Proportion and inverse proportion It is given that $x$ is directly proportional to $y^2$ and $y$ is inversely proportional to $z$. If $x=20$ and $y=2$ when $z=5$
(A) the value of $y$ when $z=20$
(B) the value of $x$ when $y=3$
(C) an equation relating $x$ and $z$
(D) the value of $z$ when $x = \frac 5 4$, given that $z>0$

if $x\propto y\implies x=ky$ (k is a constant )
$$\dfrac {x_1}{x_2}=\dfrac {{y_1}^2}{{y_2}^2}$$
partB
$$\dfrac {20}{x_2}=\dfrac {2^2}{3^2}\implies x_2=\dfrac{20\times 9}{4}\implies x_2=45$$
part A
$$\dfrac{y_1}{y_2}=\dfrac{z_2}{z_1}$$
$$\dfrac{2}{y_2}=\dfrac{20}{5}\implies y_2=0.5$$
part c
$$x\propto y^2\implies x=ky^2\implies y=\sqrt {\dfrac xk}$$ and $$y\propto \dfrac 1z\implies y=\dfrac cz\implies \sqrt{\dfrac xk}=\dfrac cz\implies \sqrt x\propto\dfrac 1z\implies x=\dfrac {1}{\sqrt z}$$
I think you can now solve part D on you own
part D
$$\dfrac{x_1}{x_2}=\sqrt{\dfrac {z_2}{z_1}}$$
$$\dfrac{20}{\frac 54}=\sqrt{\dfrac {z_2}{5}}$$
$$\dfrac{20^2}{({\frac 54})^2}={\dfrac {z_2}{5}}$$
$$z_2=\dfrac{400\times 5\times 16}{25}$$
$$z_2=1280$$
