A train approaches a tunnel AB. Inside the tunnel is a cat located at a point that is $$\frac{3}{8}$$ of the distance AB measured from the entrance A. When the train whistles, that cat runs. If the cat moves to the entrance A of the tunnel, the train catches the cat exactly at the entrance. If the cat moves to the exit B, the train catches the cat at exactly the exit. The ratio of the speed of the train to that of the cat is of the order :

Solution(By Examveda Team)

Let the length AB = x

Then, if C is the position of the cat, we have AC = $$\frac{3}{8}$$x
When the cat runs towards the entrance, the train catches it at the entrance.
This means that when the train reaches the entrance, the cat has travelled a distance of $$\frac{3}{8}$$x.
Let us now consider the case when the cat runs towards the exit.
So, when the train reaches A, the cat reaches a point D such that CD = $$\frac{3}{8}$$x
Then,
$$\eqalign{ & \text{BD} = \left[ {x - \left( {\frac{3}{8}x + \frac{3}{8}x} \right)} \right] \cr & \text{BD} = \frac{x}{4} \cr} $$
Since the train catches the cat at the exit, so the train covers distance x (= AB) in the same time in which that cat covers distance $$\frac{x}{4}$$ (= BD)
∴ Required ratio :
$$\eqalign{ & = x:\frac{x}{4} \cr & = 4:1 \cr} $$