All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > General
To the Airport (Posted on 2006-06-27) Difficulty: 2 of 5
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?

See The Solution Submitted by Charlie    
Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Another Approach | Comment 3 of 4 |
(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
  Posted by K Sengupta on 2006-06-27 12:07:28

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (14)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information