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Car B starts at point X and moves clockwise around
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26 Feb 2012, 21:51
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Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counterclockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)? A. 4pi – 1.6 B. 4pi + 8.4 C. 4pi + 10.4 D. 2pi – 1.6 E. 2pi – 0.8
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26 Feb 2012, 22:13
NYCAnalyst wrote: Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counterclockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
A. 4pi – 1.6 B. 4pi + 8.4 C. 4pi + 10.4 D. 2pi – 1.6 E. 2pi – 0.8 Step by step analyzes: 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. Also discussed here: ratesonacirculartrack86675.html#p849908Hope it helps.
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Re: Car B starts at point X and moves clockwise around
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27 Feb 2012, 04:51
NYCAnalyst wrote: Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counterclockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
A. 4pi – 1.6 B. 4pi + 8.4 C. 4pi + 10.4 D. 2pi – 1.6 E. 2pi – 0.8 It is a long question stem so make sure you analyze each statement as you read it otherwise you will waste a lot of time rereading the question again and again. Let me tell you how: Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ok, car B covers 2 miles every hour moving clockwise. We don't know the track length yet.Ten hours later, Car A leaves from point X and travels counterclockwise around the same circular track at a constant rate of 3 mph. Car A is faster and covers 3 miles every hour counter clockwise. Since their directions are opposite, their relative speed = 2+3 = 5 mph. We still don't know the track length so we cannot say where car B was when car A started. All we know is that in 10 hrs, car B traveled 20 miles If the radius of the track is 10 miles, Now we know the track length. It is \(2{\pi}r = 2{\pi}*10 = 20{\pi}.\) It is greater than 20 miles that car B covered in 10 hrs so car B had not finished one round. Of the \(20{\pi}\), it had covered 20 miles. for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)? To pass each other for the first time, cars A and B together need to cover the remaining distance on the circle i.e. \(20{\pi}  20\). To create a distance of another 12 miles, they together need to travel 12 miles more away from each other
Time taken = Total distance to be traveled/ Relative Speed Time taken = \((20{\pi}  20 + 12)/5 = (20{\pi}  8)/5 hrs = 4{\pi}  1.6 hrs\) Car B must have been traveling for \(4{\pi}  1.6 + 10 = 4{\pi} + 8.4 hrs\)
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Re: Car B starts at point X and moves clockwise around
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Re: Car B starts at point X and moves clockwise around
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28 Feb 2012, 20:45
NYCAnalyst wrote: Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counterclockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
A. 4pi – 1.6 B. 4pi + 8.4 C. 4pi + 10.4 D. 2pi – 1.6 E. 2pi – 0.8 My solution  Let us assume they meet at time t. Till that time, Car B will have travelled a distance of 2t (distance = speed *time). Car A will have travelled a distance equal to (t10)*3 by the same logic. So, we have 2t + (t10)*3 = 2 * pi * 10 (at t ) For putting in another 12 miles between them, 2t + (t10)*3 = 2 * pi * 10 + 12 which gives 2t + 3t  30 = 20*pi+12 5t  30 = 20*pi +12 5t = 20*pi + 42 t = 4*pi + 8.4 ( Answer B)  Deepti



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Cars on circular track
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03 Mar 2012, 14:18
Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counterclockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)? A) 4pi – 1.6 B) 4pi + 8.4 C) 4 pi+ 10.4 D) 2 pi – 1.6 E) 2pi – 0.8 This is how I tried solving this, but got stuck after this. The two cars travel around the circumference of a circle, the measure of which is 2 pi r = 2 pi × 10 = 20 pi miles. Car B has already traveled (2 mph)(10 hours) = 20 miles in a clockwise direction by the time Car A starts moving 10 hours later in a counterclockwise direction (and from the same starting point). At Car A’s start time, the additional distance to cover before the cars meet is (20pi – 20) miles. We are interested in the moment when the cars have passed each other and have traveled another 12 miles: (20 pi – 20) + 12 = (20 pi – 8) miles. No idea what to do after this.
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Re: Car B starts at point X and moves clockwise around
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18 Apr 2012, 01:29
i don't know where i am making the mistake ...
Car B has already traveled (2 mph)(10 hours) = 20 miles in a clockwise direction by the time Car A starts moving 10 hours later in a counterclockwise direction (and from the same starting point).
let x distance from Car A starting point the two cars meet
then distance covered by car B=2pi*1020x time taken by car B=(2pi*1020x)/2 distance covered by car A=x time taken by car A=x/3
equating the two, x/3=(2pi*1020x)/2 2x=60*pi603x x=(60*pi60)/5 x=12(pi1) car B has to travel another 12 miles,therefore time taken by car B=(2pi*1020x+12)/2
or (20*pi812*pi+12)/2=4*pi+2



