Aptitude - Problems on Trains - Discussion

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.

According to the question,

Trains length = 125m (distance).
Speed = 10sec,
Mans speed = 5km/hr.

The km the train hasn't travelled is not been given so take it as = X km/hr.
Man moves in the same direction = 5km/hr .

When two bodies are moving in the same direction then you will have to subtract it. If the person is moving on the opposite side he feels the train is moving faster hence u got to add But here it's on the same direction.

So, it's (x-5)km/hr.
When you want to convert kilometre into metres multiply with 5/18
So , (x-5) 5/18 ---> (1) i.e time

Time =distance/speed.
= 125/10 --> (2).

Therefore,
Time = distance/speed.
(X-5) 5/18 = 125/10,
(X-5)5/18 = 25/2,
(X-5) = 25/2 * 18/5 (simple cancellation).
(X-5) = 45,
X= 45+ 5,
X= 50.

Why 5 - X not taken?

If only one of the speed is given then how will we consider that which one is to be subtracted or not?

We 1st Find the speed of train relative to man Vr.
Man and train moved in the same direction = (V-U).
V = Speed of the train.
U = Speed of man.
Vr = (V-U).

It's 50 m/s unit.

x-5 = 45.
x = 45+5(because when 5 goes from the left side to right then it will be added into 45).

So,
x = 50 Answer.

Wrong solution.

Man also traveled a distance of 5*5/18*10=125/9 m in 10 sec. So this distance should be added to 125 m when calculating relative speed, so relative speed = (125+125/9)/10 = 125/9 m/s = 125/9 * 18/5 km/s = 50 km/hr.

So the speed of the train is 50+5 = 55 km/hr.

How x-5 came?

Relative speed subtracted by actual speed so we get X - 5.