Answer: Option B
:
B
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 catbe C, and let the distance of the tunnel be Y, hence thecat 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 kmand 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