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Re: Car B starts at point X and moves clockwise [#permalink]
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11 Jan 2011, 21:30
VeritasPrepKarishma wrote: Radius of track is 10 miles so circumference is 20*pi i.e. the total length of the track. B starts from X and travels for 10 hrs clockwise at 2 mph i.e. it travels 20 miles. Now car A starts from X counter clockwise. Distance between A and B is 20*pi  20. Now, to meet, they have to together cover this distance plus 12 miles more which they have to put between them. Time taken to cover this distance by them = (20*pi  20 + 12)/(3 + 2) = 4*pi  1.6 hrs Car B has been traveling for 10 + 4*pi  1.6 = (4*pi + 8.4) hrs Hi Karishma, i actually got this question wrong when i took the mgmat cat last week, i got confused on the explanation which is similar to yours (your diagram helps though), how did you derive this equation: (20*pi  20 + 12)/(3 + 2), was this manipulated from Rate x Time = Distance? thanks.



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Re: Rates on a circular track [#permalink]
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13 Jan 2011, 21:05
thank you Karishma for taking the time to explain this problem, i'll review a bit on relative speed since it is kind of new for me but your explanation is very helpful, as usual.



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Re: Car B begins moving at 2 mph around a circular track with [#permalink]
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14 Mar 2013, 12:56
EASY EQUATION: I think the easy way to calculate is.
Distance travelled by B + Distance travelled by A = Circumference + 12 Let's sat the answer is T.
2T + 3(T10) = (2 * Pi * 10 )+ 12 5t = 20 pi + 42 t= 4 pi + 8.4



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Re: Rates on a circular track [#permalink]
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28 Jul 2013, 12:59
Bunuel wrote: Car B begins moving at 2 mph around a circular track with a radius of 10 miles. Ten hours later, Car A leaves from the same point in the opposite direction, traveling at 3 mph. For how many hours will Car B have been traveling when car A has passed and moved 12 miles beyond Car B? A. 4pi – 1.6 B. 4pi + 8.4 C. 4pi + 10.4 D. 2pi – 1.6 E. 2pi – 0.8
It's possible to write the whole formula right away but I think it would be better to go step by step:
B speed: \(2\) mph; A speed: \(3\) mph (travelling in the opposite direction); Track distance: \(2*\pi*r=20*\pi\);
What distance will cover B in 10h: \(10*2=20\) miles Distance between B and A by the time, A starts to travel: \(20*\pi20\)
Time needed for A and B to meet distance between them divided by the relative speed: \(\frac{20*\pi20}{2+3}= \frac{20*\pi20}{5}=4*\pi4\), as they are travelling in opposite directions relative speed would be the sum of their rates;
Time needed for A to be 12 miles ahead of B: \(\frac{12}{2+3}=2.4\);
So we have three period of times: Time before A started travelling: \(10\) hours; Time for A and B to meet: \(4*\pi4\) hours; Time needed for A to be 12 miles ahead of B: \(2.4\) hours;
Total time: \(10+4*\pi4+2.4=4*\pi+8.4\) hours.
Answer: B. If the question was changed so that Car A starts travelling in the same direction as Car B, how will the solution be different? Do we just do a subtraction while calculcating the relative speed of the two cars? I.E. 32 instead of 3+2 in the denominator? Thanks



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Re: Car B begins moving at 2 mph around a circular track with [#permalink]
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28 Jul 2013, 13:53
That logic (32) applies to calculate the time required to keep 12 miles between car A and car B after they meet, but the 1st part is different since the distance between car A and car B when car A start is only 20 miles and not 20pi  20 miles The 1st equation will be 20 + 2t = 3t ==> t = 20 hours Hope this helps.
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Re: Car B begins moving at 2 mph around a circular track with [#permalink]
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02 Aug 2013, 10:52
Car B begins moving at 2 mph around a circular track with a radius of 10 miles. Ten hours later, Car A leaves from the same point in the opposite direction, traveling at 3 mph. For how many hours will Car B have been traveling when car A has passed and moved 12 miles beyond Car B?
R=10 c=2(pi)r Track circumference =20(pi) In 10 hours car B will have traveled 10*2=20 miles So when car A starts, car B will have a 20 mile head start on it. When A leaves, it leaves in the opposite direction. Therefore, it's not simply 20 miles behind B. For example, look at a clock. Pretend B left from where 12 is on the clock and is currently sitting on where 4 is. If A left and followed B it would be 1/3rd of the clocks circumference behind B. However, if it leaves in the opposite direction it has all the numbers between 12 and 4 between it and B, or 2/3rds of the clocks circumference between it and B. Therefore, the distance between A and B is:
20(pi)20
The time it takes for them to pass one another is the distance they must travel to do so [20(pi)20] divided by their two rates of travel (2 and 3 miles/hour)
[20(pi)20] / (2+3) [20(pi)20] / (5) Time = 4(pi)4
The time it takes for A to move 12 miles AWAY from B is their combined rate of speed: T = 12/(2+3) This caused me much confusion at first. I treated it as if A and B were moving in the same direction and I was looking for how fast A was pulling ahead of B. They are moving in opposite directions at 2 and 3 miles per hour respectively. It would be no different than if one car was moving away from point x at a speed of (2+3) The distance it would put between itself and X would be the same distance A and B put between them at 3 and 2 Miles/hour respectively!
The time it takes for A and B to move 12 miles away from one another is 12/5 = 2.4 hours.
Therefore, it takes 4(pi)4 hours for them to reach one another + another 2.4 hours for them to move another 12 miles away from one another. Keep in mind, we also need to add in the 10 hours car B traveled before car A left because the question is looking for the total number of hours car B has been on the road when car A is ten miles past it in the opposite direction.
Therefore, Car B has been traveling for 10+4(pi)4+2.4 hours
Answer: (B) 4(pi)+8.4 hours



