Calculate the center of gravity (CG) location of a loaded aeroplane in Mean Aerodynamic Chord (MAC) percentage.

The moment is calculated as mass multiplied by its arm (station).

• Basic Empty Mass (BEM) moment: $$M_1 = 1000 \text{ kg} \times 92 \text{ inches} = 92000 \text{ kg-inches}$$
• Pilot's mass moment: $$M_2 = 80 \text{ kg} \times 100 \text{ inches} = 8000 \text{ kg-inches}$$
• Fuel mass moment: $$M_3 = 120 \text{ kg} \times 130 \text{ inches} = 15600 \text{ kg-inches}$$

The 0% MAC is at station 80 inches, and the length of the MAC is 60 inches.

\text{CG}_{\%MAC} = \frac{\text{CG}_{\text{inches}} - \text{Station of 0% MAC}}{\text{Length of MAC}} \times 100\% $$\text{CG}_{\%MAC} = \frac{96.333... \text{ inches} - 80 \text{ inches}}{60 \text{ inches}} \times 100\%$$ $$\text{CG}_{\%MAC} = \frac{16.333...}{60} \times 100\%$$ $$\text{CG}_{\%MAC} = 0.27222... \times 100\%$$ $$\text{CG}_{\%MAC} \approx 27.2\%$$

Comparing this to the given options, the closest answer is D.

The final answer is \text{D) 27.2% MAC}.