Using CP = (Total distance x Ground speed home) / (Groundspeed out + Groundspeed home), if the total distance is 1500 NM, Grounds Out 600 KTS, Ground Home 450 KTS, the CP is approximately which value?

• A. 605 NM
• B. 527 NM
• C. 643 NM
• D. 700 NM

Answer: (C)

The concept being tested is how to locate a point along the route where outbound and inbound times balance given different groundspeeds. This CP comes from setting the time to the point from the home side equal to the time to the rest of the route from the origin side: x / V_home = (D − x) / V_out. Solving gives CP distance from the home to the point as CP = D × V_home / (V_out + V_home). Plug in the numbers: D = 1500 NM, V_home = 450 knots, V_out = 600 knots. CP = 1500 × 450 / (600 + 450) = 675000 / 1050 ≈ 642.857 NM, about 643 NM. This matches the option of approximately 643 NM.