Question on Combined Variations

Question

The answer I got for K is 192 but I felt that was too much. So any help clarifying would be great.

y varies directly with the square root of p and inversely with the square of q. If y=6 when p=4 and q=8. Find y when p=16 and q=4

Answer 1

Method 1:$$y \ \propto \frac{\sqrt p}{q^2} \implies y = k \frac{\sqrt p}{q^2} $$

At $y=6, p= 4 , q = 8$,

$k = 6\cdot32 = 192$

So at $p = 16, q = 4$

$ y = k \frac{\sqrt p}{q^2} = 192 \frac{4}{16} = 48$

Method 2: $$y \ \propto \frac{\sqrt p}{q^2} \implies \frac{y_2}{y_1} = \frac{\sqrt{p_2}}{q_2^2} \cdot \frac{q_1^2}{\sqrt{p_1}}$$

$$\frac{y_2}{6} = \frac{4}{16}\cdot\frac{64}{2}=8 \implies y_2 = y = 48$$