Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 64951

An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
16 Mar 2014, 23:14
Question Stats:
74% (02:56) correct 26% (03:12) wrong based on 1089 sessions
HideShow timer Statistics
The Official Guide For GMAT® Quantitative Review, 2ND EditionAn 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 Problem Solving Question: 168 Category: Arithmetic Operations on rational numbers Page: 84 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you!
Official Answer and Stats are available only to registered users. Register/ Login.
_________________




Math Expert
Joined: 02 Sep 2009
Posts: 64951

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
16 Mar 2014, 23:14
SOLUTIONAn 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.
_________________




Manager
Joined: 25 Sep 2012
Posts: 225
Location: India
Concentration: Strategy, Marketing
GMAT 1: 660 Q49 V31 GMAT 2: 680 Q48 V34

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
17 Mar 2014, 04:17
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 Time Taken 4:08 Difficulty level 650 (just because it is lengthy)



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1706
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
18 Mar 2014, 01:04
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



Manager
Joined: 04 Oct 2013
Posts: 148
Location: India
GMAT Date: 05232015
GPA: 3.45

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
18 Mar 2014, 06:37
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)



SVP
Status: Top MBA Admissions Consultant
Joined: 24 Jul 2011
Posts: 2106

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
18 Mar 2014, 11:57
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 13570 minutes = 1 hour 5 minutes of minimum waiting. (B) it is.
_________________
GyanOne [www.gyanone.com] Premium MBA and MiM Admissions Consulting
Awesome Work  Honest Advise  Outstanding Results Reach Out, Lets chat!Email: info at gyanone dot com  +91 98998 31738  Skype: gyanone.services



Manager
Joined: 30 May 2013
Posts: 143
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
20 Mar 2014, 10:30
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
Regsrds, Rrsnathan



Manager
Joined: 05 Jul 2015
Posts: 91
Concentration: Real Estate, International Business
GPA: 3.3

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
22 Nov 2015, 19:22
I made a chart a lot like Paresh did and it took me 3minutes and 50 seconds to do this problem



Retired Moderator
Joined: 29 Oct 2013
Posts: 245
Concentration: Finance
GPA: 3.7
WE: Corporate Finance (Retail Banking)

An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
24 Nov 2015, 02:22
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
_________________
Please contact me for super inexpensive quality private tutoring
My journey V46 and 750 > http://gmatclub.com/forum/myjourneyto46onverbal750overall171722.html#p1367876



Manager
Joined: 21 Jul 2013
Posts: 102
WE: Securities Sales and Trading (Commercial Banking)

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
07 Dec 2015, 16:29
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.



Manager
Joined: 03 Jan 2017
Posts: 131

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
23 Mar 2017, 10:07
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



Intern
Joined: 05 Mar 2015
Posts: 41
Location: Azerbaijan
GMAT 1: 530 Q42 V21 GMAT 2: 600 Q42 V31 GMAT 3: 700 Q47 V38

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
25 Jan 2018, 02:46
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?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10634
Location: Pune, India

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
25 Jan 2018, 04:32
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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 5010
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
25 Jan 2018, 06:29
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 [ 9.26 KiB  Viewed 11618 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)
_________________



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2799

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
29 Jan 2018, 09:36
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionAn 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
_________________
5star rated online GMAT quant self study course
See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews



VP
Joined: 09 Mar 2016
Posts: 1252

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
25 Mar 2018, 06:23
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



Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3249
Location: India
GPA: 3.12

Re: An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
25 Mar 2018, 09:04
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 dave13When you are converting 7/6 hours, you should take into consideration that 1 hour = 60 minutes. So, \(\frac{7}{6}\) hour can be rewritten 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!
_________________
You've got what it takes, but it will take everything you've got



Manager
Joined: 02 Jan 2020
Posts: 244

An airline passenger is planning a trip that involves three
[#permalink]
Show Tags
22 Apr 2020, 05:44
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. VeritasKarishmaIs it possible in such type of ques that any different combination of time gives lesser time than what we've got and if yes how do we know if we should continue checking Thanks in advance!




An airline passenger is planning a trip that involves three
[#permalink]
22 Apr 2020, 05:44




