<PRE><B>Question 558818</B>
Suppose Tracey takes x minutes to get to work (on time). Then it takes x+1 minutes going at 50 mph and x-1 minutes going at 55 mph. Since distance(D) = rate*time, we have


*[tex \LARGE D = 50(x+1) = 55(x-1)]


*[tex \LARGE 50x + 50 = 55x - 55 \Rightarrow x = 21]


Therefore Tracey takes 22 minutes going at 50 mph and 20 minutes going at 55 mph. It can be checked that the total distance is equal in both cases, 55/3 or 18 1/3 miles.</PRE>