Aptitude - Problems on Trains - Discussion

Very nice, thanks everyone for explaining the answer.

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.

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.

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.

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.

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.

125/10 = 12.5,
12.5 * 3.6 = 45,
45 + 5 = 50.

Since the man is running in the same direction as the train, the distance covered by the train relative to the man is indeed the sum of the distance travelled by the man and the length of the train.

Let's recalculate:

Step 1: Calculate the distance travelled by the man in 10 seconds
Distance traveled by man (d_m) = Speed of man × Time = 1.39 m/s × 10 s = 13.9 m.

Step 2: Calculate the total distance covered by the train relative to the man
Total distance (d_t) = Length of train + Distance traveled by man = 125 m + 13.9 m = 138.9 m.

Step 3: Calculate the relative speed of the train concerning the man
Relative speed (v_r) = Total distance / Time = 138.9 m/10 s = 13.89 m/s

Step 4: Convert the relative speed to kilometres per hour.
Relative speed (v_r) = 13.89 m/s × (3600s/1000 m) = 50 km/hr.

Step 5: Calculate the speed of the train.
Speed of the train (v_t) = Relative speed + Speed of the man = 50 km/hr + 5 km/hr = 55 km/hr.

The correct answer is:
55 km/hr.

@Hemanth sairam.

Why 60 mins and 60 seconds taken there?

Please explain me.

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.