Question

A train approaches a tunnels AB. Inside the tunnel a cat located at a point that is $\frac{3}{8}$ of the distance measured from the entrance A. when the train whistles, the cat runs. If the cat moves to the entrance of the tunnel, A, 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 speed of the train is greater than the speed of the cat by what order?

• A. 3 : 1
• B. 4 : 1
• C. 5 : 1
• D. None of these

Answer 1

The correct option is B 4 : 1

Approach 1: Conventional Approach :
Let the speed of the train be T, and let it be at a distance of X from the tunnel AB.
Let the speed of the cat be C, and let the distance of the tunnel be Y, hence the cat is at a distance of $\frac{3}{8}Y$ from A.

Now 2 conditions are given:
When the train meets the cat at B it would have travelled a distance of X+Y while the cat will travel $\frac{5}{8}Y$ at its respective speeds,
$\frac{X+Y}{T}=\left(\frac{5}{8}\right)\frac{Y}{C}.......\left(1\right)$
When the train meets the cat at A it would have travelled a distance of X while the cat will travel $\frac{3}{8}Y$ at its respective speeds.
Also, $\frac{X}{Y}=\frac{\left(\frac{5}{8}\right)Y}{C}.....\left(2\right)$
From (1)
$\frac{X}{T}+\frac{Y}{T}=\frac{5Y}{5C}$
Sub form (2)
$\frac{\left(\frac{5}{8}\right)Y}{C}+\frac{Y}{T}=\frac{5Y}{8C}$
$=\frac{Y}{T}=\frac{2Y}{8C}$
or $T:C=4:1$

Approach 2: Shortcut- Assumption

Assume the lengths of the tunnel to be 8 km and the speed of the cat be 8 km/hr

Time taken for the cat to reach A is 3 hr.

Time taken for the cat to reach B is 5 hr.

The difference in time of A and B gives the time taken by the train, which is 2 hours.

Hence the cat takes 8 hours while the train takes 2 hours,

Hence T:C = 4:1