$
\begin{aligned}
& \text { i. } p \propto \frac{1}{q} \ldots \text {..[Given] } \\
& \therefore \mathrm{p}=\mathrm{k} \times \frac{1}{q}
\end{aligned}
$
where, $\mathrm{k}$ is the constant of variation.
$
\therefore \mathrm{p} \times \mathrm{q}=\mathrm{k}
$
When $p=15, q=4$
$\therefore$ Substituting $p=15$ and $q=4$ in (i), we get
$
\begin{aligned}
& \mathrm{p} \times \mathrm{q}=\mathrm{k} \\
& \therefore 15 \times 4=\mathrm{k} \\
& \therefore \mathrm{k}=60
\end{aligned}
$
Substituting $k=60$ in (i), we get
$
\begin{aligned}
& \mathrm{p} \times \mathrm{q}=\mathrm{k} \\
& \therefore \mathrm{p} \times \mathrm{q}=60
\end{aligned}
$
This is the equation of variation.
$\therefore$ The constant of variation is 60 and the equation of variation is $p q=60$.ii. $z \propto \frac{1}{w} \ldots$ Given]
$
\therefore \mathrm{z}=\mathrm{k} \times \frac{1}{w}
$
where, $\mathrm{k}$ is the constant of variation,
$
\therefore \mathrm{z} \times \mathrm{w}=\mathrm{k}
$
When $z=2.5, w=24$
$\therefore$ Substituting $z=2.5$ and $w=24$ in (i), we get
$
\begin{aligned}
& \mathrm{z} \times \mathrm{w}=\mathrm{k} \\
& \therefore 2.5 \times 24=\mathrm{k} \\
& \therefore \mathrm{k}=60
\end{aligned}
$
Substituting $k=60$ in (i), we get
$
\begin{aligned}
& \mathrm{z} \times \mathrm{w}=\mathrm{k} \\
& \therefore \mathrm{z} \times \mathrm{w}=60
\end{aligned}
$
This is the equation of variation.
$\therefore$ The constant of variation is 60 and the equation of variation is $z W=60$.
iii. $s \propto \frac{1}{t^2} \ldots$ [Given]
$
\therefore s=k \times \frac{1}{t^2}
$
where, $k$ is the constant of variation,
$
\therefore \mathrm{s} \times \mathrm{t}^2=\mathrm{k}
$
When $\mathrm{s}=4, \mathrm{t}=5$
$\therefore$ Substituting, $\mathrm{s}=4$ and $t=5$ in (i), we get
$
\begin{aligned}
& \mathrm{s} \times \mathrm{t}^2=\mathrm{k} \\
& \therefore 4 \times(5)^2=\mathrm{k} \\
& \therefore \mathrm{k}=4 \times 25 \\
& \therefore \mathrm{k}=100
\end{aligned}
$
Substituting $k=100$ in (i), we get
$
\begin{aligned}
& \mathrm{s} \times \mathrm{t}^2=\mathrm{k} \\
& \therefore \mathrm{s} \times \mathrm{t}^2=100
\end{aligned}
$
This is the equation of variation.
$\therefore$ The constant of variation is 100 and the equation of variation is st ${ }^2=100$.
iv. $x \propto \frac{1}{\sqrt{y}} \ldots$. [Given]
$
\therefore x=\mathrm{k} \times \frac{1}{\sqrt{y}}
$
where, $\mathrm{k}$ is the constant of variation,
$
\therefore \mathrm{x} \times \sqrt{ } \mathrm{y}=\mathrm{k}
$
When $x=15, y=9$
$\therefore$ Substituting $x=15$ and $y=9$ in (i), we get
$
\begin{aligned}
& \mathrm{x} \times \sqrt{ } \mathrm{y}=\mathrm{k} \\
& \therefore 15 \times \sqrt{ } 9=\mathrm{k} \\
& \therefore \mathrm{k}=15 \times 3 \\
& \therefore \mathrm{k}=45
\end{aligned}
$
Substituting $k=45$ in (i), we get
$
\begin{aligned}
& x \times \sqrt{y}=\mathrm{k} \\
& \therefore x \times \sqrt{y}=45 .
\end{aligned}
$
This is the equation of variation.
$\therefore$ The constant of variation is $k=45$ and the equation of variation is $x \sqrt{ } y=45$