Discussion
With a mean aerodynamic chord of \( 5 \) m whose leading edge is at \( 21 \) m, a centre of gravity at \( 22.26 \) m lies at
Answer: B
Correct answer: 25 per cent MAC
Centre of gravity position is quoted as the distance aft of the leading edge of the mean aerodynamic chord, divided by that chord: \[ \frac{22.26 - 21}{5} = \frac{1.26}{5} = 0.252 \approx 25\ \text{per cent MAC.} \] The subtraction is what the distractors omit - 22.26 divided by 5 has no physical meaning, since the datum is not the leading edge.
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