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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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New post 17 Mar 2014, 00:14
3
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A
B
C
D
E

Difficulty:

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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 3/2 hours later. The second flight leaves Airport B every 20 minutes, beginning at 8:00 a.m., and arrives at Airport C 7/6 hours 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) 25 min
(B) 1 hr 5 min
(C) 1 hr 15 min
(D) 2 hr 20 min
(E) 3 hr 40 min

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Question: 168
Category: Arithmetic Operations on rational numbers
Page: 84
Difficulty: 600


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

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New post 17 Mar 2014, 00:14
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SOLUTION

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 3/2 hours later. The second flight leaves Airport B every 20 minutes, beginning at 8:00 a.m., and arrives at Airport C 7/6 hours 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) 25 min
(B) 1 hr 5 min
(C) 1 hr 15 min
(D) 2 hr 20 min
(E) 3 hr 40 min

A --> B, takes 90 minutes, at 8:00 a.m., 9:00 a.m., ...
B --> C, takes 70 minutes, at 8:00 a.m., 8:20 a.m. 8:40 a.m., ..., 9:40 a.m., ...
C --> ?, every 60 minutes, at 8:45 a.m., 9:45 a.m., 10:45 a.m., 11:45 a.m., ...

If passenger takes 8:00 a.m. flight from A:
10 minutes between flights in B from 9:30 a.m. to 9:40 a.m.
55 minutes between flights in C from 10:50 a.m. to 11:45 a.m..

Total = 10 + 55 = 65 minutes.

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

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New post 17 Mar 2014, 05:17
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The guy takes 8:00 am flight and will reach airport B in 3/2 hours i.e. \(\frac{3*60}{2}\) = 90 mins = 1 hour 30 mins (you can also just make the base 60 3/2 = 3*30/2*30 = 90/60 = 90 mins or 1 hour 20 mins)
So he reaches airport B at 9:30 am. 20 mins/flight starting from 8:00 am
8:00 8:20 8:40 9:00 9:20 9:40 and so on
So the next flight he can catch is the 9:40 am flight.
TIME WASTED = 10 mins

The guy takes 9:40 am flight will reach airport C in 7/6 hours i.e. 70 mins
So he reaches airport B at 10:50 am. 60 mins/flight starting from 8:45 am
8:45 9:45 10:45 11:45 and so on
So the next flight he can catch is the 11:45 am flight.
TIME WASTED = 55 mins

Total = 65 mins = 1 hour 5 mins

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

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New post 18 Mar 2014, 02:04
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Flights departing from A:

0800.................0900 ............... 1000 .............. 1100

Flights arriving at B after 3/2hrs i.e 90 Minutes


0930................. 1030................. 1130................... 1230

Second flight leaves Airport B every 20 minutes, beginning at 8:00 a.m
Departures from B


0800........... 0820 ............... 0840.............. 0900.......... 0920............ 0940............ 1000......... 1020............ 1040.......... 1100........ 1120..........

Best is the 0940 flight with just 10 minutes spacer in between

Arrivals at C is after 7/6 hrs i.e 70 minutes

0910............ 0930............... 0950.............. 1010.............. 1030................ 1050............... 1110................ 1130................ 1150...........

Departure at Airport C every hour, beginning at 8:45 a.m

0845................ 0945.................... 1045................ 1145............. 1245..................... 1345.......................... 1445..................

Best fit for this person is

Depart from A................. 0800
Arrive at B ..................... 0930
Depart from B................. 0940
Arrive at C...................... 1050
Depart at C.................... 1145


Total time spent between departures & arrivals at B & C = 10 + 55

= 65 Minutes = 01 Hr 05 Minutes

Answer = B

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

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New post 18 Mar 2014, 07:37
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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 3/2 hours later. The second flight leaves Airport B every 20 minutes, beginning at 8:00 a.m., and arrives at Airport C 7/6 hours 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) 25 min
(B) 1 hr 5 min
(C) 1 hr 15 min
(D) 2 hr 20 min
(E) 3 hr 40 min


Since the question asks about total time spent by the passenger between flights, it is required to find individual waiting time at Airport B and C.

Arrival at Airport A : 8:45 a.m.
Flight leaves at 9:00 a.m.

