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Re: An airline passenger is planning a trip that involves three [#permalink]
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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
Time Taken 4:08
Difficulty level 650 (just because it is lengthy)
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Re: An airline passenger is planning a trip that involves three [#permalink]
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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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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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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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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,
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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?
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Re: An airline passenger is planning a trip that involves three [#permalink]
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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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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 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)
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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.

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

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.­
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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")­
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An airline passenger is planning a trip that involves three [#permalink]
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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.­
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Re: An airline passenger is planning a trip that involves three [#permalink]
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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.   ;)
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Re: An airline passenger is planning a trip that involves three [#permalink]
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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!
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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
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