Two trains, P and Q, are moving

Solution

ATQ, Let the speed of train Q = ‘p’ m/s So, the speed of train P = p + 32.4 × (5/18) = (p + 9) m/s Also let the lengths of trains P and Q are ‘a’ m and ‘b’ m respectively. Since, both the trains cross each other in 10 seconds. So, (a+b)/(2p+9)=10 a + b = 20p + 90 ------------- (1) Since, train P crosses the 60 m long tunnel in 10 seconds. So, (a+60)/(p+9)=10 a + 60 = 10p + 90 ------------- (2) Since, train Q crosses the 60 m long tunnel in 18 seconds. So, (b+60)/p=18 p = (b+60)/18--------(3) From equations (1) and (3): 18a + 18b = 20b + 1200 + 1620 9a – b = 1410 -------------- (4) From equations (2) and (3): 18a + 1080 = 10b + 600 + 1620 9a – 5b = 570 --------------- (5) By equation (4) — equation (5): 9a - b - 9a + 5b = 1410 - 570 b = 210 From equation (4): a = 180 From equation (3): p = 15 Quantity l: Difference between the lengths of both the trains = 210- 180 = 30 m Quantity II: Speed of train P = 180 m Speed of train P = 15 + 9 = 24 m/s So, the time, in which train P will cross the 540m long bridge: (180+540)/24 = 30 seconds Hence, Quantity I = Quantity II