Arrival at Airport B : 9:00 a.m. + Flight Time = 9:00 + 3/2 hrs = 10:30 a.m.
Flight leaves every 20 minutes, beginning at 8:00 a.m.
Connecting flight to Airport C after 10:30 a.m. leaves at 10:40 a.m.
Waiting time at Airport B = 10:40 a.m. - 10:30 a.m. = 10 mins

Arrival at Airport C : 10:40 a.m. + Flight Time = 10:40 a.m. + 7/6 hrs = 11:50 a.m.
Flight leaves every hour, beginning at 8:45 a.m.
Connecting flight from Airport C after 11:50 a.m. leaves at 12:45 a.m.
Waiting time at Airport B = 12:45 a.m. - 11:50 a.m. = 55 mins

So, total waiting time = 10 mins + 55 mins = 1 hour 05 mins

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

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New post 18 Mar 2014, 12:57
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First convert the numbers to minutes. 3/2 hours = 90 minutes and 7/6 hours = 70 minutes.

Next, notice that no matter when the flight from A is taken, there is always going to be at least a 10 minute wait at B. So no point delaying at A. Take the 8:00am flight out to arrive at 9:30am at B.

Next, notice that even if you take the earliest flight (9:40am) out of B, you still arrive at C too late to get the 10:45am from C. So the earliest flight you can take from C is the 11:45am one. So you have to spend the 135 minutes of time between 9:30am and 11:45am such that 70 minutes of it is spent in the air (in the flight from B to C). No matter which way you do it, you end up with 135-70 minutes = 1 hour 5 minutes of minimum waiting.

(B) it is.
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Re: An airline passenger is planning a trip that involves three  [#permalink]

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New post 20 Mar 2014, 11:30
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I too got B

Solution:
Flight A timings: 8 AM, 9AM, 10AM . . .
Duration to reach B:90 Mins. So first flight reaches B at 9:30 AM

Flight B Timings:8:20 AM, 8:40, 9, 9:20, 9:40,. . . .
Duration to reach C: 70 Mins. So Flight reaches C at 11:50 AM
Passengers from Flight A have to wait 10Mins to catch Flight at B


Flight C timings: 8:45AM, 9:45AM, 10:45AM, 11:45AM, 12:45 . . .
So Passengers have to wait 55Mins to catch flight at C.

So total wait time = 10 Mins + 55 Mins = 1 Hr 5 Mins

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

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New post 22 Nov 2015, 20:22
I made a chart a lot like Paresh did and it took me 3minutes and 50 seconds to do this problem :-(
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An airline passenger is planning a trip that involves three  [#permalink]

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New post 24 Nov 2015, 03:22
Arrives........Waits........Leaves
A 8a.............0..............8a
B 9:30a.......10mins.......9:40a (flights every 20mins)
C 10:50a.....55mins.......11:45a (flights every hour starting at 8:45)
Total ..........65mins

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

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New post 07 Dec 2015, 17:29
DJ1986 wrote:
I made a chart a lot like Paresh did and it took me 3minutes and 50 seconds to do this problem :-(


yea, it took me 3minutes plus to figure out this problem. I wonder if there's any faster way to do questions like these, or is this one of those questions that we are expected to spend extra time.
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Re: An airline passenger is planning a trip that involves three  [#permalink]

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

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New post 25 Jan 2018, 03:46
The question for the least amount of time that the passenger must wait between flights. I agree that he spends 65 minutes between flights.

Everyone who posted started from 8:00 am.

I was wondering whether taking different flight from airport A will lead to different waiting amounts. What if he takes 12:00 AM flight from airport A. Do I have to check other starting points?
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Re: An airline passenger is planning a trip that involves three  [#permalink]

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New post 25 Jan 2018, 05:32
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kablayi wrote:
The question for the least amount of time that the passenger must wait between flights. I agree that he spends 65 minutes between flights.

Everyone who posted started from 8:00 am.

I was wondering whether taking different flight from airport A will lead to different waiting amounts. What if he takes 12:00 AM flight from airport A. Do I have to check other starting points?


Good question.

Note that the flight at airport A starts every hour.
A : 8:00, 9:00, 10:00, 11:00, 12:00 ...

Also, every hour is the same at the other two airports.
B : 8:00, 8:20:8:40, 9:00, 9:20, 9:40, 10:00, 10:20, 10:40 ...
C : 8:45, 9:45, 10:45, 11:45

So it doesn't matter at which hour you start. Your travel schedule will look exactly the same. Had the hours looked a bit different such as
B : 8:00, 8:40, 9:20, 10:00, 10:40, 11:20 ...
we would have had to give it more thought.
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Re: An airline passenger is planning a trip that involves three  [#permalink]

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New post 25 Jan 2018, 07:29
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Bunuel wrote:
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 3/2 hours later. The second flight leaves Airport B every 20 minutes, beginning at 8:00 a.m., and arrives at Airport C 7/6 hours 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) 25 min
(B) 1 hr 5 min
(C) 1 hr 15 min
(D) 2 hr 20 min
(E) 3 hr 40 min
Attachment:
Capture.PNG
Capture.PNG [ 9.26 KiB | Viewed 7754 times ]


