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Re: Car B begins moving at 2 mph around a circular track with a radius of [#permalink]
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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: Car B begins moving at 2 mph around a circular track with a radius of [#permalink]
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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: Car B begins moving at 2 mph around a circular track with a radius of [#permalink]
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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: Car B begins moving at 2 mph around a circular track with a radius of [#permalink]
Thanks a lot Bunuel. I must learn to read the question properly.
The answer is B.
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Re: Car B begins moving at 2 mph around a circular track with a radius of [#permalink]
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: Car B begins moving at 2 mph around a circular track with a radius of [#permalink]
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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: Car B begins moving at 2 mph around a circular track with a radius of [#permalink]
I also forgot to add the first 10 hours =). To reckognize all the details is sooo important!!!!
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Re: Car B begins moving at 2 mph around a circular track with a radius of [#permalink]
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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 begins moving at 2 mph around a circular track with a radius of [#permalink]
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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Car B begins moving at 2 mph around a circular track with a radius of [#permalink]
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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 57868 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.

Check this video for when to use relative speed: https://youtu.be/wrYxeZ2WsEM

Originally posted by KarishmaB on 12 Jan 2011, 20:19.
Last edited by KarishmaB on 14 Aug 2023, 00:02, edited 1 time in total.
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Re: Car B begins moving at 2 mph around a circular track with a radius of [#permalink]
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 a radius of [#permalink]
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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.
Thanks
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Re: Car B begins moving at 2 mph around a circular track with a radius of [#permalink]
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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 a radius of [#permalink]
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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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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 a radius of [#permalink]
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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,
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Re: Car B begins moving at 2 mph around a circular track with a radius of [#permalink]
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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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