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Re: Car B starts at point X and moves clockwise [#permalink]

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11 Jan 2011, 20:30

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.

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.

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]

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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]

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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]

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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 1159 times ]

Re: Car B begins moving at 2 mph around a circular track with [#permalink]

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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 1136 times ]

Re: Car B begins moving at 2 mph around a circular track with [#permalink]

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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]

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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]

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26 Dec 2014, 15:41

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Car B begins moving at 2 mph around a circular track with [#permalink]

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24 Feb 2016, 05:35

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, 13:54

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

Re: Car B begins moving at 2 mph around a circular track with [#permalink]

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28 Sep 2016, 23:02

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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If my post was helpful, feel free to give kudos!

gmatclubot

Re: Car B begins moving at 2 mph around a circular track with
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28 Sep 2016, 23:02

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