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# M08 #18 - Bus intervals

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

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23 Aug 2013, 06:43
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.

the answer is 4 here, it has asked time interval between two buses, first bus will meet at 8th min, and 2nd bus will meet at 12th min, so difference is 4 minutes...

Relative velocity concept, cyclist speed is assumed to be 0.

Buses speed are constant, so first bus will met at 8th min, second at 4th minutes. difference is 4 Min.
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Re: M08 #18 - Bus intervals [#permalink]

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26 Aug 2013, 09:20
Guess, mentioning that the two buses move with same speed would be quite necessary
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Re: M08 #18 - Bus intervals [#permalink]

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09 Aug 2014, 01:21
GMAT TIGER wrote:
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.

Could someone explain why d/b means the time interval between consecutive buses?
Thanks!
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Re: M08 #18 - Bus intervals [#permalink]

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11 Aug 2014, 07:53
Can someone explain the denominator? How come we are adding speeds of the cyclist and bus driver?
Re: M08 #18 - Bus intervals   [#permalink] 11 Aug 2014, 07:53

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# M08 #18 - Bus intervals

Moderator: Bunuel

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