Using CP = (Distance x Home) / (Out reduced + Home), if the total distance is 800 NM with Out reduced 500 KTS and Home 300 KTS, the CP is approximately which value?

• A. 400 NM
• B. 600 NM
• C. 300 NM
• D. 500 NM

Answer: (C)

The idea is to split the total distance so that the time spent on the outbound leg at its speed and the time spent on the inbound leg at its speed are equal. If you let D be the total distance, Out be the outbound speed, and Home be the inbound speed, then set D_out / Out = D_home / Home with D_out + D_home = D. Solving gives D_home = D × Home /(Out + Home). That D_home is the CP.

Plugging in the numbers: D = 800 NM, Out = 500 KTS, Home = 300 KTS. CP = 800 × 300 / (500 + 300) = 240,000 / 800 = 300 NM.

Check: outbound distance is 800 − 300 = 500 NM; outbound time 500 / 500 = 1 hour. Inbound distance 300 NM; inbound time 300 / 300 = 1 hour. Times equal, so 300 NM is the correct CP.

If the CP were 400 or 500, the outbound and inbound times wouldn’t match (e.g., 400 NM outbound at 500 KTS is 0.8 h, while 400 NM inbound at 300 KTS is 1.333 h).