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M08 #18 - Bus intervals

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M08 #18 - Bus intervals [#permalink]

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23 Mar 2008, 13:14
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A man cycling along the road noticed that every 12 minutes a bus overtakes him while every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at constant speed, what is the time interval between consecutive buses?

(A) 5 minutes
(B) 6 minutes
(C) 8 minutes
(D) 9 minutes
(E) 10 minutes

[Reveal] Spoiler: OA
B

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Re: PS - Bus intervals [#permalink]

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23 Mar 2008, 13:24
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B

$$\frac{L}{V_{bus}+V_{cyclist}}=4$$

$$\frac{L}{V_{bus}-V_{cyclist}}=12$$

$$4*(V_{bus}+V_{cyclist})=12*(V_{bus}-V_{cyclist})$$

$$V_{cyclist}=\frac12*V_{bus}$$

$$\frac{L}{V_{bus}+\frac12*V_{bus}}=4$$

$$\frac{L}{V_{bus}}=4*(1+\frac12)=6$$
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Re: PS - Bus intervals [#permalink]

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23 Mar 2008, 13:33
Walker,

How do we know that the distance L traveled by the oncoming and overtaking buses is the same, Do we need to assume that it is same and calculate or am I missing anything.
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Re: PS - Bus intervals [#permalink]

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23 Mar 2008, 13:46
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Walker,

If possible can can you dumb it down a little for us lowly earthlings . Thanks in advance.
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Re: PS - Bus intervals [#permalink]

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23 Mar 2008, 14:00
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prasannar wrote:
Walker,

How do we know that the distance L traveled by the oncoming and overtaking buses is the same, Do we need to assume that it is same and calculate or am I missing anything.

Yes. we have to assume. If the distance is not the same, the buses will collect in the one of the end stations and we have many possible solutions.

neelesh wrote:
Walker, If possible can can you dumb it down a little for us lowly earthlings . Thanks in advance.

1. I wrote two equations for the time intervals of both the oncoming and overtaking buses using formula: t=L/V
2. I divided one equation by other one in order to exclude the distance between buses.
3. I found relationship between the speed of the buses and the speed of the cyclist
4. I used the finding and got L/Vbus (the time interval between consecutive buses) from first equation.
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Re: PS - Bus intervals [#permalink]

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23 Mar 2008, 14:10
Thanks Walker....
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Re: PS - Bus intervals [#permalink]

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27 Mar 2008, 04:43
Sorry Walker,
your equation is wrong.
If all buses and the cyclist move at constant speed
Which means Vcyclist = Vbus not Vcyclist = 1/2 Vbus as you found.

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Re: PS - Bus intervals [#permalink]

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27 Mar 2008, 04:52
No sorry
I'm wrong the quote say constant not the same.

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Re: PS - Bus intervals [#permalink]

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30 Mar 2008, 08:46
this is a bit confusing..what does consecutive buses mean??

if consecutive busses means that the bus came head on to the cyclist and then the same bus passed the cyclist..then in that case I get 8 mins..

I mistook consecutive buses to mean when 2 buses back to back passed the cyclist in that case the answer was simple 4 min..

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Re: PS - Bus intervals [#permalink]

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30 Mar 2008, 11:03
walker wrote:
B

$$\frac{L}{V_{bus}+V_{cyclist}}=4$$

$$\frac{L}{V_{bus}-V_{cyclist}}=12$$

$$4*(V_{bus}+V_{cyclist})=12*(V_{bus}+V_{cyclist})$$

$$V_{cyclist}=\frac12*V_{bus}$$

$$\frac{L}{V_{bus}+\frac12*V_{bus}}=4$$

$$\frac{L}{V_{bus}}=4*(1+\frac12)=6$$

nicely done by walker. thanx.
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Re: PS - Bus intervals [#permalink]

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17 Apr 2008, 17:48
neat expln .Thanks

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Re: Gmat club M08 (#18) [#permalink]

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06 Jan 2009, 22:34
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topmbaseeker wrote:
A man cycling along the road noticed that every 12 minutes a bus overtakes him while every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at constant speed, what is the time interval between consecutive buses?

