An airline passenger is planning a trip that involves three

Question

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

Bunuel · Accepted Answer

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) 25 min
(B) 1 hr 5 min
(C) 1 hr 15 min
(D) 2 hr 20 min
(E) 3 hr 40 min

Flight A to B 
 
  Departs:  at   8:00 a.m.  , 9:00 a.m., ... 
  Duration:  2.5 hours. 
  Arrivals at B:    10:30 a.m.  , 11:30 a.m., 12:30 a.m., ...

Flight B to C 
 
  Departs:  at 8:00 a.m., 8:20 a.m. 8:40 a.m., ...,   10:40 a.m.  , ... 
  Duration:   1 \frac{1}{6}  (70 minutes). 
  Arrivals at C:  9:10 a.m, 9:30 a.m, 9:40 a.m, 9:00 a.m, ...,   11:50 a.m  , ...

Flight C onward 
 
  Departs:  at 8:45 a.m., 9:45 a.m., 10:45 a.m., 11:45 am,   12:45 a.m.  , ...

If the passenger takes the   8:00 a.m.   flight from A:
 
 Layover at B: 10 minutes (  10:30 a.m.   to   10:40 a.m.  ) 
 Layover at C: 55 minutes (  11:50 a.m.   to   12:45 p.m.  ) 
 Total layover time = 65 minutes.

Answer: B.

JeffTargetTestPrep · Answer

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

b2bt · Answer

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)

PareshGmat · Answer

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

arunspanda · Answer

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) 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)

GyanOne · Answer

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.

rrsnathan · Answer

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

kablayi · Answer

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?

KarishmaB · Answer

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.

Abhishek009 · Answer

Capture.PNG 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)

dave13 · Answer

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

pushpitkc · Answer

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!

adgarg · Answer

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?

Bunuel · Answer

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.

sheriffs · Answer

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

sirapatsam · Answer

Hey is it 10.40 not 9.40? because it take 2.30 hr right?

Bunuel · Answer

No, it should be 9:40, not 10:40.

The first flight leaves Airport A at 8:00 and takes 1 hour 30 minutes, so it reaches Airport B at 9:30.
 
Flights from Airport B depart every 20 minutes starting at 8:00 (8:00, 8:20, 8:40, 9:00, 9:20, 9:40, and so on).
 
Therefore, the next available flight after the arrival at 9:30 is the one at 9:40, not 10:40.

Usernamevisible · Answer

Shortcut for pattern to ensure all cases are covered :

This is the pattern to notice immediately:
* B flights are every 20 min
* C flights are every 60 min

And:
60 is a multiple of 20.

That means:
Every time you delay your B departure by 20 min, your arrival at C also shifts by exactly 20 min within the same repeating 60-minute cycle.

So:
* waiting at B increases by 20
* waiting at C decreases by 20

→ total waiting stays the same.
Mental shortcut:
If:
* one schedule repeats every `a` minutes
* another repeats every multiple of `a`

then delaying by `a` minutes simply redistributes waiting time.
You are not changing total waiting.

In this problem:
At B, your choices differ by:
20-minute chunks
That shifts arrival at C by:
20-minute chunks

Since C departures repeat every 60 min (= 3 chunks of 20),
all you are doing is moving waiting time between B and C.
Total waiting remains constant.
So once one valid route gives:

10 + 55 = 65 minutes
you already know every other route also gives 65 minutes.

Solution :

The real GMAT concern is:

> “How do I KNOW another flight choice won’t reduce the total waiting time?”

Here’s the clean pattern logic.

---

STEP 1: Notice repeating schedules

* Flights from A leave every 60 min
* Flights from B leave every 20 min

Since 60 is a multiple of 20:

Every time you shift the starting flight from A by 1 hour, everything downstream shifts by exactly 1 hour too.

So the waiting pattern repeats identically.

---

STEP 2: Focus only on clock endings

A → B travel time = 2 hr 30 min

A flights:

* 8:00 → 10:30
* 9:00 → 11:30
* 10:00 → 12:30

Notice:

You ALWAYS arrive at B at :30.

---

STEP 3: Next available B flight

B flights leave every 20 min:

* :00
* :20
* :40

If you arrive at :30, the next flight is always :40.

Examples:

* 10:30 → 10:40
* 11:30 → 11:40
* 12:30 → 12:40

So:

* waiting at B is ALWAYS 10 min
* departure from B is ALWAYS at :40

---

STEP 4: Arrival pattern at C

B → C travel time = 1 hr 10 min

Starting from :40:

* :40 + 1:10 = :50

Examples:

* 10:40 → 11:50
* 11:40 → 12:50
* 12:40 → 1:50

So you ALWAYS arrive at C at :50.

---

STEP 5: Waiting at C

Flights from C leave every hour at :45.

If you arrive at :50, you just missed the :45 flight.

Example:

* arrive 11:50
* next flight 12:45

So waiting at C is ALWAYS:
55 min.

---

Total waiting:

* B wait = 10 min
* C wait = 55 min

Total:
65 min = 1 hr 5 min

---

What if you intentionally skip a B flight?

Example:

* instead of taking :40, take :00

Then:

* B wait increases by 20 min
* but C wait decreases by 20 min

Why?

Because:

* B flights repeat every 20 min
* C flights repeat every 60 min (= 3 chunks of 20)

So delaying by 20 minutes simply redistributes waiting time between B and C.

The total waiting time never changes.

That is why solving ONE valid path is enough to prove the minimum total is always:

65 minutes.

An airline passenger is planning a trip that involves three

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