Motion circular track problem Amy and Amanda walk on a circular track.  Amy starts from spot A, and Amanda starts from spot B. They walk in opposite directions. After 6 minutes, they meet, 4 more minutes later, Amy arrives at spot B. 8 more minutes later, they meet again. How many minutes does it take for both to walk the full circle?
So Amy goes to B, lets say the top portion. Amanda goes to A, so the top portion since opposite directions and they meet. 6 minutes they meet, and 4 more minutes Amy arrives at spot B so 10 minutes for her to go to spot B. Then, they meet again 8 minutes later, and that's when I get stuck.
How do you know who is faster or if they are the same speed?
 A: "How do you know who is faster or if they are the same speed?'  The usual rule is that you define variables and write equations that encode what you know.  Let Amy complete a circuit in $y$ minutes and Amanda complete a circuit in $m$ minutes and so on.  
In this case there is a trick:  they meet twice, $12$ minutes apart.  If they walk the same speed, then it takes $24$ minutes for each to walk the circle.  You should be able to convince yourself that you don't have the information to solve the problem without this-there is a range of starting positions and speeds (a one parameter family, one choice of the parameter being the speed ratio) that will satisfy the problem.  As Amanda gets faster she needs to start further from Amy.  Presumably she can't be fast enough that they pass before the first noted meeting.
A: Since both are travelling at constant speed on the same circle, it should be sufficient to consider only angular displacement and angular speed. 
Let subscripts 1 and 2 represent Amy and Amanda (let's call her Betty) respectively.
Let the initial angular positions of Amy and Betty be 0 and $\beta$ respectively. 
Let $\theta$ and $\omega$ denote angular displacement and angular speed respectively.
Assume WLOG that Amy travels anticlockwise (+ve $\theta$ direction) and Betty travels clockwise (-ve $\theta$ direction). 
At time $t$, the angular positions of Amy and Betty are given as follows:
$$\begin{align}
\theta_1 &=\omega_1t\\ 
\theta_2&=\beta-\omega_2t
\end{align}$$
At $t=6$ they meet, so 
$$6\omega_1=\beta-6\omega_2 $$
At $t=10$, $\theta_1=\beta$, i.e.
$$ \beta=10\omega_1$$
The above gives
$$ \omega_2 = \frac 23 \omega_1 $$
At $t=18$ they meet again, so
$$\begin{align}
18\omega_1&=\beta-18\omega_2 \left[+2\pi\right]\\
18\omega_1&=10\omega_1-18\left(\frac 23 \omega_1\right)+2\pi\\
20\omega_1&=2\pi\\
\omega_1&=\frac \pi{10}\\
\omega_2&=\frac {2\pi}{30}\\
\end{align}$$
The angular distance of the full circle is $2\pi$.
Hence, using $t_i=\theta/\omega_i={2\pi}/\omega_i$, the time taken by Amy and Betty to walk the full circle will be 20 and 30 minutes respectively. 
A: It takes Amanda 6 minutes to walk from B to the first meeting point. It takes Amy 4 minutes to walk the same distance, from the first meeting point to B. So Amy walks 50% faster than Amanda.
They meet at 12-minute intervals, so if $\alpha$ and $\beta$ are the proportions of the circle  walked by Amy and Amanda in 12 minutes, then (i) $\alpha+\beta = 1$; and (ii) $\beta = \frac32 \alpha$.
These two equations give $\alpha = \frac25, \beta = \frac35$. Thus the times taken for Amy and Amanda to walk the whole circle are $12/\alpha = 30$ minutes and $12/\beta = 20$ minutes.
A: 
assume Amy's speed is $x$ and Amanda's speed is $y$ the distances between them are $a$ and $b$.
since they met after $6$ minutes  : 
$$\frac{a}{x+y}=6$$
$4$ minutes later Amy arrives at Amanda's first place so:
$$\frac{a}{x}=6+4=10$$
from these two equations we will have :
$$\frac{a}{y}=15$$
when Amy arrives at Amanda's place she have walked only $\frac{2}{3}a$
so when you say they meet $8$ minutes later it means this :
$$\frac{\frac{1}{3}a+b}{x+y}=8$$
$$\frac{\frac{1}{3}a}{x+y}+\frac{b}{x+y}=8$$
$$2+\frac{b}{x+y}=8$$
$$\frac{b}{x+y}=6$$
You can see that $a=b$
so we get :
$$\frac{a+b}{x}=\frac{2a}{x}=20$$
$$\frac{a+b}{y}=\frac{2a}{y}=30$$
