Given that A varies directly as B and inversely as the square of C, find the equation connecting A, B and C and then find C when A = 150 and B = 96.

A varies directly as B: $A \propto B$
A varies inversely as the square of C: $A \propto \frac{1}{C^2}$
Combining these, we get:

$$A \propto \frac{B}{C^2}$$

Introducing the constant of proportionality, $k$:

$$A = \frac{kB}{C^2}$$

Given $A = 125$, $B = 180$, $C = 6$.

$$125 = \frac{k \times 180}{6^2}$$ $$125 = \frac{180k}{36}$$ $$125 = 5k$$ $$k = \frac{125}{5}$$ $$k = 25$$

Substitute the value of $k$ back into the variation equation:

$$\boxed{A = \frac{25B}{C^2}}$$

$$150 C^2 = 25 \times 96$$ $$150 C^2 = 2400$$ $$C^2 = \frac{2400}{150}$$ $$C^2 = 16$$ $$C = \sqrt{16}$$ $$C = \boxed{4}$$

(Assuming C is a positive value in this context).

Let the original values be $B_1$ and $C_1$.
If B increases by 10%, the new value $B_2$ is:

$$B_2 = B_1 + 0.10 B_1 = 1.10 B_1$$

If C increases by 15%, the new value $C_2$ is:

$$C_2 = C_1 + 0.15 C_1 = 1.15 C_1$$

The original equation is $A_1 = \frac{25B_1}{C_1^2}$.
The new value $A_2$ is:

$$A_2 = \frac{25B_2}{C_2^2}$$

Substitute $B_2$ and $C_2$:

$$A_2 = \frac{25(1.10 B_1)}{(1.15 C_1)^2}$$ $$A_2 = \frac{25 \times 1.10 B_1}{1.15^2 C_1^2}$$ $$A_2 = \frac{1.10}{1.15^2} \times \frac{25B_1}{C_1^2}$$

Since $A_1 = \frac{25B_1}{C_1^2}$:

$$A_2 = \frac{1.10}{1.15^2} A_1$$ $$A_2 = \frac{1.10}{1.3225} A_1$$ $$A_2 \approx 0.83175886 A_1$$

Percentage increase = $\left(\frac{A_2 - A_1}{A_1}\right) \times 100\%$

$$= \left(\frac{0.83175886 A_1 - A_1}{A_1}\right) \times 100\%$$ $$= (0.83175886 - 1) \times 100\%$$ $$= -0.16824114 \times 100\%$$ $$= -16.82\% \text{ (to 2 decimal places)}$$

Since the value is negative, it represents a percentage decrease. The percentage increase is negative.

$$\boxed{-16.82\%}$$