Train X passes a pole in 18 seconds and a platform of

Solution

ATQ, Let length of train X = Lx metres, speed = Sx m/s. From “passes a pole in 18 s”: Lx / Sx = 18 ⇒ Lx = 18 Sx. From “passes 300 m platform in 30 s”: (Lx + 300) / Sx = 30 ⇒ (18Sx + 300)/Sx = 30 ⇒ 18 + 300/Sx = 30 ⇒ 300/Sx = 12 ⇒ Sx = 25 m/s. Then Lx = 18 × 25 = 450 m. Train Y: Let length Ly (m), speed Sy (m/s). From “passes a pole in 12 s”: Ly / Sy = 12 ⇒ Ly = 12 Sy. When crossing each other in opposite directions in 16 s: (Lx + Ly) / (Sx + Sy) = 16 Substitute Lx, Ly, Sx: (450 + 12Sy) / (25 + Sy) = 16 ⇒ 450 + 12Sy = 16(25 + Sy) ⇒ 450 + 12Sy = 400 + 16Sy ⇒ 12Sy − 16Sy = 400 − 450 ⇒ −4Sy = −50 ⇒ Sy = 12.5 m/s. Then Ly = 12 × 12.5 = 150 m. Convert Sy to km/h: 1 m/s = 3.6 km/h ⇒ Sy = 12.5 × 3.6 = 45 km/h. Answer: (i) Lengths: Train X = 450 m; Train Y = 150 m. (ii) Speed of train Y = 45 km/h.