Travelling round a circle A and B start running from the same point to run in opposite directions round a circular race course 4324 meters in circumference, A not starting till B has run 716 meters. They pass each other when A has run 1927 meters. Who will come first to the starting point and what distance will they then be apart?
 A: Let $t_{1}$ be the time $B$ needs to travel $716$ $\textrm{m }$ at the constant speed $v_{B}$. Then $716=v_{B}t_{1}$. If $t_{2}$ is the instant at which $A$ and $B$ pass
each other, then
\begin{cases}
4324-1927=v_{B}t_{2} \\ 
1927=v_{A}\left( t_{2}-\dfrac{716}{v_{B}}\right)  \\ 
716=v_{B}t_{1},
\end{cases}
where $v_{A}$ is the constant speed of $A$. Simplifying we get
\begin{cases}
t_{2}=\dfrac{2397}{v_{B}} \\ 
\dfrac{v_{A}}{v_{B}}=\dfrac{1927}{1681}=\dfrac{47}{41} \\ 
716=v_{B}t_{1}.
\end{cases}
If $t_{A},t_{B}$ is the additional time needed, respectively, by $A$ and $B$
to reach the starting point, then
\begin{cases}
1927=v_{B}t_{B} \\ 
2397=v_{A}t_{A}=\dfrac{47}{41}v_{B}t_{A},
\end{cases}
which implies that $\dfrac{t_{A}}{t_{B}}=\dfrac{51}{47}>1$. So $B$ comes first
to the starting point. When $B$ reaches this point, $A$ still needs to travel
\begin{align*}
2397-v_{A}t_{B} &=2397-v_{A}\frac{1927}{v_{B}}=2397-\frac{47}{41}1927 \\
&=2397-2209=188\text{ }\mathrm{m}.
\end{align*}
A: Hint:  how far has B run when they meet?  How far did B run while A was running?  What is the ratio of their speeds?  How far does B still have to run?  How far will A run in the time it takes B to run that far?
