212
5
10
80
Correct answer is D
y \propto \frac{1}{x^2} \Rightarrow y = \frac{k}{x^2} \Rightarrow k = yx^2\\ k = 1\frac{1}{4} \times 4^2 \Rightarrow k=\frac{5}{4}\times \frac{16}{1}=20\\ ∴y = 20 \div \left(\frac{1}{2}\right)^2=20\times \frac{4}{1}=80
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