Answer
$p=\dfrac{1000}{9}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use $
p=kqr^2
$ and solve for the value of $k$ with the given $
p,q
$ and $
r
$ values. Then use the equation of variation to solve for the value of the unknown variable.
$\bf{\text{Solution Details:}}$
Since $p$ varies jointly as $q$ and $r^2,$ then $
p=kqr^2
.$ Substituting the given values, $
p=100, q=2
$ and $
r=3
,$ then the value of $k$ is
\begin{array}{l}\require{cancel}
p=kqr^2
\\\\
100=k(2)(3)^2
\\\\
100=k(2)(9)
\\\\
100=k(18)
\\\\
\dfrac{100}{18}=k
\\\\
k=\dfrac{\cancel2\cdot50}{\cancel2\cdot9}
\\\\
k=\dfrac{50}{9}
.\end{array}
Hence, the equation of variation is given by
\begin{array}{l}\require{cancel}
p=kqr^2
\\\\
p=\dfrac{50}{9}qr^2
.\end{array}
If $q=5$ and $r=2,$ then
\begin{array}{l}\require{cancel}
p=\dfrac{50}{9}(5)(2)^2
\\\\
p=\dfrac{50}{9}(5)(4)
\\\\
p=\dfrac{1000}{9}
.\end{array}
