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Re: Rates on a circular track [#permalink]
13 Jan 2011, 20:05
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.
Re: Rates on a circular track [#permalink]
28 Jul 2013, 11:59
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: \(20*\pi-20\)
Time needed for A and B to meet distance between them divided by the relative speed: \(\frac{20*\pi-20}{2+3}= \frac{20*\pi-20}{5}=4*\pi-4\), 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: \(4*\pi-4\) hours; Time needed for A to be 12 miles ahead of B: \(2.4\) hours;
Total time: \(10+4*\pi-4+2.4=4*\pi+8.4\) 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. 3-2 instead of 3+2 in the denominator?
Re: Car B begins moving at 2 mph around a circular track with [#permalink]
28 Jul 2013, 12:53
That logic (3-2) 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
Re: Car B begins moving at 2 mph around a circular track with [#permalink]
02 Aug 2013, 09:52
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
Re: Car B begins moving at 2 mph around a circular track with [#permalink]
15 Sep 2013, 01:24
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.
Attachments
Hours_travelled_By_Car_B.png [ 13.24 KiB | Viewed 861 times ]
Re: Car B begins moving at 2 mph around a circular track with [#permalink]
19 Sep 2013, 06:06
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.
Attachments
Hours_travelled_By_Car_B.png [ 13.24 KiB | Viewed 837 times ]
Re: Car B begins moving at 2 mph around a circular track with [#permalink]
19 Sep 2013, 20:54
Expert's post
mfabros wrote:
Ugh. That's so sleazy to call the first car "Car B" and the second car "Car A". That's what tripped me up.
Yes, actual GMAT questions will not try to trick you in such an uncool manner. If you get tricked by something in GMAT, it will be conceptual such that when you see the explanation you will go 'oh wow!' _________________
Re: Car B begins moving at 2 mph around a circular track with [#permalink]
25 Dec 2013, 02:24
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.
Re: Car B begins moving at 2 mph around a circular track with [#permalink]
25 Dec 2013, 02:25
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.
Re: Car B begins moving at 2 mph around a circular track with [#permalink]
26 Dec 2014, 15:41
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