Answer
$\dfrac{20}{3} \text{ } mph$
Work Step by Step
Let $x$ be the speed of the current. Using $D=rt,$ then, going upstream,
\begin{array}{l}\require{cancel}
54=(24-x)t
\\\\
\dfrac{54}{24-x}=t
.\end{array}
Using $D=rt,$ then, going downstream,
\begin{array}{l}\require{cancel}
90=(24+x)t
\\\\
\dfrac{90}{24+x}=t
.\end{array}
Equating the equations of $t,$ then
\begin{array}{l}\require{cancel}
\dfrac{54}{24-x}=\dfrac{90}{24+x}
.\end{array}
By cross-multiplication and using the properties of equality, then
\begin{array}{l}\require{cancel}
54(24+x)=90(24-x)
\\\\
1200+54x=2160-90x
\\\\
54x+90x=2160-1200
\\\\
144x=960
\\\\
x=\dfrac{960}{144}
\\\\
x=\dfrac{20}{3}
.\end{array}
Hence, the speed of the current, $x,$ is $
\dfrac{20}{3} \text{ } mph
.$
