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Two cars run in opposite directions on a circular track. Car A travels at a rate of \(6\pi\) miles per hour and Car B runs at a rate of \(8\pi\) miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

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08 Apr 2015, 05:13

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Lets say t = meeting time, obviously cars won't meet in the starting point right away so they will definitely do some rounds first, lets say a = amount of rounds first car does, b - second car. We get an equation. \(t = \frac{2*pi*6*a}{6} = \frac{2*pi*6*b}{8}\) after solving the 2nd equation I get the relation between a and b: a/b = 3/4. So basically their first meeting at starting point happens when first car does 3 laps and 2nd car does 4 laps, therefore after we put a into our equation for t: \(t = \frac{2*pi*6*3}{6} = 6*pi\) hours. At this point I'm not sure what I'm doing wrong. edit: with the amendments to the task, result is 6, answer is D

Last edited by Zhenek on 08 Apr 2015, 07:37, edited 1 time in total.

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08 Apr 2015, 07:18

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Zhenek wrote:

Lets say t = meeting time, obviously cars won't meet in the starting point right away so they will definitely do some rounds first, lets say a = amount of rounds first car does, b - second car. We get an equation. \(t = \frac{2*pi*6*a}{6} = \frac{2*pi*6*b}{8}\) after solving the 2nd equation I get the relation between a and b: a/b = 3/4. So basically their first meeting at starting point happens when first car does 3 laps and 2nd car does 4 laps, therefore after we put a into our equation for t: \(t = \frac{2*pi*6*3}{6} = 6*pi\) hours. At this point I'm not sure what I'm doing wrong.

I think there is misprint in task and this question mark after speeds it's actually pi, so speeds equal 6 pi and 8 pi
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Lets say t = meeting time, obviously cars won't meet in the starting point right away so they will definitely do some rounds first, lets say a = amount of rounds first car does, b - second car. We get an equation. \(t = \frac{2*pi*6*a}{6} = \frac{2*pi*6*b}{8}\) after solving the 2nd equation I get the relation between a and b: a/b = 3/4. So basically their first meeting at starting point happens when first car does 3 laps and 2nd car does 4 laps, therefore after we put a into our equation for t: \(t = \frac{2*pi*6*3}{6} = 6*pi\) hours. At this point I'm not sure what I'm doing wrong.

I think there is misprint in task and this question mark after speeds it's actually pi, so speeds equal 6 pi and 8 pi

Yes, you are right. Edited the typo. Thank you for noticing this.
_________________

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08 Apr 2015, 07:22

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Bunuel wrote:

Two cars run in opposite directions on a circular track. Car A travels at a rate of 6? miles per hour and Car B runs at a rate of 8? miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

This is case when we can't use sum of speeds because we need know when cars meet at the start of the circle distance. So we should not sum speeds but subtract it 8pi - 6pi = 2pi Circumference of circle equal to 2piR = 12pi 12pi / 2pi = 6 hours

Re: Two cars run in opposite directions on a circular track. Car A travels [#permalink]

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08 Apr 2015, 07:28

Bunuel wrote:

Two cars run in opposite directions on a circular track. Car A travels at a rate of \(6\pi\) miles per hour and Car B runs at a rate of \(8\pi\) miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

Assuming it is 6*Pi and not 6? as written in question stem . same with 8?.

circumference of track is \(12*Pi\) Car B will cover 8*pi in 1hour and car A has covered \(6*Pi\) in 1 hour time. Car B has covered \(2*pi\) miles more than car A . after how many hours of ride Car B will be 12*Pi ahead of Car A --> \(\frac{12*pi}{2*pi}\)=6hr

Answer :D
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08 Apr 2015, 09:27

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Bunuel wrote:

Two cars run in opposite directions on a circular track. Car A travels at a rate of \(6\pi\) miles per hour and Car B runs at a rate of \(8\pi\) miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

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08 Apr 2015, 09:57

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marinab wrote:

could you plz explain why finding the difference of speeds helps us to reach to the solution? And in what cases do we need to find it?

Our cars drive by circle. One clockwise and another counterclock wise (actually it doesnt' matter). And we need to find when this two cars intersect on the start.

First car driving with speed 6pi and another car with speed 8pi Circle equal to 12 pi

Hand calculating approach (for better understanding what's going on): After two hours, first car (6pi) will be on start but another car will drive 16 pi because it have speed 8pi and by two hours it'll be 16pi After the next hours first will drive 24pi and will be again on the start but second car will be on 32pi so it's not the start position And finally after six hours both cars will be simultaneously on the start 6pi * 6 = 36 (3 times 12) and 8pi * 6 = 48 (4 times 12)

Formulas approach: Why we should subtract speeds? Second car drives faster. So we want to know when second car make on 1 circle more than first car which drives slowly. We should find how much second car faster than first car. 8pi - 6pi = 2pi And for now we should divide distance 12pi on 2pi and we get 6 hours.

So we know that after 6 hours second car will drive on 1 circle more.
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Two cars run in opposite directions on a circular track. Car A travels at a rate of \(6\pi\) miles per hour and Car B runs at a rate of \(8\pi\) miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

What would we usually do in such a question? Two cars start from the same point and run in opposite directions – their speeds are given. This would remind us of relative speed. When two objects move in opposite directions, their relative speed is the sum of their speeds. So we might be tempted to do something like this:

Perimeter of the circle = 2\(\pi\)r = 2\(\pi\)*6 = 12? miles

Time taken to meet = Distance/Relative Speed = 12\(\pi\)/(6? + 8?) = 6/7 hrs

But take a step back and think – what does 6/7 hrs give us? It gives us the time taken by the two of them to complete one circle together. In this much time, they will meet somewhere on the circle but not at the starting point. So this is definitely not our answer.

