total distance 1200NM, GROUNDSPEED OUT 404KTS, REDUCED GROUNDSPEED HOME 396KTS. CP = ?

• A. 400 NM
• B. 600 NM
• C. 300 NM
• D. 594 NM

Answer: (D)

The key idea is to locate the turnaround point so that the time spent on the outbound leg equals the time spent on the inbound leg, given fixed total distance and different groundspeeds.

Let the outbound distance be x NM. Then the inbound distance is 1200 − x NM. Time outbound = x/404 hours, time inbound = (1200 − x)/396 hours. If these times are equal:

x/404 = (1200 − x)/396

Solve: 396x = 404(1200 − x) → 800x = 484800 → x = 606 NM.

So you fly 606 NM outbound, leaving 1200 − 606 = 594 NM to travel back home. If CP is the distance from the turnaround point back to home, CP = 594 NM.