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.

Question

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

Answer 4

To solve this problem, let's break it down into smaller steps:

Step 1: Let's assume that the motorist's original rate is R miles per hour.

Step 2: The time taken to travel the first 24 miles at a rate of R miles per hour is 24/R hours.

Step 3: For the next 6 miles, the motorist averaged 10 miles per hour less than the original speed, so the rate for this portion is (R - 10) miles per hour. The time taken to travel this distance is 6/(R - 10) hours.

Step 4: The remaining distance of 80 - 24 - 6 = 50 miles is traveled at a rate 50% greater than the original rate. So the rate for this portion is 1.5R miles per hour. The time taken to travel this distance is 50/1.5R = 100/(3R) hours.

Step 5: The total time taken for the entire trip is the sum of the times taken for each portion. So the total time is 24/R + 6/(R - 10) + 100/(3R).

Step 6: According to the problem, the motorist arrived 22 minutes earlier than he would have if he traveled the whole distance at his original rate. This means the total time taken for the original rate should be 22 minutes (or 22/60 hours) more than the total time we calculated in Step 5.

Step 7: Set up an equation to solve for R:
Total time with original rate + 22/60 = Total time calculated in Step 5.

Step 8: Solve the equation to find the value of R (the original rate).
Total time with original rate = 24/R + 6/(R - 10) + 100/(3R)
Total time calculated in Step 5 = Total time with original rate + 22/60

Solving the equation will give us the value of R, which is the original rate.