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An airline passenger is planning a trip that involves three

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An airline passenger is planning a trip that involves three

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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

OA B

Source: Official Guide

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BTGmoderatorDC 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
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. Choosing different flights out of Airport A will not change the waiting time either, since every flight out of Airport A arrive at Airport B at half hours. 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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Scott Woodbury-Stewart Founder and CEO

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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

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BTGmoderatorDC 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

Source: Official Guide
FOCUS: minimum (sum of between-flights times)

Let´s say now is 8:00 a.m., perfect moment for the passenger to catch his/her first flight.

He/She will be "ready" for next flight from 9:30 a.m. on, therefore he/she MUST wait 10min to catch at 9:40 a.m.

One hour and 10min later (hence 10:50 a.m.) he/she will be "ready" for the third flight, that will be caught only 11:45 a.m.

Hence He/She must wait 55min to be on that flight.

? = (10+55)min = 1h5min.

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.

_________________
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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