recall your d = rt, distance = rate * time
so, let's say Fritz travel by Car at a speed of "r", if the Train runs faster then his car, then the Train runs at "r+32".
Bear in mind that, he travels from home to work by Car or Train, so, the distance if the same for either vehicle, let's say is "d" miles.
[tex]\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{minutes}{time}\\
&-----&-----&-----\\
Car&d&r&36\\
Train&d&r+32&20
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=36r\\
d=20(r+32)\\
----------\\
\boxed{36r}=20(r+32)
\end{cases}
\\\\\\
\cfrac{36r}{20}=r+32\implies \cfrac{9r}{5}=r+32\implies 9r=5r+160
\\\\\\
4r=160\implies r=\cfrac{160}{4}\implies \boxed{r=40}[/tex]
so... he travels at 40mph for 36 minutes, now, 36minutes is not even an hour, is 36/60 or 3/5 hr, so... [tex]\bf 40\cdot \cfrac{36}{60}\implies 40\cdot \cfrac{3}{5}\implies \stackrel{miles}{24}[/tex]
