A motorist was to travel from town A to town B, a distance of 80 miles. He traveled the first 24 minutes at a certain rate; traffic then increased and for the next 6 minutes 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 minutes at a certain rate; traffic then increased and for the next 6 minutes 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. 
 PLEASE PROVIDE SOME EXPLANATIONS

Steve · Answer

let x be the original rate (mi/min)
Let y be the time traveled at 50% faster speed.
Since distance = time*speed,

24x + 6(x-1/6) + y(1.5x) = 80
24+6+y = 80/x - 22

x = 13/16
y = 604/13

original speed = 13/16 mi/min = 48.75 mi/hr

check:
24 min @ 13/16 mi/min = 39/2 mi
6 min @ 31/48 mi/min = 31/8 mi
remainder 453/8 mi @ 39/32 = 604/13 min
total time: 994/13 min

80/(13/16) = 1280/13

1280/13 - 994/13 = 286/13 = 22 min

strange answer, but it works.