Time spent at Airport B for going to Airport C = 10 Minutes
Time spent at Airport C for going to destination = 55 Minutes

So, The total time spent waiting at airports is 65 minutes or 1 Hour and 5 Minutes, Answer must be (B)

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

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New post 29 Jan 2018, 10:36
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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 3/2 hours later. The second flight leaves Airport B every 20 minutes, beginning at 8:00 a.m., and arrives at Airport C 7/6 hours 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) 25 min
(B) 1 hr 5 min
(C) 1 hr 15 min
(D) 2 hr 20 min
(E) 3 hr 40 min


Notice that the flight from Airport A to Airport B takes 3/2 hours = 90 minutes (or 1 hour 30 minutes) and the flight from Airport B to Airport C takes 7/6 hours = 70 minutes (or 1 hour 10 minutes).

Let’s say the first flight leaves A at 8 am; it would then arrive at B at 9:30 am. Then we have to wait 10 minutes since the next flight would leave B at 9:40 am and arrive at airport C 1 hour and 10 minutes later, or at 10:50 am. However, we have to wait 55 minutes since the next flight would leave C at 11:45 am. Thus, in this case we have to wait a total of 10 + 55 = 65 minutes or 1 hour 5 minutes. This is answer choice B.

Looking at the answer choices, we see there is only one answer choice with a waiting time less than 1 hour 5 minutes. However, there is no way we can wait for the connecting flight for as little as 25 minutes (which is choice A). For example, when we arrive at Airport B at 9:30 am, if we don’t take the 9:40 am flight, then the next flight leaving from B is at 10 am, which means we have to wait 30 minutes already. Thus, the least total amount of time we have to wait for the connecting flights is 1 hour 5 minutes (described in the previous paragraph).

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

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New post 25 Mar 2018, 07:23
Bunuel wrote:
SOLUTION

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 3/2 hours later. The second flight leaves Airport B every 20 minutes, beginning at 8:00 a.m., and arrives at Airport C 7/6 hours 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) 25 min
(B) 1 hr 5 min
(C) 1 hr 15 min
(D) 2 hr 20 min
(E) 3 hr 40 min

A --> B, takes 90 minutes, at 8:00 a.m., 9:00 a.m., ...
B --> C, takes 70 minutes, at 8:00 a.m., 8:20 a.m. 8:40 a.m., ..., 9:40 a.m., ...
C --> ?, every 60 minutes, at 8:45 a.m., 9:45 a.m., 10:45 a.m., 11:45 a.m., ...

If passenger takes 8:00 a.m. flight from A:
10 minutes between flights in B from 9:30 a.m. to 9:40 a.m.
55 minutes between flights in C from 10:50 a.m. to 11:45 a.m..

Total = 10 + 55 = 65 minutes.

Answer: B.



I had some confusion about converting fractions into timing.... 7/6 hours later ..if I divide 7 by 6 I get 1.1666 so thought I should round and got 1.2 :?
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Re: An airline passenger is planning a trip that involves three  [#permalink]

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New post 25 Mar 2018, 10:04
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dave13 wrote:
Bunuel wrote:
SOLUTION

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 3/2 hours later. The second flight leaves Airport B every 20 minutes, beginning at 8:00 a.m., and arrives at Airport C 7/6 hours 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) 25 min
(B) 1 hr 5 min
(C) 1 hr 15 min
(D) 2 hr 20 min
(E) 3 hr 40 min

A --> B, takes 90 minutes, at 8:00 a.m., 9:00 a.m., ...
B --> C, takes 70 minutes, at 8:00 a.m., 8:20 a.m. 8:40 a.m., ..., 9:40 a.m., ...
C --> ?, every 60 minutes, at 8:45 a.m., 9:45 a.m., 10:45 a.m., 11:45 a.m., ...

If passenger takes 8:00 a.m. flight from A:
10 minutes between flights in B from 9:30 a.m. to 9:40 a.m.
55 minutes between flights in C from 10:50 a.m. to 11:45 a.m..

Total = 10 + 55 = 65 minutes.

Answer: B.



I had some confusion about converting fractions into timing.... 7/6 hours later ..if I divide 7 by 6 I get 1.1666 so thought I should round and got 1.2 :?


Hi dave13

When you are converting 7/6 hours, you should take into consideration that 1 hour = 60 minutes.

So, \(\frac{7}{6}\) hour can be re-written as \(1 + \frac{1}{6}\) hour. The \(\frac{1}{6}\)th part of an hour is \(\frac{1}{6}*60\) = 10 minutes.

Time \(\frac{7}{6}\) hour translates to 1 hour and 10 minutes

Hope this helps you!
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