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Re: Car B starts at point X and moves clockwise around
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18 Apr 2012, 03:33
mrinal2100 wrote: i don't know where i am making the mistake ...
Car B has already traveled (2 mph)(10 hours) = 20 miles in a clockwise direction by the time Car A starts moving 10 hours later in a counterclockwise direction (and from the same starting point).
let x distance from Car A starting point the two cars meet
then distance covered by car B=2pi*1020x time taken by car B=(2pi*1020x)/2 distance covered by car A=x time taken by car A=x/3
equating the two, x/3=(2pi*1020x)/2 2x=60*pi603x x=(60*pi60)/5 x=12(pi1) car B has to travel another 12 miles,therefore time taken by car B=(2pi*1020x+12)/2
or (20*pi812*pi+12)/2=4*pi+2 What does the red part even mean? If it means the distance between B and A by the time, A starts to travel then we know exactly what it is. It equal to: \(20*\pi20\) ( carbstartsatpointxandmovesclockwisearound128215.html#p1050349).
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Re: Car B starts at point X and moves clockwise around
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18 Apr 2012, 04:28
i have explained the statement marked in red "let x distance from Car A starting point the two cars meet" by means of diagram.Both the cars meet at the point.Car A travels x mile while car B travels a distance of 2pi*1020x.Since the time taken would be equal,we are equating the equation x/3=(2pi*1020x)/2
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Re: Car B starts at point X and moves clockwise around
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18 Apr 2012, 09:24
mrinal2100 wrote: i don't know where i am making the mistake ...
Car B has already traveled (2 mph)(10 hours) = 20 miles in a clockwise direction by the time Car A starts moving 10 hours later in a counterclockwise direction (and from the same starting point).
let x distance from Car A starting point the two cars meet
then distance covered by car B=2pi*1020x time taken by car B=(2pi*1020x)/2 distance covered by car A=x time taken by car A=x/3
equating the two, x/3=(2pi*1020x)/2 2x=60*pi603x x=(60*pi60)/5 x=12(pi1)
Everything is correct till here. Now, x is the distance traveled by A so time taken = x/3 = 12(pi1)/3 = 4(pi1) Now notice that they have to together put 12 miles distance between them. They are moving in opposite directions so time taken to cover 12 miles together = 12/(2+3) = 2.4 hrs (relative speed concepts) Total time taken = 4(pi1) + 2.4 = 4*pi  1.6 hrs Car B has traveled 10 hrs extra so it must have been traveling for 4*pi  1.6 + 10 = 4*pi + 8.4 hrs car B has to travel another 12 miles,therefore time taken by car B=(2pi*1020x+12)/2
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Re: Car B starts at point X and moves clockwise around
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18 Apr 2012, 21:32
Hi Karishma,
I have been fan of your logical approach, which I keep applying wherever possible. More so when I am not very comfortable with algebra part of word problems. I am attempting to solve this question based on logic.
I will consider Pi as 3. So the distance round the circle is 20pi or 60 mi approx. Out of this Car B has already traveled 20 mi in 10 hours. Thus we are left with 40 mi to be covered by A and B together but driving in opposite direction @ 3km/hr and 2km/hr respectively. Thus to cover balance distance plus 12km i.e. 52miles, I need 52/5=10.4 hours.
Thus answer is 10hours plus 10.4 hours i.e. 20.4 hours and answer choice B (4pi+8.4) is correct.



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Re: Car B starts at point X and moves clockwise around
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13 May 2012, 02:29
I think there is a flaw in this question... moving on a circular track could not be CONSTANT RATE! it has variable velocity So you could not solve the problem with Rate= Distance/ time Formula. Because this formulation is just for straight distances not for circular ones...
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Re: Car B starts at point X and moves clockwise around
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14 May 2012, 07:58
manjeet1972 wrote: Hi Karishma,
I have been fan of your logical approach, which I keep applying wherever possible. More so when I am not very comfortable with algebra part of word problems. I am attempting to solve this question based on logic.
I will consider Pi as 3. So the distance round the circle is 20pi or 60 mi approx. Out of this Car B has already traveled 20 mi in 10 hours. Thus we are left with 40 mi to be covered by A and B together but driving in opposite direction @ 3km/hr and 2km/hr respectively. Thus to cover balance distance plus 12km i.e. 52miles, I need 52/5=10.4 hours.
Thus answer is 10hours plus 10.4 hours i.e. 20.4 hours and answer choice B (4pi+8.4) is correct. Yes, that's absolutely fine. I followed the exact same approach above (but with pi). I hope you understand that the approximation of pi worked because the answer was also in terms of pi. If you had a pure numerical answer, you would not have assumed pi = 3. Here, pi acted like a variable. Say, if you put pi = 10, you will still get the answer (of course you will need to substitute pi = 10 in the options too)
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Re: Car B starts at point X and moves clockwise around
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14 May 2012, 08:02
omidsa wrote: I think there is a flaw in this question...
moving on a circular track could not be CONSTANT RATE! it has variable velocity So you could not solve the problem with Rate= Distance/ time Formula. Because this formulation is just for straight distances not for circular ones... Velocity has two components  speed and direction. Velocity is variable here only because of changing direction. Speed = Distance/Time is applicable for circular distances as well since speed is not a vector. GMAT only deals with speed, not with velocity. Constant rate implies a constant speed.
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Re: Car B starts at point X and moves clockwise around
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11 Jul 2012, 19:13
"for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them" i have a little problem with the language of the question stem ..isint the question asking that we have to calculate the time of travel since they met for the first time and to the point when the distance between them is 12 miles .in that case the answer wud have been straight 12 /5 hrs . i feel the language is little faulty .karisma ,buneul plz claify



