Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.2 9/18/84; site brl-tgr.ARPA
Path: utzoo!watmath!clyde!burl!ulysses!allegra!bellcore!decvax!genrad!panda!talcott!harvard!seismo!brl-tgr!gwyn
From: gwyn@brl-tgr.ARPA (Doug Gwyn <gwyn>)
Newsgroups: net.puzzle
Subject: Re: a good interview question (the solution)
Message-ID: <9289@brl-tgr.ARPA>
Date: Sun, 17-Mar-85 14:10:01 EST
Article-I.D.: brl-tgr.9289
Posted: Sun Mar 17 14:10:01 1985
Date-Received: Wed, 20-Mar-85 03:52:37 EST
References: <302@ssc-bee.UUCP> <463@petsd.UUCP> <807@loral.UUCP>
Organization: Ballistic Research Lab
Lines: 17

> >>	You have to travel from town A. to town B.   The distance
> >>	between the two towns is 2 miles.  If you drive 30 mph for the
> >>	first mile, how fast do you have to drive the second mile in
> >>	order to average 60 mph?

> Question:  The original question says nothing about time.  Why can't you go
> 30mph for the first mile and 90mph for the second mile, therefore averaging
> out to 60mph.  If you are driving along going 30mph for 1 mile and then
> 90mph in the next, you are averaging 60mph for the 2 miles.  Am I missing
> something here?

Yes.  Just because you can add two numbers and divide by two does not
mean that you have thereby produced a physically meaningful "average".
You might as well say that the "average distance" is (1 + 1) / 2 = 1
mile!

Average speed = Total path length / Total travel time