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

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11 Nov 2009, 02:34

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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.

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;

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

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. _________________

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

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10 Jan 2011, 19:48

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

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

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10 Jan 2011, 20:01

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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)?

Make the diagram. You will be able to see how to solve it.

Attachment:

Ques1.jpg [ 16.63 KiB | Viewed 19248 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 _________________

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

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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.

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

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12 Jan 2011, 20:19

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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 [ 14.88 KiB | Viewed 19093 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. _________________

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

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17 Mar 2013, 23:07

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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 ?

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. _________________

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

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21 Mar 2013, 00:13

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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.

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

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21 Mar 2013, 01:35

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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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