The actual time taken to meet at point S will be given by 12\(\pi\)/(8\(\pi\) – 6\(\pi\)) = 6 hrs

This is what we mean by unexpected! The relative speed should be the sum of their speeds. Why did we divide the distance by the difference of their speeds? Here is why:

For the two objects to meet again at the starting point, obviously they both must be at the starting point. So the faster object must complete at least one full round more than the slower object. In every hour, car B – the one that runs at a speed of 8\(\pi\) mph covers 2\(\pi\) miles more compared with the distance covered by car A in that time (which runs at a speed of 6\(\pi\) mph). We want car B to complete one full circle more than car A. In how much time will car B cover 12\(\pi\) miles (a full circle) more than car A? In 12\(\pi\)/2\(\pi\) hrs = 6 hrs.
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Re: Two cars run in opposite directions on a circular track. Car A travels [#permalink]

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04 Nov 2016, 07:57

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Bunuel wrote:

Two cars run in opposite directions on a circular track. Car A travels at a rate of \(6\pi\) miles per hour and Car B runs at a rate of \(8\pi\) miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

Circumference of the track - \(12\pi\) Car A, with a speed of \(6\pi\), will get back to point S in two hours. Car B, with a speed of \(8\pi\), will get back to point S in 1h3m.

so let's say they start at 12:00 PM... 1:30PM car B full lap 2:00PM car A full lap

3:00PM car B full 2 lap 4:00PM car A full 2 lap

4:30PM car B full 3 lap 6:00PM car A full 3 lap

6:00PM car B full 4 lap 8:00PM car A full 4 lap

so..we have some overlaps here.. when Car A finishes 3 laps, car B finishes 4 laps, and both cars are at the same point..S... therefore, it takes 6 hours for them to meet again at the point S.

Re: Two cars run in opposite directions on a circular track. Car A travels [#permalink]

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04 Nov 2016, 08:02

Zhenek wrote:

Lets say t = meeting time, obviously cars won't meet in the starting point right away so they will definitely do some rounds first, lets say a = amount of rounds first car does, b - second car. We get an equation. \(t = \frac{2*pi*6*a}{6} = \frac{2*pi*6*b}{8}\) after solving the 2nd equation I get the relation between a and b: a/b = 3/4. So basically their first meeting at starting point happens when first car does 3 laps and 2nd car does 4 laps, therefore after we put a into our equation for t: \(t = \frac{2*pi*6*3}{6} = 6*pi\) hours. At this point I'm not sure what I'm doing wrong. edit: with the amendments to the task, result is 6, answer is D

awesome approach...didn't think about it...but i believe that it's the most easiest one...

Re: Two cars run in opposite directions on a circular track. Car A travels [#permalink]

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04 Nov 2016, 10:56

Bunuel wrote:

Two cars run in opposite directions on a circular track. Car A travels at a rate of \(6\pi\) miles per hour and Car B runs at a rate of \(8\pi\) miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

Re: Two cars run in opposite directions on a circular track. Car A travels [#permalink]

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24 Nov 2016, 18:08

Bunuel wrote:

Two cars run in opposite directions on a circular track. Car A travels at a rate of \(6\pi\) miles per hour and Car B runs at a rate of \(8\pi\) miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

Ans is A, Please correct if I am wrong, (A) says, it is less than one hour, means, A covers 6 miles and B covers 8 miles, the track distance is 12. It will meet less than a hour. SO the answer is A

Two cars run in opposite directions on a circular track. Car A travels at a rate of \(6\pi\) miles per hour and Car B runs at a rate of \(8\pi\) miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

Ans is A, Please correct if I am wrong, (A) says, it is less than one hour, means, A covers 6 miles and B covers 8 miles, the track distance is 12. It will meet less than a hour. SO the answer is A

The correct answer is D, not A. It is given under the spoiler in the original post. Several posts above present different solution to the problem. Please re-read the discussion above.
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Two cars run in opposite directions on a circular track. Car A travels [#permalink]

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02 Jul 2017, 05:55

Harley1980 wrote:

Bunuel wrote:

Two cars run in opposite directions on a circular track. Car A travels at a rate of 6? miles per hour and Car B runs at a rate of 8? miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

This is case when we can't use sum of speeds because we need know when cars meet at the start of the circle distance. So we should not sum speeds but subtract it 8pi - 6pi = 2pi Circumference of circle equal to 2piR = 12pi 12pi / 2pi = 6 hours

Answer is D

Subtracting the two speeds will not always give the correct answer.

Suppose the values of the speed given in the question are changed to 4 \pi and 16 \pi, and radius is kept same to 12 \pi.

Subtracting the two speeds will give 12 \pi which when divided by the circumference 12 \pi will result into 1 hr. Note that when time elapsed is 1 hr both cars are at diametrically opposite ends instead of meeting at the starting point.

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21 Jul 2017, 07:02

Since the track is circular so if the cars are driving in opposite direction it can be considered they meet twice So to calculate the hours taken there can be two scenarios 1) after how many hours the two cars will meet again. Find the sum of the speeds 2) after how many hours the two cars will meet at the starting point again . Find the difference of the speeds. Now that we know the circumference is 12*pi. Every hour, car B that runs at a speed of 8π mph covers 2π miles more compared with the distance covered by car A in that time (which runs at a speed of 6ππ mph). So car B to cover 12π miles (a full circle) more than car A should be given by 12π/2π hrs = 6 hrs.

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