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?

The airport is 90 miles away.

The problem gives us three equations for the distance:

d = rt
d = (r + 5)(t - 1/5)
d = (r - 5)(t + 1/4)

Setting equations 2 and 3 equal to each other gives:

rt - r/5 + 5t - 1 = rt + r/4 - 5t - 5/4
10t = r/4 + r/5 - 5/4 + 1
10t = 9r/20 - 1/4
10t = (9r - 5)/20
t = (9r - 5)/200

Setting equations 1 and 2 equal to each other gives:

rt = rt - r/5 + 5t - 1
5t - r/5 = 1
(9r - 5)/40 - r/5 = 1
(r - 5)/40 = 1
r - 5 = 40
r = 45 miles/hour

Which means t = 2 hours and d = 90 miles.

We can do a quick check for r by setting equations 1 and 3 equal to each other:

rt = rt + r/4 - 5t - 5/4
r/4 - 5t = 5/4
r/4 - (9r - 5)/40 = 5/4
(r + 5)/40 = 5/4
r + 5 = 50
r = 45 miles/hour

The point is, Joe probably wishes he lived closer to the airport.

Posted by Jyqm on 2006-06-27 07:05:46