Question

A train, an hour after starting, meets with an accident which detains it a half hour, after which it proceeds at $\frac{3}{4}$ of its former rate and arrives $3\frac{1}{2}$ hours late. Had the accident happened $90$km farther along the line, it would have arrived only $3$ hours late. The length of the trip in km was:

• A. $400$
• B. $465$
• C. $600$
• D. $640$
• E. $550$

Answer 1

Answer: (C)

$600$

Let x be the distance from the point of the accident to the end of the trip and R the former rate of the train.
Then the normal time for the trip, in hours, is $\frac{x}{R}+1=\frac{x+R}{R}$.
Considering the time for each trip, we have
$1+\frac{1}{2}+fracx\left(3/4\right)R=\frac{x+R}{R}+3.$
$\therefore x=540andR=60$; $\therefore$ the trip is $540+60=600\left(km\right).$