I imagine this post is going to appeal to very few, so if you find yourself yawning do not feel bad. I like to compute arithmetic in my head, and it is always a source of entertainment while driving. I do not have any special aptitudes, so a parallel exercise is to find tricks. This is one ... I drove to the store today at 130 mpg, filled up with petrol and reset the mpg meter. Return home was 50 mpg. The combined MPG was a little over 72 mpg. I'll disclose how this can be computed without paper, pencil or calculator in a subsequent post -- after giving everybody a shot at finding this trick, or better ones. The technique used by the way is adapted from a method my son's Algebra I teacher taught his students 10 years ago. Hat is off to Mr. Prince.
If it were me doing this in my head... first, I'd drop the trialing 0's, remembering to multiply by 10 at the end - so 13mpg and 5 mpg. Next, some simple multiplication - 13x5 = 65. So, lets just say the distance traveled was 65 miles each way. That means you used 5 gallons going, and 13 gallons returning - a total of 18 gallons of gas, over a total of 130 miles. 130/18 = 7 R4. 4/18 =2/9=0.2222. So we have 7.2222... now remember those trailing 0's we took off at the beginning? Multiply by 10 and you've got 72.222 as the answer. All told, you do a single multiplication, a single addition, and a single division. Removing the trailing 0's at the beginning and adding them back on at the end just makes that easier (who wants to keep track of numbers like 6500 and 130 when you don't have to?)
I get as far as (2 * 130 * 50) / (130 + 50) and then do the arithmetic in my head. Others might have a different algebra vs arithmetic balance point.
My expression ends up like Eagle's, although I think of it a bit differently: x = fuel to go from A to B at 130 mpg; So 13x/5 is fuel from B to A. Round trip fuel is 18x/5, or 9x/5 to go one way. This leaves (5*130/9) to calculate, 650/9 or 70 and 20/9ths MPG.
Wouldn't you just add 130 to 50, then divide by 2? :huh: 130+50= 180 180 / 2 = 90 Unless you went different distances for each way (different routes). Or, unless your 130 was on the entire tank, in which case, I'm not playing.
Bisco: ScanGauge Rae: Nope. Not very intuitive, is it ? Consider the numbers x, y, and a number z halfway between x and y. x/z does not equal z/y
Not quite, although that is a very common mistake. Since the distance is the same each way, what you really care about is how much gas is used each (gallons per mile) way. 130 MPG is actually 130 miles / 1 gallon - to find out how much gas is used, you would formally do it like this: X miles * (1 / 130 MPG) = X/130 gallons X miles * (1 / 50 MPG) = X/50 gallons. Lets assume X=1 mile, just to get rid of the variable. For the total amount of gas used, you have to add them... but first get common denominators: 50/(130*50) = 50/6500 130/(50*130) = 130/6500 50/6500 + 130/6500 = 180/6500 gallons Now, we need this back in MPG, right? So take our total distance traveled (we said X=1, so each direction was 1 mile - total distance traveled was 2 miles) and divide by the amount of gas used: 2 / (180/6500) = 2*6500/180 = 72.22222 The math becomes odd because we use Miles Per Gallon, and not Gallons Per Mile. It's the same kind of problem you see most often with speed - a guy drove 20 miles to work going 45MPH, and 20 miles home going 60MPH. What was his average speed? Everyone hears that and knows immediately it's a "trick question", since the answer isn't (45+60)/2. But most people only know it's a trick question because they had to solve it back in school... not necessarily because they understand the logic behind it.
Here is another way to see the problem of adding the two MPG numbers and saying the average is half that sum: I drive 30 miles at 30 mph, and then drive 60 miles at 60 mph. My average speed is 45 mph. But this is different than driving 30 mph for 30 miles and then driving back at 60 mph. In the first case I drove 90 miles in two hours while in the second case I drove 60 miles in 1.5 hours = 40 mph average.
Can one of you quote the math rule that 'splains why you can't average mpg like that? I instinctively know you can't average two rates like that but it's been so long since I took Algebra that I don't remember the rule.
If the elements of the set you are averaging have different weights, the weights cannot be ignored. Or something like that
The rule of common denominators. You're dealing with Miles / Gallon... your miles stay the same in both directions, but your gallons change. It helps to understand if you assign some arbitrary distance - lets say 5 miles: 130mpg going there, 50mpg coming back, 5 miles traveled in each direction. MPG = total miles traveled / total gas used We know the total miles traveled is 10 for this example... but how much gas was used? 5 miles / 130 mpg + 5 miles / 50 mpg Since we now have our mpg's on the bottom, we have to use common denominators to do the addition: 250 miles / 6500 mpg + 650 miles / 6500mpg = 900 miles / 6500 mpg = 9/65 So, we can plug that in now: 10 / (9/65) = 650/9 = 72.222 Taking that whole process into a single equation: Code: 2 * distance per leg ------------------------------------------------ distance per leg distance per leg --------------- + --------------- mileage for leg 1 mileage for leg 2 = (This is the common denominator step!) 2 * distance per leg ----------------------------------------------------------------------------------- distance per leg * mileage for leg 2 distance per leg * mileage for leg 1 --------------------------------- + ---------------------------------- mileage for leg 1 * mileage for leg 2 mileage for leg 1 * mileage for leg 2 = 2 * distance per leg ----------------------------------------------------------------------------------- distance per leg * mileage for leg 2 + distance per leg * mileage for leg 1 ------------------------------------------------------------------- mileage for leg 1 * mileage for leg 2 = 2 * distance per leg * mileage for leg 1 * mileage for leg 2 ----------------------------------------------------------------------------------- distance per leg * mileage for leg 2 + distance per leg * mileage for leg 1 = 2 * mileage for leg 1 * mileage for leg 2 -------------------------------------------- mileage for leg 2 + mileage for leg 1 The key is simply to understand that you need to break it down into miles traveled and gas used, then recombine them to make MPG. At least, back in school that's the process my teacher would have wanted to see in order to get full credit for "showing my work"... even though I'd write down the answer first!
If I drive 50 miles for one hour and then drive 100 miles for one hour, my average speed is 75 mph. If I drive 50 miles for one hour, and then 500 miles for 5 hours, my average speed is not 75 mph -- even though I averaged 100 mph for the second part of the journey. This is intuitive, since I spent a lot more time at 100 mph than at 50 mph. The original problem defined equal distances, but these are not equal times. The times are weighted.
