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
 A: 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 $W$.
A: Think of the park as a circle. We can think of the distance as degrees.
The distance A has run is 6 times longer than B's: A=6B.  
When they meet, they are at the same point, but A has run significantly more degrees. Luckily, A's advantage can only consist of full rounds, so it can be said that they meet when their difference is a multiple of 360:  
A-B=360k  
Solving equation system (with the equation at the top): B=72k
In other words, they meet every 72 degrees B has run.
(Just take k=1 and you will see how that makes sense.)
The 27th time, B has run 1994 degrees, which corresponds to 5.4 rounds (you have to divide by 360).
$\frac{4}{10}$ of a round correspond to 144 degrees, and that is between S and W.
A: Another approach : Consider the unit circle $x^2+y^2=1$ and set
$$E(1,0),S(0,1),W(-1,0),N(0,-1).$$
(Note that I set these points differently.)
Then, let $6\theta,\theta$ be the counterclockwise angle of $A, B$ respectively. 
$$6\theta-\theta=2\pi\times 27\Rightarrow \theta=10\pi+\frac{4}{5}\pi$$ tells us the answer is between $S$ and $W$.
Here, note that for the $m$-th meet, $6\theta-\theta=2\pi\times m$ holds.
A: I would set it up similarly to Darth Geek, but would call $x$ the number of laps that $B$ has covered when they meet for the 27th time. $A$ will have covered $x+27$ laps, so 
$$6x=x+27,$$
and hence $x=\frac{27}5=5+\frac25$. Now noting that $\frac14<\frac25<\frac12$, we see that they will be between S and W.
A: 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}\\$
