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# An airline passenger is planning a trip that involves three

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An airline passenger is planning a trip that involves three [#permalink]

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06 Feb 2011, 22:16
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25% (medium)

Question Stats:

78% (02:39) correct 22% (03:19) wrong based on 208 sessions

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An airline passenger is planning a trip that involves three connecting flights that leave from airports A, B and C, respectively. The First flight leaves Airport A every Hour, beginning at 8:00 a.m., and arrives at Airport B 2nd a half hours later. The second flight leaves Airport B every 20minutes, beginning at 8:00a.m, and arrives at Airport C 1h 10 minutes later. The third flight leaves Airport C every hour, beginning at 8.45 a.m. What is the least total amount of time the passenger must spend between flights if all flights keep to their schedules?

A. 25min
B. 1hr 5min
C. 1hr 15min
D. 2hr 20min
E. 3hr 40min
[Reveal] Spoiler: OA

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Last edited by Bunuel on 28 Mar 2012, 10:14, edited 1 time in total.
Edited the question
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Re: Least total amount of time [#permalink]

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07 Feb 2011, 01:41
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Following way may not be a pure mathematical approach to solve it but it got me the result within a minute.

A: Flight leaves every hour; 8,9,10,11,12,1,2,3
B: Flight leaves every 20min; 8:20,8:40,9,9:20,9:40,10,10:20,10:40,11
C: Flight leaves every hour: 8:45,9:45,10:45,11:45,12:45

Passenger leaves at 8:00AM from A; reaches B@ 10:30am - he will catch 10:40am flight from B{10 minutes spent at airport B}; reaches c after 1hour10minutes @11:50AM;his next flight at C is at 12:45 - he waits another 55minutes;

Total wait = 10+55= 65 minutes = 1hour 5 mins

Ans: "B"
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Re: Least total amount of time [#permalink]

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07 Feb 2011, 02:12
I would go with Ans b - 1 hr 5 min.

I would say one easy way to do this is to Imagine one self having booked a ticket(s) for the same flight schedule and go by -

'' I would have my first flight from A at 8:00 AM , reach B by 2 1/2 Hrs, so 10:30 AM , but darn I would miss the flight from B by 10 minutes and would have to WAIT another 10 minutes Cos a flight starts EVERY 20 minutes from there and then would take a flight at 10:40 AT B and reach C at exactly at 11:50 AM, and now I could watch a short movie , cos I missed the flight by just 5 min and would have about 55 Minutes to kill until the next flight, as it is 11:50 now and I have until 12:45 to wait for the next flight.

Best.
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Re: Least total amount of time [#permalink]

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08 Feb 2011, 06:42
DaVagabond wrote:
I would go with Ans b - 1 hr 5 min.

I would say one easy way to do this is to Imagine one self having booked a ticket(s) for the same flight schedule and go by -

'' I would have my first flight from A at 8:00 AM , reach B by 2 1/2 Hrs, so 10:30 AM , but darn I would miss the flight from B by 10 minutes and would have to WAIT another 10 minutes Cos a flight starts EVERY 20 minutes from there and then would take a flight at 10:40 AT B and reach C at exactly at 11:50 AM, and now I could watch a short movie , cos I missed the flight by just 5 min and would have about 55 Minutes to kill until the next flight, as it is 11:50 now and I have until 12:45 to wait for the next flight.

Best.

Thanks I agree with b.just got confused in time constraints.
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Re: Least total amount of time [#permalink]

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28 Mar 2012, 10:10
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A: 8,9,10,11,12,1,2,3
B: 8:20,8:40,9,9:20,9:40,10,10:20,10:40,11
C: 8:45,9:45,10:45,11:45,12:45

10 mins to catch the flight in city B
55 mins to catch the flight in city C

total 65 mins 0 1hr 5 mins
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Re: An airline passenger is planning a trip that involves three [#permalink]

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05 Sep 2012, 08:21
The best way to solve this problem is to work backwards i.e. from Flight C->B->A.

Think about it. If it's not clear let me know
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Re: An airline passenger is planning a trip that involves three [#permalink]

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28 May 2013, 05:46
fameatop wrote:
The best way to solve this problem is to work backwards i.e. from Flight C->B->A.

Think about it. If it's not clear let me know

Is this always the best way for these type of bus schedule / flight schedule problems?
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Re: Least total amount of time [#permalink]

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04 Aug 2013, 04:47
DaVagabond wrote:
I would go with Ans b - 1 hr 5 min.

I would say one easy way to do this is to Imagine one self having booked a ticket(s) for the same flight schedule and go by -

'' I would have my first flight from A at 8:00 AM , reach B by 2 1/2 Hrs, so 10:30 AM , but darn I would miss the flight from B by 10 minutes and would have to WAIT another 10 minutes Cos a flight starts EVERY 20 minutes from there and then would take a flight at 10:40 AT B and reach C at exactly at 11:50 AM, and now I could watch a short movie , cos I missed the flight by just 5 min and would have about 55 Minutes to kill until the next flight, as it is 11:50 now and I have until 12:45 to wait for the next flight.

Best.

hi, i don't understand, what's "imagine oneself having booked a ticket for the same flight schedule"?

thx.
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Re: An airline passenger is planning a trip that involves three [#permalink]

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10 Nov 2014, 05:21
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Re: An airline passenger is planning a trip that involves three [#permalink]

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17 Mar 2016, 11:11
Hello from the GMAT Club BumpBot!

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Re: An airline passenger is planning a trip that involves three [#permalink]

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23 Mar 2017, 11:07
I approached it by step by step calculation, let's say
we leave from A at 8:00, then we get to B at 10:30, the next flight to C is in 10 minutes
asnwe leave in 10:40 we arrive at C at 11:50, the next flight is at 12:45, we have to stay for 55 minutes

10+55 minutes = 1 hour 5 minutes
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Re: An airline passenger is planning a trip that involves three [#permalink]

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24 Mar 2017, 12:02
I doubt i could do this is less than 2 minutes.
Reading and understand only can take more than one and half minute.
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Re: An airline passenger is planning a trip that involves three   [#permalink] 24 Mar 2017, 12:02
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