Two places A and B are 506 km apart.  Train X leaves from A to B and Train Y leaves from B to A at the same time.  Both trains meet after 5\(\frac{1}{2}\) hour. The speed of Y is 10 km/h more than X.  What is the speed of X (in km/h)?

41

Let speed of X = X and Y = Y

X + Y = \(\frac{506}{11}\) × 2 = 46 × 2 = 92

X + Y = 92 .......(i)

and Y - X = 10 (given).....(ii)

Now,

(i) + (ii)

⇒ 2Y = 102 

⇒ Y = 51 

from (ii), if Y = 51, then

⇒ X = 41