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Car B begins moving at 2 mph around a circular track with

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

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New post 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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New post 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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New post 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(T-10) = (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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New post 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: \(20*\pi-20\)

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?

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

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New post 28 Jul 2013, 13:53
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

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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New post 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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New post 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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New post 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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New post 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 ? :| :scared :scared


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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New post 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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New post 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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New post 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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New post 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
Car B begins moving at 2 mph around a circular track with   [#permalink] 25 Feb 2016, 14:54

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