p∝1qp \propto \frac{1}{\sqrt{q}}p∝q1
⟹ p=kq\implies p = \frac{k}{\sqrt{q}}⟹p=qk
when p = 3, q = 16.
3=k163 = \frac{k}{\sqrt{16}}3=16k
k=3×4=12k = 3 \times 4 = 12k=3×4=12
∴p=12q\therefore p = \frac{12}{\sqrt{q}}∴p=q12
when p = 4,
4=12q ⟹ q=1244 = \frac{12}{\sqrt{q}} \implies \sqrt{q} = \frac{12}{4}4=q12⟹q=412
q=3 ⟹ q=32\sqrt{q} = 3 \implies q = 3^2q=3⟹q=32
q=9q = 9q=9
