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