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M08 #18 - Bus intervals

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Re: M08 #18 - Bus intervals [#permalink] New post 03 Jun 2010, 03:46
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ykaiim wrote:
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


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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.


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


When I asked the above question I wanted to show that your approach is not correct. It worked for the given numbers but it won't work for another ones. Answer to my question would be 24/7 minutes not 12/5 minutes.

This question was posted in PS forum as well below is my solution for it:

Let's say the distance between the buses is d. We want to determine Interval=\frac{d}{b}, where b is the speed of bus.

Let the speed of cyclist be c.

Every 12 minutes a bus overtakes cyclist: \frac{d}{b-c}=12, d=12b-12c;

Every 4 minutes cyclist meets an oncoming bus: \frac{d}{b+c}=4, d=4b+4c;

d=12b-12c=4b+4c, --> b=2c, --> d=12b-6b=6b.

Interval=\frac{d}{b}=\frac{6b}{b}=6

Answer: 6 minutes.

Hope it helps.
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Re: M08 #18 - Bus intervals [#permalink] New post 03 Jun 2010, 03:58
Bunuel,

I dont deny this mathematical approach but how the answer is 24/7 in that case?
I tried this by consedring an interval of 12 min for overtake and 2 min for cross-over.
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Re: M08 #18 - Bus intervals [#permalink] New post 03 Jun 2010, 04:22
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ykaiim wrote:
Bunuel,

I dont deny this mathematical approach but how the answer is 24/7 in that case?
I tried this by consedring an interval of 12 min for overtake and 2 min for cross-over.


ykaiim, please do the math: substitute new numbers (12 min and 2 min) in the above approach.
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Re: PS - Bus intervals [#permalink] New post 19 Jun 2010, 19:09
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


I have question regarding the equation. if T=L/V then L should be the distance. How come L ended up becoming the time interval between each bus?
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Re: M08 #18 - Bus intervals [#permalink] New post 21 Jul 2010, 01:03
Can someone please explain this:

Every 12 minutes a bus overtakes cyclist: , d/(b-c) = 12

Every 4 minutes cyclist meets an oncoming bus: , d/ (b+c) = 4

b- avg speed of bus and c- avg speed of cyclist. Why are we adding and subtracting avg speeds???
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Re: M08 #18 - Bus intervals [#permalink] New post 18 Aug 2010, 19:36
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Here is some additional explanations that might help:

We are looking for the time between two conseutive buses.
Since the bus speed is constant, the time between two consecutive buses will be constant and so will be the distance between two consecutive buses as distance = time*speed.

Let's l be the distance between two buses. The time between two buses is equal to l/b where b is the speed of the bus. We need to express l in terms of b in order to have the result of this ratio.

The information provided allows us to do so.
1/we know that the cyclist crosses a bus every four minutes. It means that the bus plus the cyclist run the distance between two buses in 4 minutes. Hence we have l=4(b+c) (1) where c is the speed of the cyclist.
(The relative speed of U and V with speed equal u and v going in the same direction is u+v)

2/ we know that the cyclist is caught up by a bus every twelve minutes. During these twelve minutes the bus will do more than the distance between two buses as the cyclist is going forward. We know that the relative speed between U and V running in opposite direction is u-v (where u>v).
Hence to get the distance l we need to calculate the distance covered by the bus in twelve minutes and minus the distance covered by the cyclist. This means l =12b-12c--> l=12(b-c) (2)

From 1 and 2 we get 4b+4c=12b-12c-->16c=8b-->c=1/2b. We can now replace c by 1/2 b in (1) and we get l=4b+2b=6b. The time between two buses is l/b=6b/b=6
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Re: M08 #18 - Bus intervals [#permalink] New post 19 Aug 2010, 01:21
This whole question seems a bit confusing to me. I really think the current wording is not clear. I mean we can assume that we have equal distance, really that is the only way to solve this problem.

I want to recommend the following question as a replacement for this question:

A man cycling around a circular track notices that every 12 minutes a bus overtakes him while every 4 minutes he meets an oncoming bus. If both buses and the cyclist move at constant speeds, what is the time interval between the two buses?

(A) 5 minutes
(B) 6 minutes
(C) 8 minutes
(D) 9 minutes
(E) 10 minutes
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Re: M08 #18 - Bus intervals [#permalink] New post 30 Aug 2010, 09:19
using the MGMAT "RTD overtake" method, v = 1/4 - 1/12 = (3-1)/12 = 2/12 = 1/6 therefore 6 minutes....
don't even udnerstand how it works in this case but seems to be working.........please someone explain if understand.....
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Re: M08 #18 - Bus intervals [#permalink] New post 20 Oct 2010, 09:51
if the relative speeds are in AP then the relative time intervals are in HP

Since the speeds b+c , b , b-c are in AP

The relevant times are in HP.......

speed is b when the cyclist is standing. We are asked to find the time interval between the buses. That time interval will be the time when cyclist is stationary.
Time interval when bus is moving on the same side is 12 and on the other side is 4.