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Re: Car B starts at point X and moves clockwise around
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aditya8062 wrote: "for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them" i have a little problem with the language of the question stem ..isint the question asking that we have to calculate the time of travel since they met for the first time and to the point when the distance between them is 12 miles .in that case the answer wud have been straight 12 /5 hrs . i feel the language is little faulty .karisma ,buneul plz claify The language of the question isn't very good. This does lead to different interpretations by different people sometimes. But most people will interpret it as 'from the starting point to right now (which is the point when the cars have passed each other for the first time and put another 12 miles between them)' In GMAT, the language is always very clear. e.g. you could be given "After some time, the cars had put a distance of 12 miles between them after meeting for the first time. For how many total hrs did car B travel till this time?"  In this case, your answer is the one given above. or you could have "After meeting for the first time, the cars continued on their respective paths and put a distance of 12 miles between them (they have met only once). How long has car B been traveling since meeting car A for the first time?"  In this case, your answer is 12/5
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Re: Car B starts at point X and moves clockwise around
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NYCAnalyst wrote: Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counterclockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
A. 4pi – 1.6 B. 4pi + 8.4 C. 4pi + 10.4 D. 2pi – 1.6 E. 2pi – 0.8 It is easier to understand the question using a drawing: \(A_1\) and \(B_1\) are the points A and B are respectively after their first time pass. If we denote by T the time B travels from X to \(B_1\), then the total distance the two cars travel until the situation depicted in the figure is one whole circumference of the circle plus the overlap of 12 miles. Car A travels \(T10\) hours. We can write the following equation: \(2T+3(T10)=2\pi10+12\) or \(5T=20\pi+42\), from which we get \(T=4\pi+8.4\). Answer: B
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Car B starts at point X and moves clockwise around a circular tr
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23 Sep 2012, 00:21
Let rA and rB denote the speeds of A and B, respectively rA = 3mph, rB= 2mph C = 20*π miles (circumference of the circle) In 10h B will have covered 20 miles (10h * 2mph) The remaining distance 20*π  20 miles and an extra 12 miles (20*π 8 ) got to be covered both by A and B with relative speed of 5mph. Thus B will have been racing a total time of 10h + ((20*π 8 )/5) h = 4*π + 8.4 hours Brother Karamazov



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Re: Car B starts at point X and moves clockwise around
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Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counterclockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
This problem deals with the distance the cars have moved over a certain period of time so we should find the distance (circumference) of the track: r=10 c=(pi)d c=(pi)(20) c=20(pi)
Car B leaves 10 hours before car A. When car A starts traveling B has traveled: distance=rate*time: distance=2miles/hour*10 hours: D=20 miles.
We need to know how long it takes for them to reach one another then how long it takes for them to put another to miles between the two of them.
When car A leaves B has already traveled for 10 hours and 20 miles. Therefore, car A and B have [(pi)20  20] miles to travel before they reach one another.
Their combined rate is 2miles/hour and 3miles/hour = 5 miles/hour
Time = distance/rate Time = [(pi)20  20]/5 Time = [(pi)4  4]
It takes [(pi)4  4] hours for them to reach one another.
We now have to figure out how long it takes for them to travel past one another and put 12 miles between them:
Distance = r*t 12 = (2+3)*t 12/5 = time
It takes another 2.4 hours for the two cars to travel a combined 12 hours away from one another.
So, Car B has been on the circuit for 10 hours + (pi)4  4 hours + 2.4 hours: (pi)4+8.4 hours total.
ANSWER: B. 4pi + 8.4



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Re: Car B starts at point X and moves clockwise around
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NYCAnalyst wrote: Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counterclockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
A. 4pi – 1.6 B. 4pi + 8.4 C. 4pi + 10.4 D. 2pi – 1.6 E. 2pi – 0.8 Hi Guys, here's my approach. I just don's like using constant Pi here and will replace it with ~ 3 Radius = 10 so the circumference = 20*Pi = 20*3=60 Car B = has already traveled 10 Hours x 2 mph = 20 and Rate A + Rate B = 5. They must travel 40 miles to meet and additional 12 miles as stated in the question = 52 Miles \(5*t=52\) so t=10,4 Hours + 10 hours previously traveled by Car B = 20,4 = 4pi + 8.4 or 4*3+8,4=20,4
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