The aircraft is currently at 36 000 ft and is required to descend to 12 000 ft which means that it must descend 24 000 ft.
With a rate of descent of 2000 ft/min, the aircraft will take 12 minutes to descend 24 000 ft (24 000 ft ÷ 2000 ft/min).
We must now calculate the distance the aircraft will cover in 12 minutes when flying at a GS of 420 kt: (420 kt x 12 min) ÷ 60 min = 84 NM
Since the aircraft should reach FL120 at 40 NM from the next waypoint, it should start descent at a minimum distance of 124 NM (84 NM + 40 NM).
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