A train approaches a tunnel AB. Inside the tunnel a cat located at a point i.e. 5/12 of the distance AB measured from the entrance A. When the train whistles the Cat runs. If the cat moves to the exit B, the train catches the cat exactly the exit. The speed of the train is greater than the speed of the cat by what order ?

Explanation:-

Train(T)__________ A_____5k____CAT__________B

T<-------x--------------><----12k-------------------------->

Let the speed of train be u and the speed of Cat be v and train whistles at a point T, X km away from A. Let AB = 12k and Cat was 5k distance away from A. Time was constant for both, then
$$\eqalign{ & \Rightarrow \frac{v}{u} = \frac{x}{{5k}} = \frac{{\left( {x + 12k} \right)}}{{7k}} \cr & \Rightarrow 7x = 5\left( {x + 12k} \right) \cr & \Rightarrow \frac{x}{k} = \frac{{30}}{1} \cr & {\text{Thus}}, \cr & \Rightarrow \frac{u}{v} = \frac{{30}}{5} = \frac{6}{1} \cr}$$