(C) 2008 GMAT Club - m08#18

* 5 minutes
* 6 minutes
* 8 minutes
* 9 minutes
* 10 minutes

d = distance the bus and cyclist togather run
c = cyclist's speed
b = bus's speed

d/(c+b) = 4
d/(b-c) = 12
d/(c+b) = d/3(b-c)
3c+3b = b-c
4c = 2b
b = 2c, or
c = c/2

we need bus's time to complet d distance. so replace c in terms of b:

d/(c+b) = 4
d/(b/2 + b) = 4
d/b = 4x3/2
d/b = 6

Thats the time taken to pass d for a bus. after every 6 minuets another bus comes.

So it is B.

I know this is another hard question but when the concept is clear, it wont be difficult to get the result..
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Re: Gmat club M08 (#18) [#permalink]

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08 Jan 2009, 10:46
Why is the distance the same for both the cases?

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Re: Gmat club M08 (#18) [#permalink]

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08 Jan 2009, 16:05
topmbaseeker wrote:
Why is the distance the same for both the cases?

Hope the drawing helps understand how and in which direction the buses and cyclist are running.
Attachments

Distance.doc [27 KiB]

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Re: M08 #18 - Bus intervals [#permalink]

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28 May 2009, 23:38
I still don't get this question and just wondering if someone could explain.

I basically don't understand the question at all. Is it saying that one bus is traveling on (say) a clockwise route and it passes the cyclist (who is also moving clockwise) every 12 minutes. If you assume distance of the route is d then d / vb = 12.

Same for the other bus?? It's moving in the opposite direction.
AND we should assume SAME speed for the other bus, AND same distance??

If so, then d / vb = 4.
But this doesn't make any sense.

How does the distance relate to the cyclist. Are you assuming the cyclist is on the same route and traveling distance d?
But just at speed vc?

Hopefully someone can explain in really simple terms , and lay out all of their assumptions.

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Re: M08 #18 - Bus intervals [#permalink]

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08 Nov 2009, 06:40
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I believe that d is the distance of the bus not both.

d/(c+b) = 4
d/(b/2 + b) = 4
d/b = 4x3/2
d/b = 6

If d is the distance of the bus run after another one, the answer B will make sense.

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Re: M08 #18 - Bus intervals [#permalink]

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05 Dec 2009, 08:39
Hi,
This might be overly simplistic - but can someone comment on my logic here?

Tb-Tc=4
Tb+Tc=8
2Tb=12
Tb=6

Does that work? Or only in this case?

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Re: M08 #18 - Bus intervals [#permalink]

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02 Jun 2010, 06:01
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My approach was different.

If in 12 min a bus overtakes the cyclist, while in 4 min they both cross each other. So, at the end of 12th min, the overtaking bus will meet the 3rd bus coming from the opposite side, which means there are 2 buses gone in the 12 min time period.

We are given speeds of buses and cyclist constant. So, in 12 min, if two buses passes at an interval of 6 min.
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Re: M08 #18 - Bus intervals [#permalink]

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02 Jun 2010, 13:14
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ykaiim wrote:
My approach was different.

If in 12 min a bus overtakes the cyclist, while in 4 min they both cross each other. So, at the end of 12th min, the overtaking bus will meet the 3rd bus coming from the opposite side, which means there are 2 buses gone in the 12 min time period.

We are given speeds of buses and cyclist constant. So, in 12 min, if two buses passes at an interval of 6 min.

What would be the answer with your approach if it were 2 minutes instead of 4?
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Re: M08 #18 - Bus intervals [#permalink]

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03 Jun 2010, 04:24
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In every 2 min, cyclist is meeting bus from opposite side. So, at the end of the 12th min, he will meet the 6th bus from opposite end. That means there are 5 intervals passed at the end of 12th min. So, frequesncy = 12/5 = 2 min 24 sec

Bunuel wrote:
ykaiim wrote:
My approach was different.

If in 12 min a bus overtakes the cyclist, while in 4 min they both cross each other. So, at the end of 12th min, the overtaking bus will meet the 3rd bus coming from the opposite side, which means there are 2 buses gone in the 12 min time period.

We are given speeds of buses and cyclist constant. So, in 12 min, if two buses passes at an interval of 6 min.

What would be the answer with your approach if it were 2 minutes instead of 4?

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Re: M08 #18 - Bus intervals   [#permalink] 03 Jun 2010, 04:24

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