Algebra word problem about trains. I have came across this problem and after trying to answer it for some time, I thought my solution was correct but apparently it is not. Can you please explain to me what I have done wrong ?
Problem:
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 its usual rate, arriving at Beetown 2 hours late. If the train had covered 80 miles before the accident, it would have been just one hour late. What is the usual rate of the train ?
My Attempt:
Let $d$ be the total distance of the trip, $t$ the usual time and $x$ the usual rate. We know that 
$$x\cdot t = d$$
The first part of the question tells us that
$$x\cdot 1+ \frac 45 x(t_2) = d$$
where $t_2$ is the time traveled at four-fifths the usual speed and $$t_2 = (t+2)-\frac 12 - 1 = t+\frac 12.$$ The $(t+2)$ is the "two hours late" part and the $(-1-\frac 12)$ is the break and the one hour before the accident. The second part tells us that
$$80 + \frac 45 x(t_3) = d$$
where, again, $t_3$ is the time traveled at four-fifths the usual speed and
$$t_3 = (t+1)-\frac 12 - \frac{80}x.$$
Similar to the first equation, $(t+1)$ is the one-hour late part and the $(-\frac 12 - \frac{80}x)$ is the break along with the time spent travelling before the accident. Solving for $x$ I got $16$, however, the answer is $20$. What have I done wrong ?
 A: I think you got the right answer.  You should have continued, solving for the other unknowns, so you could check.  I get $x=16, d=112,$ so that the usual trip time is $7$ hours, and $4/5$ of the normal speed is $12.8$ mph.
If the accident occurs after one hour, the train has traveled $16$ miles and has $96$ miles to go, which will take $7.5$ hours at $12.8$ mph.  So the length of the trip is $1 + .5+7.5=9,$ two hours longer than usual.
If the accident occurs after $80$ miles, the train has traveled for $5$ hours, and has $32$ miles to go, which will take $2.5$ hours at $12.8$ mph.  The total trip time is $5+.5+2.5=8,$ an hour longer than usual.
The $20$ in the book appears to be an error. 
A: The difference of answer is due to you have answered a different question. You have answered your question correctly. Good job.
This is the question from the book:

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 its usual rate, arriving at Beetown $2$ hours late. If the train had covered $80$ miles $\color{blue}{\text{more}}$ before the accident, it would have been just one hour late. What is the usual rate of the train ?

In that case, the $80$ miles that are being travelled using different speeds is responsible for the difference of $1$ hour.
$$\frac{80}{4x/5}-\frac{80}{x} =1$$
$$\frac{80}{x}\left(\frac14 \right)=1$$ 
$$x=20$$
