An airline passenger is planning a trip that involves three

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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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)

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

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)

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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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

I made a chart a lot like Paresh did and it took me 3minutes and 50 seconds to do this problem

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

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.

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

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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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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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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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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VeritasKarishma, BrentGMATPrepNow

Is 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!

Something else to keep in mind is that the problem could be solved another way:

Take 8 AM Flight, Arrive at B 9:30
Layover for 50 mins at B, take 10:20
Arrive at C at 11:30, layover of 15 mins

=65 mins

Still the same, but did it other in case to see if there is a difference.

First Flight leaves Airport A at \(8:00\) am and arrive at Airport B after 1 hour 30 minutes at 9:30 am

The second flight leaves Airport at 8:20, 8:40, 9:00, 9:20, \(9:40\) So, the passenger waits \(10\) minutes in Airport B

The passenger arriving airport C after \frac{7}{6}*60=70 minutes or in 1 hour 10 minutes at 10:50 am

The Third Flight leaves Airport C at 8:45, 9:45, 10:45, \(11:45\), So the passenger waits 55 minutes to flight at Airport C.

Total Waiting time \(55+10=65\) minutes or \(1\ hour \ 5 \ minutes. \)

The Answer is \(B.\)
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