Train 'C' can cross a pole in 6 seconds and a 120 metres

Solution

Let the length and the speed of the train 'C' be 'x' metres and 's' m/s.
ATQ,
(x/s) = 6
So, 'x' = 6s.........(i)
Also, {(x + 120) ÷ s} = 10
Or, 6s + 120 = 10s [from equation (i)]
Or, 120 = (10s - 6s)
Or, 's' = (120/4) = 30
Length of train 'C' = (6 × 30) = 180 metres
So, length of train 'D' = (180 × 0.25) = 45 metres
Let the speed of train 'D' be 'd' m/s.
ATQ,
(180 + 45) ÷ (d - 30) = 15
Or, 225 = 15 × (d - 30)
Or, (d - 30) = 15
So, 'd' = 15 + 30 = 45
Therefore, speed of train 'D' = 45 m/s