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Two runners start running laps at the same time, from the starting position. George runs a lap in $50$ seconds; Sue runs a lap in $30$ seconds. When will the runners next be side by side?

George's speed: $\dfrac{1}{50}$ lap per second

Sue's speed: $\dfrac{1}{30}$ lap per second

We are looking for the time $t$ when they next meet.

How many laps George has run after time $t$: $\dfrac{t}{50}$ laps

How many laps Sue has run after time $t$: $\dfrac{t}{30}$ laps

This is where I get confused. The solution says: They will next be even when Sue has run exactly one more lap, that is, when $\dfrac{t}{30}-\dfrac{t}{50}=1$.

What is the justification to say they will be next to each other when Sue has run exactly one more lap? To me this just came out of the blue.

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    $\begingroup$ "What is the justification to say they will be next to each other when Sue has run exactly one more lap?" It's trivial. Consider the distances run by them both. To meet at the same point their distances run must have a difference which is some multiple of the lap distance and that happens first when it is just one lap distance. $\endgroup$
    – Simon Goater
    Commented May 15 at 8:27
  • $\begingroup$ It probably makes it clearer to express it as "...when Sue has run exactly one more lap than George", i.e. when she's lapped him - as opposed to when she's run exactly one more lap than she has now! $\endgroup$
    – psmears
    Commented May 15 at 21:43

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Assuming the runners run in the same direction, when they meet again, the slower runner is being "lapped".

So the faster runner has completed $n+1$ laps (where $n$ is likely to be a fraction since 50 is not a multiple of 30) and the slower runner has completed $n$ laps.

The rest of the solution is then found by solving $\frac{t}{30}-\frac{t}{50}=1$ for $t$:

$20t=1500$

$t=75$ seconds.

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  • $\begingroup$ Yes, my mistake. It was meant to say $20t=1500$ because $30\times 50=1500$. Fixing now. Thanks for spotting it. $\endgroup$
    – Red Five
    Commented May 15 at 21:14
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Think about this physically. When they are running the laps, they will only meet when one person has run some integer multiple of lap distance more than another. But this happens for the first time when the faster person is exactly one lap ahead. You can model it physically to help yourself understand. Ask if you have any more questions.

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I think there may be a misconception here about how the "extra lap" is measured. The extra lap that Sue has run is measured relative to however far George has run by that time, not relative to the starting line. The observation doesn't mean that Sue and George must pass at the starting line, it means they must pass when the difference in distances is one lap. Wherever George happens to be at the time they pass, he can return to the same spot by traveling exactly one lap. Sue can't be anything other than exactly an integer number of laps ahead of or behind George and be standing next to him.

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    $\begingroup$ When I wrote "n laps" I did not mean relative to each other, but relative to the starting position. Yes, the extra needs to be an integer (and the first time it happens that integer is 1). I also think this answer should have been a comment, perhaps. $\endgroup$
    – Red Five
    Commented May 15 at 21:14

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