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!linus!philabs!cmcl2!seismo!brl-tgr!ron From: ron@brl-tgr.ARPA (Ron Natalie ) Newsgroups: net.puzzle Subject: Re: a good interview question (the solution) Message-ID: <9276@brl-tgr.ARPA> Date: Sat, 16-Mar-85 20:49:31 EST Article-I.D.: brl-tgr.9276 Posted: Sat Mar 16 20:49:31 1985 Date-Received: Mon, 18-Mar-85 03:40:57 EST References: <302@ssc-bee.UUCP> <463@petsd.UUCP> <807@loral.UUCP> Organization: Ballistic Research Lab Lines: 34 > 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? Your average speed would be: Distance Traveled (miles) -------------------------- Time Taken (hours) which is to say (First Mile) + (Second Mile) average speed = ---------------------------------------------- (Time for first mile) + (Time for second mile) Putting in some numbers 2 60 = ------------------------------------------------- (1/30) + (time for second mile) since the total distance is two miles, you want the average speed to be 60 mph and it took you 2 minutes (1/30 of an hour) to do the first mile. Solve and find that (time for second mile) = 0 Which means you must travel one mile in no time at all. -Ron