Question

a train crosses a tunnel half of its length at 72kmph in 1 minutes then find how much time it will take to for another train of double the length which is standing on the platform with 60% of its speed

Answer 1

Answer:

Case 1:

Let’s assume,

The length of the train be “L1” meter

the length of the tunnel be “L2” = L1/2 meter

Speed of the train be “S1” = 72km/hr = 72*5/18 = 20 m/s

Time taken be “t” = 1 minute = 60 seconds

∴ Time = Distance/Speed = [L1 + L2] / S1  

Or, 60 = [L1 + (L1/2)]/20

Or, (3/2) L1 = 1200

Or, L1 = 1200 * (2 / 3) = 800 meter

Case 2:

Let the another train be “L3” = 2 * L1 = 2 * 800 = 1600 meter

Since another train is standing on the platform i.e., it is stationary therefore in this case the speed that comes into consideration is the speed “S1” of the first train crossing the stationary train. So, here the speed S1 is given as,  

= 60% of 20 m/s = [60/100] * 20 = 12 m/s

∴ Time taken

= distance/speed

= [L1 + L3] / 12

= [1600+800] / 12

= 2400 / 12

= 200 sec ≈ 3.33 minutes.

Hence, the train takes 3.33 minutes to cross the other train.