Respuesta :
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]
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]
