If P varies inversely as the square root of q, where p = 3 and q = 16, find the value of q when p = 4.
12
8
9
16
Correct answer is C
\(p \propto \frac{1}{\sqrt{q}}\)
\(\implies p = \frac{k}{\sqrt{q}}\)
when p = 3, q = 16.
\(3 = \frac{k}{\sqrt{16}}\)
\(k = 3 \times 4 = 12\)
\(\therefore p = \frac{12}{\sqrt{q}}\)
when p = 4,
\(4 = \frac{12}{\sqrt{q}} \implies \sqrt{q} = \frac{12}{4}\)
\(\sqrt{q} = 3 \implies q = 3^2 \)
\(q = 9\)