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.

Answer 1

What part of Steve's answer, which he gave you yesterday, did you not like?

http://www.jiskha.com/display.cgi?id=1382654399

Answer 2

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

Answer 3

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