Joe is driving to the airport to pick up his wife. If he drives at his current speed, he'll arrive on time. If he drives 5 miles per hour faster, he'll arrive 12 minutes early. If he were to drive 5 miles per hour slower instead, he'd arrive 15 minutes late.
How far away is the airport?
(In reply to
Solution by Jyqm)
My final result is in conformity with that of Jyqm, and the methodology culminating in the said result is furnished hereunder as follows:
Let (Distance of the airport, Rate of drive) = (D, R)
Then, in terms of provisions of the problem under reference:
(i) D/(R+5) = D/R – 1/5
(ii) D/(R-5) = D/R +1/4
So, 25*D/(R(R+5)) = 20*D/(R(R-5)) = 1,
giving, (R-5)/(R+5) = 4/5, so that R = 45
Hence, 25*D = 45*50; giving, D = 90.
Consequently, the airport is 90 MILES AWAY and :
Joe's current speed is 45 miles per hour .
(Verification:
Substituting (D,R) = (90,45) :
in (i) 90/ 50 = 1.8= 90/45- .2
in (ii) 90/40 = 2.25 = 90/45 + .25)
Edited on June 27, 2006, 12:13 pm
Edited on June 27, 2006, 12:17 pm
Edited on June 27, 2006, 12:18 pm
Another Approach
