Last visit was: 21 Jul 2024, 19:25 It is currently 21 Jul 2024, 19:25
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# An airline passenger is planning a trip that involves three

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94441
Own Kudos [?]: 642826 [104]
Given Kudos: 86716
Math Expert
Joined: 02 Sep 2009
Posts: 94441
Own Kudos [?]: 642826 [31]
Given Kudos: 86716
Target Test Prep Representative
Joined: 04 Mar 2011
Affiliations: Target Test Prep
Posts: 3036
Own Kudos [?]: 6609 [5]
Given Kudos: 1646
General Discussion
Manager
Joined: 25 Sep 2012
Posts: 203
Own Kudos [?]: 566 [4]
Given Kudos: 242
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]
2
Kudos
2
Bookmarks
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

Time Taken 4:08
Difficulty level 650 (just because it is lengthy)
SVP
Joined: 27 Dec 2012
Status:The Best Or Nothing
Posts: 1558
Own Kudos [?]: 7300 [5]
Given Kudos: 193
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Re: An airline passenger is planning a trip that involves three [#permalink]
3
Kudos
2
Bookmarks
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

Manager
Joined: 04 Oct 2013
Posts: 127
Own Kudos [?]: 304 [1]
Given Kudos: 55
Location: India
GMAT Date: 05-23-2015
GPA: 3.45
Re: An airline passenger is planning a trip that involves three [#permalink]
1
Kudos
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

CEO
Joined: 24 Jul 2011
Status: World Rank #4 MBA Admissions Consultant
Posts: 3189
Own Kudos [?]: 1600 [3]
Given Kudos: 33
GMAT 1: 780 Q51 V48
GRE 1: Q170 V170
Re: An airline passenger is planning a trip that involves three [#permalink]
3
Kudos
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.
Manager
Joined: 30 May 2013
Posts: 126
Own Kudos [?]: 371 [2]
Given Kudos: 72
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82
Re: An airline passenger is planning a trip that involves three [#permalink]
2
Kudos
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
Intern
Joined: 05 Mar 2015
Posts: 26
Own Kudos [?]: 18 [0]
Given Kudos: 111
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]
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?
Tutor
Joined: 16 Oct 2010
Posts: 15126
Own Kudos [?]: 66769 [4]
Given Kudos: 436
Location: Pune, India
Re: An airline passenger is planning a trip that involves three [#permalink]
4
Kudos
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.
Board of Directors
Joined: 11 Jun 2011
Status:QA & VA Forum Moderator
Posts: 6049
Own Kudos [?]: 4767 [2]
Given Kudos: 463
Location: India
GPA: 3.5
Re: An airline passenger is planning a trip that involves three [#permalink]
2
Kudos
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 42985 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)
VP
Joined: 09 Mar 2016
Posts: 1142
Own Kudos [?]: 1027 [0]
Given Kudos: 3851
Re: An airline passenger is planning a trip that involves three [#permalink]
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.

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
CEO
Joined: 26 Feb 2016
Posts: 2865
Own Kudos [?]: 5328 [2]
Given Kudos: 47
Location: India
GPA: 3.12
An airline passenger is planning a trip that involves three [#permalink]
1
Kudos
1
Bookmarks
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

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!­
Intern
Joined: 02 Feb 2023
Posts: 28
Own Kudos [?]: 2 [0]
Given Kudos: 29
Location: India
An airline passenger is planning a trip that involves three [#permalink]
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

Flight From B to C takes 7/6 hours which is 1.16 hours.

so when we add 1.16 hours to 9:40 am we get, 10:56 am, but you wrote 10:50am. Can you please tell what is wrong in this calculation which i did?­
Math Expert
Joined: 02 Sep 2009
Posts: 94441
Own Kudos [?]: 642826 [0]
Given Kudos: 86716
An airline passenger is planning a trip that involves three [#permalink]
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

Flight From B to C takes 7/6 hours which is 1.16 hours.

so when we add 1.16 hours to 9:40 am we get, 10:56 am, but you wrote 10:50am. Can you please tell what is wrong in this calculation which i did?

First of all, 7/6 hours = 1.16666... hours, not 1.16 hours.

