Train 'X' can cross a pole in 15 seconds and a 180 metres

Solution

Let the length and the speed of the train 'X' be 'x' metres and 's' m/s.
ATQ,
(x/s) = 15
So, 'x' = 15s.........(i)
Also, {(x + 180) ÷ s} = 18
Or, 15s + 180 = 18s [from equation (i)]
Or, 180 = (18s - 15s)
Or, 's' = (180/3) = 60
Length of train 'X' = (15 × 60) = 900 metres
So, length of train 'Y' = (900 × 0.80) = 720 metres
Let the speed of train 'Y' be 'y' m/s.
ATQ,
(900 + 720) ÷ (y - 60) = 12
Or, 1620 = 12 × (y - 60)
Or, (y - 60) = 135
So, 'y' = 135 + 60 = 195
Therefore, speed of train 'Y' = 195 m/s