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There was a question in my math text book the other day that stated:

$2$ cars each travelling at a constant velocity around a ring , complete exactly $4$ and $7$ rounds in one hour. If they start at exactly the same time from the same place but travel in opposite directions around the track, how many times will they pass each other in the one hour?

I tried various things using $\displaystyle v = \frac{d}{t}$ but couldnt get very far. Any help is appreciated, thanks.

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    $\begingroup$ Is the question still in your text book today? $\endgroup$ – Will Jagy May 16 '13 at 21:07
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    $\begingroup$ No, we use electronic textbooks on our laptops that generate problems, so no. $\endgroup$ – simplton May 17 '13 at 7:00
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Let's look from the driver in the slower car's point of view. If he is traveling at $4$ laps per hour and the other car is traveling at $7$ laps per hour in the opposite direction, it looks like the other car is traveling at $11$ laps per hour. From the slower car's viewpoint, he seems to be sitting still while the other car has passed him $11$ times.

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The two cars are essentially travelling towards each other; They will collide when the sum of the distances they each cover is equal to one round. Consider how far each car travels in $t$ hours to get an idea.

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You're on the right track. In the time the faster driver completes his first lap, they've obviously crossed paths. In time $t$ hours, one has completed $4t$ laps and the other $7t$. Remembering they're going in opposite directions, when do they cross first?

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  • $\begingroup$ so in one hour do they cross three times? $\endgroup$ – simplton May 16 '13 at 19:16
  • $\begingroup$ No. They would cross the first time when $$4t=1-7t\,.$$ Why? $\endgroup$ – Ted Shifrin May 16 '13 at 19:29
  • $\begingroup$ so 11 times? i think. $\endgroup$ – simplton May 16 '13 at 19:36
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Suppose the slower car stands still for one hour. How often will the faster car pass it? Then stop the faster car and start the slower car for another hour. How often will the slow car pass the stopped car? Add.

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Consider alternative case when cars complete exactly $4$ and $8$ rounds. It's easily seen that the number of times they pass is

$$2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 = 12$$

So for $4$ and $7$ it would be one less than that which is $11$.

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  • $\begingroup$ so for 11 and 14 it would be 25? $\endgroup$ – simplton May 16 '13 at 19:56

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