Choice A is the correct answer. Find half-life from remaining percentage.

1. Model: $A(t) = A_0(0.5)^{t/h}$ where $h$ is half-life.
2. Given: After 10 years, 40% remains: $0.40 = (0.5)^{10/h}$.
3. Solve: Take log: $\log(0.40) = \frac{10}{h} \log(0.5)$.
4. Calculate: $h = \frac{10 \log(0.5)}{\log(0.40)} \approx \frac{10(-0.301)}{-0.398} \approx 7.56$ years.

💡 Strategic Tip: Use logarithms or test values: if $h=7.5$, then $(0.5)^{10/7.5} \approx 0.39$ ✓.

Choice B is incorrect because after 10 years (2 half-lives), only 25% would remain. Choice C is incorrect because this would leave 50%, not 40%. Choice D is incorrect because this would leave about 18%.