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

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Car B begins moving at 2 mph around a circular track with [#permalink] New post 11 Nov 2009, 01:34
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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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Mar 2012, 09:49, edited 2 times in total.
Edited the question and added the OA
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Re: Rates on a circular track [#permalink] New post 11 Nov 2009, 02:10
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Alright, first of all, let's determine the distance of the circular track:

Circumference = pi*d = 20*pi

Now, let's represent each car's distance from the starting point (along the track), t hours from when Car A starts:

B(t) = 20 + 2t [Distance traveled after 10 hours, + 2mph)
A(t) = 20*pi - 3t [Starting at starting point (20pi), -3mph)

We need to determine the time it takes for car A to be 12 miles past car B.

B(t) - A(t) = 12
20 + 2t - (20*pi -3t) = 12
t = (-8 + 20pi)/5
t = 4pi - 1.6

Therefore, car A has been traveling (4pi - 1.6) hours before the criterion is satisfied.The question, however, asks how long car B has been traveling.

t + 10 = 4pi - 1.6 + 10
= 4pi + 8.4

Therefore the answer is B: 4pi + 8.4.
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Re: Rates on a circular track [#permalink] New post 11 Nov 2009, 02:12
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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.
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Re: Rates on a circular track [#permalink] New post 11 Nov 2009, 02:54
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Where I am going wrong? Please tell me

The total distance the cars need to travel at relative speed of 2+3=5 mph
is 2piX10-20 + 12 miles
time required is
(2piX10 -20 + 12)/5 = 4pi -1.6
Answer -- A
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Re: Rates on a circular track [#permalink] New post 11 Nov 2009, 03:13
Expert's post
pradhan wrote:
Where I am going wrong? Please tell me

The total distance the cars need to travel at relative speed of 2+3=5 mph
is 2piX10-20 + 12 miles
time required is
(2piX10 -20 + 12)/5 = 4pi -1.6
Answer -- A


You are calculating time Car A have been traveling when car A has passed and moved 12 miles beyond Car B. And we are asked about the time for car B. As car B was travelling 10 more hours before A started, so you just should add 10 to your calculations.
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Re: Rates on a circular track [#permalink] New post 11 Nov 2009, 03:28
Thanks a lot Bunuel. I must learn to read the question properly.
The answer is B.
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Re: Rates on a circular track [#permalink] New post 17 Nov 2009, 12:59
Also, instead of using "pi" just use "3" and multiply. Instead of 20pi the track is 20*3= 60 miles, and the total time for B will be 20.4 = 4pi +8.4
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Re: Rates on a circular track [#permalink] New post 19 Sep 2010, 02:19
Solved the equation (20pi-20)-3t-2t=-12, but forgot to add the first 10 hours as well, so I got A at first.
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Re: Rates on a circular track [#permalink] New post 01 Oct 2010, 08:36
I also forgot to add the first 10 hours =). To reckognize all the details is sooo important!!!!
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Re: Car B starts at point X and moves clockwise [#permalink] New post 10 Jan 2011, 18:48
1. Circumference = 2(pi)r = 2(pi)10 = 20(pi)
2. Car B travels for 10 hours @ 2 miles = 20 miles
3. Car A starts at same location but travels counter clock wise (so A and B approaching each other = add speed)
4. Distance [Remaining] between two cars [when car A starts] = 20(pi) [total] - 20 [traveled by B]
5. cars need to travel additional 12 miles so total distance to travel is 20(pi) - 20 + 12
6. time = distance / speed = (20(pi)-8) / 5 = 4(pi) - 1.6
7. We have been asked to find how much time car B is travelling = 10 + 4(pi) - 1.6 = 4(pi) + 8.4

Hope This Helps
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Re: Car B starts at point X and moves clockwise [#permalink] New post 10 Jan 2011, 19:01
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ajit257 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 counter-clockwise 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

did not get this one.



Make the diagram. You will be able to see how to solve it.
Attachment:
Ques1.jpg
Ques1.jpg [ 16.63 KiB | Viewed 10341 times ]

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
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Re: Car B starts at point X and moves clockwise [#permalink] New post 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.
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Re: Car B starts at point X and moves clockwise [#permalink] New post 12 Jan 2011, 19:19
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gtr022001 wrote:
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.

