I was going to take a trip to my friend's house in San Jose by airplane. If my airplane had arrived on time as scheduled, my friend would drive out to pick me up from San Jose's airport. She would get to the ariport just as I would be leaving it, and drive me to her house. This way, I would come to her house at 12:00.

But as it happened, my airplane arrived at the airport 1 hour earlier. So I walked from the airport directly to my friend's house. On the way to her house, I met her driving to pick me up. I got into the car, and we drove me to the house. This way, I arrived at the house at 11:40.

How long did I spend walking? (The only thing known about my walking speed and my friend's driving speed is that they are different and they are constant.)

All variables are minutes prior to 12:00
x = original plane arrival time
y = friend's departure time
z = time spent walking

Since the friend's travel speed is constant, we know that time to and from airport are equal.

x - y = 0 - x

0 being 12:00

From the actual travel info, we know that

(x - 60 + z) - y = -20 - (x - 60 + z)
(pick up time)

2x = y solves out both x and y and leaves z = 50

There are an infinite pair of x and y values consistent with the scenario, only constrained by realistic bounds on car and foot travel speeds.

I applaud the intuitive method, I wish it would have occurred to me.

Posted by michael on 2003-04-09 06:51:38