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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?

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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*}

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  • $\begingroup$ @panav2000k You are welcome. Glad to help. $\endgroup$ – Américo Tavares Jun 19 '14 at 12:06
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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?

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