Attachment:
Ques1.jpg
Ques1.jpg [ 14.88 KiB | Viewed 10198 times ]


The red distance is what B has already covered at 2 mph in 10 hrs. This distance is 20 miles.
A and B are now moving towards each other (as shown by green arrows). To meet for the first time, they have to cover the remaining circumference of the track i.e. a distance of 20pi - 20. (20pi is the circumference of the circle out of which 20 has already been covered by B). They need to create a further 12 miles distance between them. So together they need to cover (20pi - 20 + 12) miles in all.
Since, A and B are moving towards each other, their relative speed (i.e. combined speed here) will be (3 + 2) mph.
So time taken for them to meet = D/S = (20pi - 20 + 12)/(3 + 2)

- Here, we are using the concept of Relative Speed. When two objects (speeds S1 and S2) move in opposite directions (towards each other or away from each other), they cover the distance between them (or create distance between them) at the rate of (S1 + S2). Here they are moving in opposite directions towards each other so their relative speed is sum of their speeds. After meeting, they are moving away from each other but their relative speed is still sum of their speeds.
When two objects move in same direction, their speeds get subtracted.
If this is unclear, I would suggest looking up the theory of relative speed for details.
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Re: Rates on a circular track [#permalink] New post 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.
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Re: Rates on a circular track [#permalink] New post 28 Mar 2012, 09:47
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ok.

SO Since they are travelling in opposite directions then the sum of their individual distances should be equal to the total distance.

Therefore,
Let T be the time they meet
Distance travelled by A = 3x(t-10), since it started 10 hours after B
Distance travelled by B = 2xt,

Total distance =20Pi + 12

hence
Distance of A + Distance of B=Total distance
2t+3(t-10)=20pi+12
Solving.....
5t=20pi+42

t=4pi+8.4

the interesting thing is, the object which starts later, say "A", in this case is subtracted from time T by the number of hours it starts late.
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Re: Car B begins moving at 2 mph around a circular track with [#permalink] New post 14 Mar 2013, 11:56
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: Car B begins moving at 2 mph around a circular track with [#permalink] New post 15 Mar 2013, 11:40
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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
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Re: Car B begins moving at 2 mph around a circular track with [#permalink] New post 17 Mar 2013, 22:07
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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


If you are hoping for a high Quant score then you can certainly come across such a question. The effect it will have on your score is the effect you let it have - if you put in 5 mins to solve it, do not still get it, guess on it and get all bogged down, it will have a big effect on your score. If you try to work it out for a couple of mins but are not able to so guess and move on and just take it in your stride, it will not have much impact. One question doesn't decide your score.
But, if you already know that you don't know how to handle such questions, put in the effort to learn right now rather than worry in the test.
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Re: Car B begins moving at 2 mph around a circular track with [#permalink] New post 20 Mar 2013, 23:13
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Hi,
Can u plz tell me where i am going wrong:
Rate of Car B Rb=2mph,Ra=3mph
time taken by car B = t+10
Time taken by A = t

since both will meet at some point,so they cover entire distance which is 20pi miles
so equating:
2(t+10)= 3t
t=20hrs
so they meet in 20hrs time
Now we need to check how much time B wouldve spent to cover the additional 12 miles.
Its speed is 2mph so to cover 12miles it will take 6 hrs.
How does pi come into the answer choices?
Can u explain the correct approach using 20hrs meeting time as the starting point.
Plz help.

Thanks,
Shreeraj
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Re: Car B begins moving at 2 mph around a circular track with [#permalink] New post 21 Mar 2013, 00:35
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shreerajp99 wrote:
Hi,
Can u plz tell me where i am going wrong:
Rate of Car B Rb=2mph,Ra=3mph
time taken by car B = t+10
Time taken by A = t

since both will meet at some point,so they cover entire distance which is 20pi miles
so equating:
2(t+10)= 3t


From where do you get this equation? You are assuming that the distances covered by them are equal. That is not the case. They together covered the entire circumference of the circle which is 20\pi. We can't say that they covered equal distances of 10\pi each.
Check the diagram and explanation given in my post above.
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Re: Car B begins moving at 2 mph around a circular track with   [#permalink] 21 Mar 2013, 00:35
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