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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 2 \frac{1}{2} hours later. The second flight leaves Airport B every 20 minutes, beginning at 8:00 a.m., and arrives at Airport C 1 \frac{1}{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?

Explanation	Calculations	Help
$We convert the mixed numbers into decimal hours.$	$2 + \frac{1}{2} = 2.5 1 + \frac{1}{6} \approx 1.167$	Theory & Tactics Method Card FDPR1-A Click to view full details
$We note that flights from A depart every hour starting at 8:00 and take 2.5 hours to reach B.$	$t_{A, dep} = 8 + n t_{B, arr} = t_{A, dep} + 2.5$
$We observe that arrivals at B occur 30 minutes past the hour and that the next departure from B is at 40 minutes past the hour, so the wait is 10 minutes.$	$arrival minutes = 30 next departure minutes = 40 wait_{B} = 40 - 30 = 10 minutes = \frac{1}{6} hour \approx 0.167$
$We calculate the flight time from B to C as 1.167 hours, which is 1 hour 10 minutes.$	$1 + \frac{1}{6} \approx 1.167 hours = 1 hour 10 minutes$
$We find that an arrival at C happens 40 minutes past the hour plus 10 minutes of flight, giving 50 minutes, and the next departure from C is at 45 minutes past the hour, so the wait is 55 minutes.$	$arrival minutes at C = 40 + 10 = 50 departure minutes at C = 45 wait_{C} = (60 - 50) + 45 = 55 minutes \approx 0.917 hours$
$We sum the waits at B and C to get the total layover time.$	$10 + 55 = 65 minutes = 1 hour 5 minutes$

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1 hour 5 minutes

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 2 \frac{1}{2} hours later. The second flight leaves Airport B every 20 minutes, beginning at 8:00 a.m., and arrives at Airport C 1 \frac{1}{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?

Explanation	Calculations	Help
$We convert the mixed numbers into decimal hours.$	$2 + \frac{1}{2} = 2.5 1 + \frac{1}{6} \approx 1.167$	Theory & Tactics Method Card FDPR1-A Click to view full details
$We note that flights from A depart every hour starting at 8:00 and take 2.5 hours to reach B.$	$t_{A, dep} = 8 + n t_{B, arr} = t_{A, dep} + 2.5$
$We observe that arrivals at B occur 30 minutes past the hour and that the next departure from B is at 40 minutes past the hour, so the wait is 10 minutes.$	$arrival minutes = 30 next departure minutes = 40 wait_{B} = 40 - 30 = 10 minutes = \frac{1}{6} hour \approx 0.167$
$We calculate the flight time from B to C as 1.167 hours, which is 1 hour 10 minutes.$	$1 + \frac{1}{6} \approx 1.167 hours = 1 hour 10 minutes$
$We find that an arrival at C happens 40 minutes past the hour plus 10 minutes of flight, giving 50 minutes, and the next departure from C is at 45 minutes past the hour, so the wait is 55 minutes.$	$arrival minutes at C = 40 + 10 = 50 departure minutes at C = 45 wait_{C} = (60 - 50) + 45 = 55 minutes \approx 0.917 hours$
$We sum the waits at B and C to get the total layover time.$	$10 + 55 = 65 minutes = 1 hour 5 minutes$

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1 hour 5 minutes