Three quantities x, y, z are such that x varies directly as the square root of y and inversely as the square of z. Find the % dec. in x if y dec. by 44% when z inc. by 4%.

The problem states that P varies directly as Q and inversely as the square root of R. This can be written as:

$$P = \frac{kQ}{\sqrt{R}}$$

where $k$ is the constant of proportionality.

Let the initial values be $P_1, Q_1, R_1$.
Q decreases by 4%, so the new value of Q, $Q_2$, is:

$$Q_2 = Q_1 - 0.04 Q_1 = 0.96 Q_1$$

R increases by 44%, so the new value of R, $R_2$, is:

$$R_2 = R_1 + 0.44 R_1 = 1.44 R_1$$

Using the relationship from Step 1 with the new values $Q_2$ and $R_2$:

$$P_2 = \frac{kQ_2}{\sqrt{R_2}}$$

Substitute the expressions for $Q_2$ and $R_2$:

$$P_2 = \frac{k(0.96 Q_1)}{\sqrt{1.44 R_1}}$$ $$P_2 = \frac{k(0.96 Q_1)}{1.2 \sqrt{R_1}}$$

Rearrange the terms to relate $P_2$ to $P_1$:

$$P_2 = \left(\frac{0.96}{1.2}\right) \left(\frac{kQ_1}{\sqrt{R_1}}\right)$$

Since $P_1 = \frac{kQ_1}{\sqrt{R_1}}$, we have:

$$P_2 = 0.8 P_1$$

The percentage decrease is given by the formula:

$$\text{Percentage Decrease} = \frac{P_1 - P_2}{P_1} \times 100\%$$

Substitute $P_2 = 0.8 P_1$:

$$\text{Percentage Decrease} = \frac{P_1 - 0.8 P_1}{P_1} \times 100\%$$ $$\text{Percentage Decrease} = \frac{0.2 P_1}{P_1} \times 100\%$$ $$\text{Percentage Decrease} = 0.2 \times 100\%$$ $$\text{Percentage Decrease} = 20\%$$

The percentage decrease in P is 20%.

$$\boxed{20\%}$$