Discussion
A crack takes \( 12\,000 \) flights to grow from detectable to critical. With two inspections required in that period, the interval is
Answer: B
Correct answer: 6000 flights
If two inspections must fall within the interval during which a crack grows from detectable to critical, the interval can be no more than half that period: \[ \frac{12{,}000}{2} = 6{,}000\ \mathrm{flights}. \] The requirement for two rather than one is deliberate redundancy - a single missed detection must not lose the aircraft.
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