Train Traveling from Aytown to Beetown Accident (AHSME 1955) I was working on the question: A train traveling from Aytown to Beetown meets with an accident after 1 hour. The train is stopped for 30 minutes, after which it proceeds at four-fifths of its usual rate, arriving at Beetown 2 hours late. If the train had covered 80 miles more before the accident, it would have been just one hour late. What is the usual rate of the train?
I thought my answer was correct, but it isn't. What was wrong with my work?
For the first scenario, I came up with this equation:
$d=r+(\frac{4}{5}r \cdot \frac{3}{2})$
The r represents the distance travelled in one hour, since the train meets the accident after one hour. The $\frac{3}{2}$ shows the time left after the accident (30 min); 2-(1/2) = 3/2.
For the second scenario, I came up with the equation $d=r+80+(\frac{4}{5}r \cdot \frac{1}{2})$. The $\frac{1}{2}$ is because the train was only 1 hour late and the accident took up 30 minutes, leaving 30 minutes (half an hour) to travel with $\frac{4}{5}$ speed.
I set th equations equal to each other and solved for r. I got that r is 100, but r is actually 20. What did I do wrong?

There is a very simple way of doing such problems, not always taught.
The difference in "lateness" decreases by $1$ hr. if the train covers $80$ more miles before the accident, thus
$\dfrac{80}{\frac{4}{5}v} - \dfrac{80}{v} = 1,\; \rightarrow v = 20\; mph$
