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Dear Mitch,

A few weeks ago you put up a Thanksgiving word problem about someone who drove to a Thanksgiving dinner at forty miles per hour and then immediately turned around without having dinner and drove straight back at 60 miles per hour. And you asked what the driver's average rate of driving was? It was multiple choice, but you warned against going immediately for the obvious answer.

The choices were:

A) 45

B) 48

C) 50

D) 52

E) 54

I was wondering, now that the holiday is actually here, can you reveal the answer and explain it a little?

Thank you,

Doug Craft,

Bethesda, MD



Dear Doug,

Of course!

First, as you might guess, the most commonly chosen wrong answer is 50 miles per hour. Why? Because people simply average the two rates of movement. BUT, this famous problem is a little bit trickier than that.

It asks about the average rate of movement, and here you have to think how that is figured. Basically, it is the rate of travel per hour or per minute. HOWEVER, here you do not have the same number of minutes being traveled at both rates. Even though the same distance is being covered, as is usually the case in a 'round trip,' you have to consider that more time is spent driving at the slower rate. After all, when one travels at a slower rate than a fast rate, he or she takes more minutes to get to the same destination. So, since more minutes are spent traveling the slower rate, this becomes what is called a 'weighted average' question. (Think of an obvious example: If you are told there are two teams in a game of tug-of-war, and one team has members who have an average weight of seventy-five pounds and the other team has members who average fifty pounds, one's first inclination is to bet on the team whose members average seventy-five pounds. BUT, one might wish to ask how many members there are on each team. If the fifty pound member team has one hundred members and the seventy-five pound person team has only ten members, then it is not a straightforward average question but is a weighted average question. In other words, one has to account for more characters at one of the weights than the other and average in each and every character involved...

Here, since there are more minutes of 40 miles per hour than 60 miles per hour, we know immediately that the 'average' has to be less than halfway between the two. That means less than fifty miles per hour. So, in this multiple choice question we can immediately cross off fifty miles per hour, fifty-two miles per hour, and fifty-four miles per hour, which were choices C, D, and E. That leaves choice A and B.

By the end of Thanksgiving I will give the final answer, but until then, one last cliffhanger on this famous question: Which of those two choices is correct?

HINT: Pick a certain distance for the trip and see how many hours it will take at each rate, then plug in the rate that each hour is traveled, to and from. And remember to pick a nice number that works easily with forty and sixty, without worrying about being 'realistic'!

Until tomorrow... or, at least tonight, good luck!

Happy Thanksgiving!

Hope this helps,





NOVEMBER 23, 2007:

Here's the final answer and how to get it:

As I mentioned, pick a number for the distance, without worrying about picking a realistic one; rather, your focus should be on selecting a number that is easy to try with your possibilities.

Let's say the distance from the start point to the finish is 120 miles, which would make the round trip 240 miles. One way the driver drove 40 miles per hour, and the other way 60 miles per hour. So, one 120-mile trip at 60 miles per hour would take two hours, and the other 120-mile trip at 40 miles per hour would take three hours. So, to calculate miles per hour you have three hours at 40 miles per hour, and 2 hours at 60 miles per hour:

40 + 40 + 40 + 60 + 60 = 240 miles traveled (seems correct), and it was traveled in a total of five hours. So, 240 divided by 5 =

240/5 =

48 miles per hour. That is the average of all five hours, and that is the correct answer, which was choice B.

Safe travels on the Holiday!


- Mitch