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Re: Car B begins moving at 2 mph around a circular track with [#permalink]
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15 Sep 2013, 02:24
yangsta8 wrote: Car B begins moving at 2 mph around a circular track with a radius of 10 miles. Ten hours later, Car A leaves from the same point in the opposite direction, traveling at 3 mph. For how many hours will Car B have been traveling when car A has passed and moved 12 miles beyond Car B?
A. 4pi – 1.6 B. 4pi + 8.4 C. 4pi + 10.4 D. 2pi – 1.6 E. 2pi – 0.8
The OA is pretty long and even solving it that way takes me +2 mins. Hopefully someone can offer a fast solution. It can also be solved by Tabular form as is suggested in the GMAT Club Math Book.
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Re: Car B begins moving at 2 mph around a circular track with [#permalink]
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19 Sep 2013, 06:44
Ugh. That's so sleazy to call the first car "Car B" and the second car "Car A". That's what tripped me up.



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Re: Car B begins moving at 2 mph around a circular track with [#permalink]
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19 Sep 2013, 07:06
Perhaps wrote: these type of ques can really come in gmat????? if v r not able to do these type of ques...how much it cud effect our scores ? This is actually not that hard if you have your basics right!! I learnt this tabular format in the Math GMAT Club book. Might help you out with such questions. It has helped me for sure.
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Re: Car B begins moving at 2 mph around a circular track with [#permalink]
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25 Dec 2013, 03:24
the length of the circular track is ~63 miles(2*pi*r). B and A are travelling in opp directions B started earlier at 2 mph, travelling for 10 hrs=dist. covered 20 miles. now A starts from opp direction at 3 mph from same point(the key clue) and both A and B will cover ~43 miles at the combined speed of 5 mph which give time as 8.6 hrs for each A and B. question also involves additional travel of 12 miles in opp direction which results in additional 2.4 hrs for each A and B. so car B has been travelling for 10 hrs+8.6 hrs +2.4 hr=21 hrs.
option B) 4pi + 8.4 = 20.97 hrs = ~21 hrs



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Re: Car B begins moving at 2 mph around a circular track with [#permalink]
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25 Dec 2013, 03:25
the length of the circular track is ~63 miles(2*pi*r). B and A are travelling in opp directions B started earlier at 2 mph, travelling for 10 hrs=dist. covered 20 miles. now A starts from opp direction at 3 mph from same point(the key clue) and both A and B will cover ~43 miles at the combined speed of 5 mph which give time as 8.6 hrs for each A and B. question also involves additional travel of 12 miles in opp direction which results in additional 2.4 hrs for each A and B. so car B has been travelling for 10 hrs+8.6 hrs +2.4 hr=21 hrs.
option B) 4pi + 8.4 = 20.97 hrs = ~21 hrs



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Re: Car B begins moving at 2 mph around a circular track with [#permalink]
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Car B begins moving at 2 mph around a circular track with [#permalink]
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24 Feb 2016, 06:35
The problem gets easier when we get rid of unnecessary abstraction caused by presence of pie. Just take the length of the lap as 60 miles (2*pie*10). Second, understand that the cars are moving towards each other/from each other hence you need to add up the individual speeds (3+2). So in first 10 hours B covered 20 miles (2 *10) and when A started off only 40 miles separated them on the lap. This will be covered in 8 hours (40/5). Finally after they meet and go in opposite directions again 12 miles will be covered in 2.4 hours (12/5). So in total this sums up to 10+8+2.4 = 20.4 hours for B. Answer B only fits if we take pie for 3.
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Car B begins moving at 2 mph around a circular track with [#permalink]
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25 Feb 2016, 14:54
there are 3 legs to B's trip: leg 1=10 hours before A starts leg 2=(20⫪20)/(2+3)=4⫪4 hours before meeting A leg 3=12/(2+3)=2.4 hours before A moves 12 miles beyond B total time for B's trip=4⫪+8.4 hours



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Re: Car B begins moving at 2 mph around a circular track with [#permalink]
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29 Sep 2016, 00:02
yangsta8 wrote: Car B begins moving at 2 mph around a circular track with a radius of 10 miles. Ten hours later, Car A leaves from the same point in the opposite direction, traveling at 3 mph. For how many hours will Car B have been traveling when car A has passed and moved 12 miles beyond Car B?
A. 4pi – 1.6 B. 4pi + 8.4 C. 4pi + 10.4 D. 2pi – 1.6 E. 2pi – 0.8
The OA is pretty long and even solving it that way takes me +2 mins. Hopefully someone can offer a fast solution. d = 2*pi*r = 20pi let pi = 3 (roughly), hence d would be 66 after 10 hrs B would have travelled 20 miles, so remaining d = 46 A travels in opposite direction, so inorder to meet time taken would be 46/2+3 = 9.2 time for further 12 miles would be 12/2+3 = 2.4 B has already been travelling for 10 hours, so total time = 9.2+2.4+10 = 21 (approx) from answer choices B and C are closest B is 4pi +8.4 => 12+8.4 = 20.4 C is 4pi + 10 => 12+10.4 = 22.4 Since B is closer, I went with B
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Re: Car B begins moving at 2 mph around a circular track with [#permalink]
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13 Jun 2017, 17:36
This is still a very challenging question for me, particularly because the distance of the track can loop
For example, what if B was going at a rate of 8 miles per hour for 20 hours before A started  how then would you calculate the distance between them?
If the track is only 20*pi miles (roughly 60.2 miles) and car B has driven 160 miles, how would that change the set up to find the distance between A and B when A is starting?




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