Asked by ashley
A motorist drove 180 mi before running out of gas and walking 6 mi to a gas station. The motorist's driving rate was twelve times the walking rate. The time spent walking was 3 h less than the time spent driving. How long did it take for the motorist to drive the 180 mi?
Answers
Answered by
Steve
since time = distance/speed, if the man's walking speed was w, then we have
180/12w - 3 = 6/w
w = 3
so, he drove at 36 mi/hr, doing the 180 miles in 5 hours.
180/12w - 3 = 6/w
w = 3
so, he drove at 36 mi/hr, doing the 180 miles in 5 hours.
Answered by
Henry
Walking rate = r mi/h.
Time = (T-3) hrs.
r*(T-3) = 6, r = 6/(T-3).
Driving rate = 12r mi/h.
Time = T hrs.
12r*T = 180, Replace r with 6/(T-3): (12^6/(T-3))*T = 180,
72T/(T-3) = 180, 72T = 180*(T-3),
72T = 180T-540, T = 5 hrs.
r = 180mi/5h = 36 mi/h.
Time = (T-3) hrs.
r*(T-3) = 6, r = 6/(T-3).
Driving rate = 12r mi/h.
Time = T hrs.
12r*T = 180, Replace r with 6/(T-3): (12^6/(T-3))*T = 180,
72T/(T-3) = 180, 72T = 180*(T-3),
72T = 180T-540, T = 5 hrs.
r = 180mi/5h = 36 mi/h.
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