Asked by Jane

A motorist was to travel from town A to town B, a distance of 80 miles. He traveled the first 24 miles at a certain rate; traffic then increased and for the next 6 miles he averaged 10 miles per hour less than his original speed; then traffic eased up and he traveled the remaining distance at a rate 50% greater than his original rate. He arrived 22 earlier than he would had he traveled the whole distance at his original rate. Find his original rate.
Answered by Reiny
What part of Steve's answer, which he gave you yesterday, did you not like?

http://www.jiskha.com/display.cgi?id=1382654399
Answered by Jane
It's not his problem it's just that I stated the question wrong instead of 24 and 6 minutes, they're supposed to be 24 and 6 miles
Answered by Reiny
let rate in 1st leg be x mph
time takenfor 1st leg = 24/x hrs

rate for 2nd leg = x-10
time for 2nd leg = 6/(x-10)

remaining distance = 80-24-6 = 50 miles
time for last leg = 50/(1.5x)

time for whole trip at x mph = 80/x

difference in times = 22 minutes = 22/60 hrs

80/x - 24/x - 6/(x-10) - 50/(1.5x) = 22/60

solving this on paper (don't feel like typing all that algebra)
I got x = 40 or x = 15.4545..

the original rate was 40 mph or 15.45 mph

let's check x = 40
time for first leg = 24/40 = .6
time for 2nd leg = 6/30 = .2
time for last part = 50/60 = 5/6
total time = .6+.2+5/6 hrs = 49/30 hrs or 98 minutes

time at 40 mph for whole trip = 80/40 = 2 hrs or 120 minutes
difference = 22 minutes, YEAHHHH

surprisingly x = 14.4545 also works, I will let you check it the same way