If \(y \propto \frac{1}{\sqrt{x}}\) and x = 16 when y = 2, find x when y = 24

A.

\(\frac{1}{9}\)

B.

\(\frac{1}{6}\)

C.

\(\frac{1}{3}\)

D.

\(\frac{2}{3}\)

Correct answer is A

\(y \propto \frac{1}{\sqrt{x}}\)

\(y = \frac{k}{\sqrt{x}}\)

When x = 16, y = 2.

\(2 = \frac{k}{\sqrt{16}} \implies 2 = \frac{k}{4}\)

\(k = 8\)

\(y = \frac{8}{\sqrt{x}}\)

When y = 24,

\(24 = \frac{8}{\sqrt{x}}\)

\(\sqrt{x} = \frac{8}{24} = \frac{1}{3}\)

\(\therefore x = (\frac{1}{3})^2\)

\(x = \frac{1}{9}\)