Thus the times 12 , t, 4 are in HP

=> t = 2*12*4/ (12+4) = 6

The above result is very useful and time saving.
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Re: M08 #18 - Bus intervals [#permalink] New post 01 Nov 2010, 22:19
I second TallJTinChina's suggestion
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Re: M08 #18 - Bus intervals [#permalink] New post 23 Aug 2011, 09:20
answer is 6 mins...
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Re: M08 #18 - Bus intervals [#permalink] New post 23 Aug 2011, 09:39
i like the HP approach , easy to get answer using this approach
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Re: M08 #18 - Bus intervals [#permalink] New post 27 Aug 2011, 09:00
My Understanding is as follows

Cyclist & Buses are at constant speed, every time there is fixed interval so Distance is also constant

Now when going towards relative speed is Vb-Vc, & oncoming is Vb+Vc

distance assumed to constant because it is not mentioned that oncoming & going buses have different speeds

so D/Vb-Vc = 12......1
D/Vb+Vc=4......2

4(Vb+Vc)=12(Vb-Vc) --> Vb+Vc=3Vb-3Vc ---> 2Vb=4Vc Vb=1/2Vc

Put this value in Eq 1
D/Vb-1/2Vb=12
2D/Vb=12
D/Vb=6
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Re: M08 #18 - Bus intervals [#permalink] New post 03 Sep 2011, 12:50
I need a fresh opinion here. :p
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Re: M08 #18 - Bus intervals [#permalink] New post 20 Apr 2012, 15:23
Here, we have to assume that Vb > Vc. Otherwise, I will get a different answer.

For instance, (Vc > Vb ; this is the case in India especially where I used to live)

Vc+ Vb = d/4
Vc- Vb = d/12

Therefore, if Vc=12 (Say); Vb = 6 (solving above equations)


Therefore Tc = 6 and Tb = 12.

If I reverse the equation, i.e. Vb-Vc = d/12; I get Vb = 12; Vc =6 and Tb=6; Tc = 12.


Am I wrong? Please help me


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Re: M08 #18 - Bus intervals [#permalink] New post 22 Apr 2012, 04:25
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voodoochild wrote:
Here, we have to assume that Vb > Vc. Otherwise, I will get a different answer.

For instance, (Vc > Vb ; this is the case in India especially where I used to live)

Vc+ Vb = d/4
Vc- Vb = d/12

Therefore, if Vc=12 (Say); Vb = 6 (solving above equations)


Therefore Tc = 6 and Tb = 12.

If I reverse the equation, i.e. Vb-Vc = d/12; I get Vb = 12; Vc =6 and Tb=6; Tc = 12.


Am I wrong? Please help me


Thanks


We are told that "every 12 minutes a bus overtakes the cyclist", so the speed of the bus is more than the speed of the cyclist.
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Re: M08 #18 - Bus intervals [#permalink] New post 23 Aug 2012, 05:49
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?


Bunuel,

Can you please explain this problem. I get so confused with the speed problems...
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Re: M08 #18 - Bus intervals [#permalink] New post 23 Aug 2012, 05:58
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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?


Bunuel,

Can you please explain this problem. I get so confused with the speed problems...


My solution: m08-18-bus-intervals-74253-20.html#p732805

This question is also discussed here: a-man-cycling-along-the-road-noticed-that-every-12-minutes-88723.html

Hope it helps.
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Re: M08 #18 - Bus intervals [#permalink] New post 23 Aug 2012, 08:12
Hopefully this many help.

Rate * Time = Distance

Let B = velocity of bus
Let C = velocity of cyclist
Let D = distance

The question is asking for the time inverval between consecutive buses. AKA Distance/Rate of Bus (D/B).

D/B=?

(B+C)*4=D Bus and Cyclist are moving in same direction
(B-C)*12=D Bus and Cyclist are moving in opposite directions

Combine these formulas--> ( B+C)*4=(B-C)*12
4B+4C=12B-12C
16C=8B
C=B/2

Plug this back into one of the original formulas:
(B+B/2)*4=D
(3/2)*B*4=D
6B=D
D/B=?
D/B=6

Answer is B
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Re: M08 #18 - Bus intervals [#permalink] New post 29 Aug 2012, 23:26
I did end up getting the answer as 6 but I am not sure if my approach was correct and can be used again for similar problems.

Personally I did not prefer approaching the equations way of solving this problem. They might be way more easier but instead I took a particular scenario wherein the total time was from zero to 12 mins. In this time duration, one bus overtook the cyclist and he faced 3 oncoming buses.

i.e. total time = 12 mins + 12 mins = 24 mins
Total buses passed = 1 (overtaking) + 3 (oncoming)

Interval = 24/4 = 6.

There was particular response bu Bunuel where he asked what would be the solution if the interval was 2 mins for the oncoming buses. In that case the interval between the buses is 24/7 as he has mentioned.

Will dig in further to see if my approach is correct or whether it was just apt for this particular set of numbers and was a fluke.
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Re: M08 #18 - Bus intervals   [#permalink] 29 Aug 2012, 23:26
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