Where will they meet? A and B are running along the wall of a square park. The corners of the park are facing North, South, East and West and are named N, S, E and W respectively. They start at E and run towards S. If the speed of A is 6 times than of B, where do they meet for the 27th time?

a) between S and W
b) at S
c) between W and N
d) at N
Since $A$ runs faster, every time he meets $B$, he will have outruned him by an extra lap. Every lap measures $4$ times the length of the square, so if $x$ is the distance $B$ runs between two encounters with $A$ then:

$6x = 4+x \Rightarrow x = \frac{4}{5}$. So when they meet, $B$ would have ran $4/5$ times the side of the square.

So the $27$th time, $B$ would have run $27·\frac{4}{5} = 21 + \frac{3}{5}$ times the length of the side of the square.

Since every $4$ times, he's back on $E$, He'll be $1+\frac{3}{5}$ lenghts after $E$, i.e. between $S$ and $W$.
For $1$ complete round by $B$, $A$ completes 6 rounds of the square park.



Now for $B$ to meet $A$ for $27th$ time, $B$ must be completing $\frac {27} {6}$
rounds that is $4{\frac{1}{2}}$ rounds or 
$\text{starting from E towards S, he will be at a position -}\\$

$\text{$a)$ between $S$ and $W$ side of the park}\\$