let the original rate be x miles/hr
distance covered on first leg = (24/60)x = 2x/5 miles
distance covered on his second leg = (6/60)(x-10) = (x-10)/10
remaining distance = 80 - 2x/5 - (x-10)/10
time for remaining distance = (80 - 2x/5 - (x-10)/10 )/1.5x
time had he gone the whole distance at x mph = 80/x
sum of times of 3 legs + 22/60 = time at one speed
24/60 + 6/60 + (80 - 2x/5 - (x-10)/10 )/(1.5x) + 22/60 = 80/x
2/5 + 1/10 + (80 - 2x/5 - (x-10)/10 )/1.5x + 11/30 = 80/x
13/15 + (80 - 2x/5 - (x-10)/10 )/1.5x ) = 80/x
times 15x
13x + 10(80 - 2x/5 - (x-10)/10 ) = 1200
13x + 800 - 4x - x +10 = 1200
8x = 390
x = 390/8 = 48.75 mph
check:
distance of 1st leg = (24/60)(48.75) = 19.5 miles
distance of 2nd leg = (6/60)(38.75) = 3.875
distance left = 80-19.5-3.875 = 56.625 miles
at speed of 50% more or 73.125 mph, time = 56.625/73.125 hrs or 46.46 min.
total time = 24+10+46.46 or 80.46 min
time going at one speed of 48.75 = 1.6410 hr or 98.46 min
difference = 98.46 - 80.46 = 18
ARGGGG, should have been 22
Can't find my arithmetic error, see if you can find it.
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 AND A TABLE WOULD BE BEST THANKS!!
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