiply the denominator by 1/3600.

This conversion part becomes 3600/1000 thus simplifying this gives an std form of 18/5.. which is always directly used in converting m/sec conversion to km/hr.. helps in just multiplying this 18/5, instead of separate conversion of numerator and denominator to km and hr separately which is time-consuming..

Similarly, the reciprocal of this is just the vice versa case from km/hr to m/sec just multiply with 5/18.

Let X km/hr be the speed of the train.
Length = 125m = 0.125km.
Relative Speed= (X-5)km/hr (as train and man moving in the same direction hence it is subtracted).
0125/(X-5) = 10/3600.
X = 50km/hr.

HERE,
1hr = 3600sec then, 1sec = 1/3600hr
1km =1000m then, 1m = 1/1000km
So,
km/hr to m/s is : 1/1000km bigger divided by 1/3600hr.
So, they put 18/5.