(a) 30 km/hr
Let the speed of the faster train be x km/hr and that of the slower train be y km/hr.
Relative speed when both move in same direction = (x – y) km/hr
Relative speed when both move in opposite directions = (x + y) km/hr
Total distance travelled = Sum of lengths of both the trains = 200 m
Given, \(\frac{200}{(x-y)\times\frac{5}{18}}=60\) and \(\frac{200}{(x+y)\times\frac{5}{18}}=10\)
⇒ \(\frac{3600}{5(x-y)}=60\) and \(\frac{3600}{5(x+y)}=10\)
⇒ x - y = \(\frac{3600}{300}=12\) ....(i)
and x + y = \(\frac{3600}{50}=72\) ....(ii)
Adding eqn (i) and eqn (ii), we get
2\(x\) = 84 ⇒ \(x\) = 42 km/hr
∴ From (i), y = 30 km/hr.