I just sent a puzzler idea to Car Talk and since they probably won't use it (too simple, perhaps?) I thought I'd give you guys a crack at it. No, I don't mean that I think you guys are simple minded... It's just a fun little mind-teaser.

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Yesterday I drove my new car, a Honda Fit, from San Jose to Reno and back, a round-trip distance of 500 miles. On the outbound leg, according to the mileage gauge I got 43.5 MPG. I didn't reset the gauge for the return trip, and when I arrived home the gauge showed an average for the entire trip of 46.5 MPG. The difference between outbound and return trip is because it is mostly uphill going out, downhill coming back. (Reno is a mile higher than San Jose)

So here's the question. If I got 43.5 MPG on the way out and 46.5 overall, what was my MPG on the return leg?
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Try and work it out before you look at the spoiler. It is not difficult math, but you do need to know how to approach the problem.



These are real numbers by the way - I really did make the trip, and with two traverses over Donner Pass (elevation ~7300 feet) 46.5 MPG average is respectable. Speed (cruise control) was 60 MPH with occasional bursts to 70 or 75 passing trucks (I needed to get out of the passing lane quickly to avoid 11 MPG SUVs that wanted to pass me), and occasional periods of 0--10 MPH in traffic jams. Overall average speed for time spent on the road was 59 MPH. (9.5 hours door to door with an hour of down-time in Reno. Purpose of trip was to pick up an Ekornes recliner chair purchased on eBay.)

tanstaafl.

When I looked at the question, I thought to myself "about 50 mpg" before looking at the answer.

It's basically the number that along with 43.5 will divide by two to equal approximately 46.5.

Key word there being "approximately". Your solution gives an incorrect result, although it does indeed provide a quick and reasonably accurate (+/- 2%) answer. Nicely done.

However, if the variance in the MPG on the trip out and the overall trip had been greater, your method would yield less accurate results. For instance, if I had gotten 25 MPG on the out leg, and 40 MPG overall, your averaging method would have had considerable error. The correct answer would be 82% higher than the answer your method provides.

tanstaafl.

I'm sorry, my math must be really off, but how is the answer to this one not 55 MPG? Isn't the total distance the key? If the distance there and back was the same, wouldn't Bruno's math work?

This is really just another version of the old puzzle which uses MPH instead of MPG to ask essentially the same question: if I averaged 25MPH on the way there, and want to average 50MPH for the whole journey, how fast do I need to go on the way back? (Like your question, the answer is actually independent of the distance.)

If you use "litres per 100km", like in continental Europe, the figures do average linearly as per hybrid8's reply.

Peter

The difference between 43.5 & 46.5 is 3. Add that to 46.5 for 49.5 on the homeward leg. It really is too simple.

Except you're wrong. Same mistake/estimation Bruno made.

Was that a joke? Because it's not true. It doesn't really matter what the units are. The issue is that, despite the fact that you're travelling the same distance each way, the fuel midpoint occurs a little before you reach your distance midpoint. (Or after, if the fuel efficiency is lower on the return trip.)

No.

25 MPG on the way out would have used exactly 10 gallons of gasoline to get there. [250 miles per 10 gallons = 25 miles per one gallon]

40 MPG for the whole trip would have used exactly 12.5 gallons of gasoline. [500 miles per 12.5 gallons = 40 miles per one gallon]

Using 10 gallons on the trip out, and 12.5 gallons total, that would leave just 2.5 gallons for the 250 miles return trip, requiring 100 miles per gallon, not 55.

As Peter pointed out earlier, it's like the classic, non-intuitive average speed problem: Average 25 MPH for the first half of the trip, how fast do you have to go to average 50 MPH overall? The answer? It can't be done because to average 50 MPH you have to do the entire trip in a certain amount of time, and you've already used up that entire time allotment in the first half of the trip.

So, by analogy, if I'd averaged 20 Miles per Gallon on the trip out, and 40 MPG on the overall trip, 40 MPG still means a total fuel consumption of 12.5 gallons, but I would have used all 12.5 of those gallons in the first half of the trip, requiring an infinite MPG for the return. Maybe if the return was all downhill...

I'll let Peter and Bitt fight it out over the Liters per 100 KM vs. Miles per Gallon concept, but here's a clue: Division is not a commutative function.

tanstaafl.

Um, it is true, because it matters critically what the units are. In particular, it matters whether the distance term is in the numerator or the denominator: as Tanstaafl said, it's about commutativity -- or, put differently, it's because the reciprocal of the sum is not the same as the sum of the reciprocals, but a sum of sums works both ways round.

Try it: I go somewhere 100km away and come back, my fuel consumption on the way there is 2 litres/100km, for the whole 200km trip it's 10 litres/100km. That means I use 20 litres for the whole trip, 2 litres on the way there, must have been 18 on the way back -- for a consumption of 18 litres/100km. And 18 is exactly what simple linear averaging would also have come out with.

Peter

Oh oh oh. I missed that you measure the reciprocal ratio. I was thinking that you were claiming that there was some special relationship between liters and kilometers that doesn't exist between miles and gallons.

Thank you Peter, that was beautifully put. I tried to express that last night, but it was after midnight and my brain was fuzzy. All I could figure out how to say was it was a commutativity issue, so I tossed the ball back into your court.

tanstaafl.

On top of that, Peter was the first to use the word commutativity on the forum. (Since I know you like to track things like that.)

Sure, but I was the first (and so far only) one to use the word "commutative" -- dating all the way back to July 2000.

tanstaafl.

It's exponential because you can only ever (at best) double your MPG on the return leg of a trip. If you use 0 gallons on the return leg, you're doubling your distance, but the total gallons consumed stays the same, and so the MPG doubles.

100 MPG on first leg and infinite MPG on return leg != (infinity+100)/2 total MPG.
It equals 200 total MPG.

Same goes for average MPH problems.

If we're going to be picky, I'll point out that the property we're discussing isn't actually commutativity, it's distributivity...

Peter

Well... yes. But the basis of the discussion in this case centers around the inversion of numerator and denominator (as you pointed out to Bitt), which carries an implicit reference to division and thus the claim of commutativity or lack thereof.

I love this bbs and the chance to engage in dialog with people who are smarter than I am!

tanstaafl.

Yes, Paul (you are Paul, aren't you?) but that is the answer to a question not asked. We already know the overall MPG (that's stated in the question), the discussion is about MPG on the return leg.

Nonetheless, your reply is nicely put.

tanstaafl.