Next, 1.16666... hours does not mean 1 hour and 16... minutes. 1.16666... hours = 1 hour and 1/6 of an hour, and 1/6 of an hour is 10 minutes. So, 1.16666... hours is 70 minutes. Or, as shown in the solution, we can directly calculate that 7/6 hours = 7/6*60 = 70 minutes.­
Manager
Joined: 07 Feb 2024
Posts: 81
Own Kudos [?]: 8 [0]
Given Kudos: 101
Re: An airline passenger is planning a trip that involves three [#permalink]

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

­There is a mistake in the stem Bunuel in the highlighted text since from airport A to B is 2 hours and a half which means that he arrives at the second airport at 10.30am.

(the official question "The first flight leaves Airport A every hour, beginning at 8:00 a.m., and arrives at Airport B 2 ​​ hours later")­
Math Expert
Joined: 02 Sep 2009
Posts: 94441
Own Kudos [?]: 642826 [0]
Given Kudos: 86716
An airline passenger is planning a trip that involves three [#permalink]
Gmatguy007 wrote:
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

­There is a mistake in the stem Bunuel in the highlighted text since from airport A to B is 2 hours and a half which means that he arrives at the second airport at 10.30am.

(the official question "The first flight leaves Airport A every hour, beginning at 8:00 a.m., and arrives at Airport B 2 ​ 1/2 ​ hours later")­

­Yes, there was a typo. However, it does not affect the answer at all. Thank you though for spotting this.­
Manager
Joined: 07 Feb 2024
Posts: 81
Own Kudos [?]: 8 [1]
Given Kudos: 101
Re: An airline passenger is planning a trip that involves three [#permalink]
1
Kudos
Bunuel wrote:
Gmatguy007 wrote:
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

­There is a mistake in the stem Bunuel in the highlighted text since from airport A to B is 2 hours and a half which means that he arrives at the second airport at 10.30am.

(the official question "The first flight leaves Airport A every hour, beginning at 8:00 a.m., and arrives at Airport B 2 ​ 1/2 ​ hours later")­

­Yes, there is a typo. However, it does not affect the answer at all. Thank you thoigh for spottin this.­

­Yeah, I know this doesn't change the outcome, just so some members don't get tangled up later.
Math Expert
Joined: 02 Sep 2009
Posts: 94441
Own Kudos [?]: 642826 [1]
Given Kudos: 86716
Re: An airline passenger is planning a trip that involves three [#permalink]
1
Kudos
Gmatguy007 wrote:
­Yeah, I know this doesn't change the outcome, just so some members don't get tangled up later.

­
Fixed the typo. Thank you!
Intern
Joined: 05 Feb 2024
Posts: 7
Own Kudos [?]: 3 [0]
Given Kudos: 0
Re: An airline passenger is planning a trip that involves three [#permalink]
Let's break down the travel time for each leg of the trip:

1. First flight: leaves Airport A at 8:00 a.m. and arrives at Airport B 2.5 hours later, which is 10:30 a.m.
2. Second flight: leaves Airport B every 20 minutes, starting at 8:00 a.m. To find the first departure time of the second flight that arrives at Airport C 1.16 hours later, we add 1.16 hours to the arrival time of the first flight: 10:30 a.m. + 1.16 hours = 11:46 a.m. The first departure time of the second flight is 11:20 a.m. (20 minutes before 11:46 a.m.).
3. Third flight: leaves Airport C every hour, starting at 8:45 a.m.

The passenger must spend time between flights as follows:

* 10:30 a.m. (arrival at Airport B) to 11:20 a.m. (departure of the second flight) = 50 minutes (connection time at Airport B)
* 11:46 a.m. (arrival at Airport C) to 12:00 p.m. (departure of the third flight) = 14 minutes (connection time at Airport C)

The total time spent between flights is:

50 minutes (connection time at Airport B) + 14 minutes (connection time at Airport C) = 64 minutes

Converting 64 minutes to hours and minutes, we get:

1 hour 4 minutes

The correct answer is not among the options, but the closest answer is:

(B) 1 hr 5 min
Re: An airline passenger is planning a trip that involves three [#permalink]
Moderator:
Math Expert
94441 posts