Aptitude - Problems on Trains - Discussion

@Kapil.

Your method is easy. Thanks for it. But at last step it is:
((55/9)-(5/9))*9=50m.

CNA finish off this way.

What is find out by 5/9 and 2/9.

I mean is it speed of train ?

2 kmph = 2*(5/18) = 5/9 (m/s).
4 kmph = 4*(5/18) = 10/9 (m/s).

Let the speed of the train be x.

(x-5/9)9 = (x-10/9)10 (because the distance is same).

On solving we get.

x = 55/9 (m/s).

Train length = (x-5/9)9.

= (55/9-5/9)9.
= (50/9)*9.
= 50 m.

Take Length of Train = X meter.

And Speed of train = Y m/s.

2kmph = 5/9m/s, 4kmph = 10/9 m/s.

For first train X/(Y-5/9) = 9.

That is X-9Y = -5.....eq1.

For Second train X/(Y-10/9) = 10.

That is 9X-90Y = -100.....eq2.

Solving eq1 and eq2.

X = 50 meter.

And speed Y = 22 kmph or 55/9 m/s.

At last why it is multiplied by 9 not by 10?

There are two cases:

In the first case, speed of man = 2 km/h => 2x5/18 => 5/9 m/sec.

Therefore, relative speed of train we know that first train (let speed of train be 'V') ,

V-5/9 = L/9....(1) here 'L' is length of train.

Now, for 2nd case, speed of man = 4 km/h => 4x5/18 => 10/9 m/sec.

Therefore, V-10/9 = L/10....(2).

From equation (1) we get,

9V-5/9 = L/9.

Cut denominator 9 on both side we get.

9V-5 = L or 9V = L+5 or V = (L+5) /9.

Now substituting the value of V in equation --- (2).

(L+5) /9 - 10/9 = L/10.
L+5-10/9 = L/10.
L+5-10 = 9L/10.

On cross multiplication,

10L+50-100 = 9L.

On rearranging.

10L-9L = 100-50.
L = 50 meters.

Please any one can simply this step by step x/(y-10/9) = 10.

How can we get 10(9y-10) = 9x?

What is the speed of a train if it overtakes two persons who are walking in the same direction at the rate of a m/s, (a+1) m/s, and passes them completely in b second and (b+1) seconds respectively?

Please anyone can simplify this step?

9y-x=5 and 90y-9x=100. I can't understand this. How will we get x=50?

@Kabil.

How is 55/9 is coming?