The other thread here reminds me that I have had this question for many years. Several puzzles relate to the issue and I’ve never known enough about math to figure it out. E.g. - a person travels for three hours at x mph. Then that person travels for four hours at y mph. Is it possible to derive an average just using x and y? Or would you have to calculate total distance and total time and then calculate the rate?

Well, you could do it that way, and any other way to get the same result will (tautologically) be equivalent to doing it that way. You certainly can’t get the answer *just* using x and y; you need to use some of the information contained in the “three hours” and “four hours” as well.

Specifically, if you already know the speeds x and y, the additional information you need to know, one way or another, is that 3/7 of the total time was spent at speed and 4/7 of the total time was spent at speed y. This lets you conclude that the average speed over the entire journey was 3/7 * x + 4/7 * y (in the sense that this is the total distance divided by the total time). It doesn’t matter if it was specifically 3 hours at x and 4 hours at y, or 6 hours at x and 8 hours at y, or 30 hours at x and 40 hours at y, but as long as you know what proportion of the total time was spent at each speed, you can take the appropriately weighted average.

Nevermind. **Indistinguishable** got it.

Certainly, that’s something you could call an average of the two speeds, **DCnDC**, but it won’t be the same as the total distance/total time, which is what most people would call the average speed over the whole trip, and presumably what the OP was looking for. Specifically, total distance/total time is always the *time-weighted* average speed, whereas (speed1 + speed2)/2 is clearly just the average giving both speeds equal weight.

On edit: **DCnDC**’s post disappeared, so none of this reply makes sense anymore. But I’ll keep it up as a warning to others.

If you know how much **time** is pent at each speed, it’s relatively easy to give a weighted average. For example, A% of the time at x speed, and the rest of the time at y speed, gives:

A% * x + (100-A)% * y

However, typically the problems are given in **distances** travelled at each speed. That means that either you work out the time spent at each speed, and use the formula above, or you just work out total distance and total time, which will generally be easier.

Weighted averages almost always result in more useful results than simple averages.

And as an example as to why it’s always good practice to quote what you’re replying to, even if you expect to be the very next post.

**Indistinguishable**, I posted, then realized I forgot to factor in the 3 and 4 hours, and when I went back to edit I saw you had answered the question pretty thoroughly so I just erased my moot response.

The original puzzle, one that’s familiar to many, was this: A man travels a mile at 30 mph. How fast must he make the return trip in order to average 60 mph for the whole trip? The initial impulse is to say 90 mph, of course. But the answer is zero, because a two mile trip at 60 mph would take two minutes, and those two minutes are shot in the first leg of the trip. However, I don’t really know how to solve that problem mathematically.

No problem, **DCnDC**; I figured that’s what happened.

The reasoning you gave *is* a mathematical solution to that problem. It correctly establishes that the answer is that the return trip must be made instantaneously. Any correct way of establishing the answer is, of course, correct.

(Note, lest any confusion arise, that the answer is that the return trip must be made in zero time, which is of course a very different thing from zero speed.)

A huge hint, which a lot of not-math-people overlook, is hidden in the words themselves. Remember, “is” means “equals”, “of” means “times”, and “per” means “divided by”. My 7th grade math teacher used to chant “Of means multiply, per means divide! Of means multiply, per means divide!”

You could say “My speed is X miles per hour”, right? So you can translate those literal english words directly into an equation: speed = miles / hours, aka s= d/t.

From there, you get more useful equations:

t = d/s

d = st

So you said a person goes “3 hours at X” and “4 hours at Y”. Those are Ts and Ss. So we can find the Ds and figure some stuff out:

Average speed for the trip is total d / total t, or d1+d2/t1+t2. Using substitution, we can replace d1 and d2 with their equivalent above, and get:

st1 + st2 / t1+t2

Subbing in the X, Y, 3, and 4 in the appropriate place, we get:

3X + 4Y / 3+4 = (3x+4Y)/7

And there’s your answer. You’ll notice that you can split this fraction into the same one mentioned above. I just took you the long, “prove it” way of getting there.