[morgaine300]

Interchanging them at what point in time?

[Unknown008]

Well, there's no more information. I guess taking interchanging whenever we like.

[morgaine300]

Oh, I get it now. The maximum... i.e. if you change them at a time to achieve the maximum. Duh. Well, now I have to think again.

[Unknown008]

Lol! Take your time. A more toughie afterwards ;)

[morgaine300]

It's 48,000. I can tell you how I figured it out, but it was a lot of trial and error. When they get to 24,000 km switch them. I have no idea how to set it up in some equation to just get there. It took me a while to think through how it works, let alone how to make an equation.

The longer you leave them on, the more switching the back to the front will increase their mileage, but the front ones switched to the back will decrease. 24,000 is the point at which they both get exactly the same, If you switch them there, they each get another 24,000.

I had a feeling it was correct when they both came out to 48,000, the same number. But I did check switching at like 23,500 and 24,500 just to be sure.

What I actually did in my trial & error was thus:
Just using 20,000 miles as a starting point... er, kilometers... so how do you say mileage then? :confused:

The tires on the front at 20,000 will be halfway worn. If I move those to the back, they will have half their life left. But on the back, half the life is 30,000. For a total of 50,000. The ones on the back at 20,000 are 1/3 worn, leaving 2/3 life. Putting those on the front, 2/3 life is now 26,667 for a total of 46,667.

If we make it 30,000... That's .75 on the front, leaving .25 life. Move to the back with .25 life and that's 15,000 left, for a total of 30,000. (That one goes down as you increase the life before switching.) Then 30,000 on the back is .50 so switch to the front and you have .50 left and that's 20,000, for a total of 50,000. (That one increases as you go further cause it's taking advantage of the life on the back first.)

I just kind of tried a few intervals and plotted it on a graph and saw what happened. So I was able to home in closer. When I got to 25,000 I went to 26,000 and realized it got worse. So I went down to 24,000, and bingo, they both ended up with the same life at 48,000, which was the most I got.

So now it's up to you to make an equation for that. I can maximize something with a set of parameters, but have no clue how that's done when switching something off like that.

Oh, that's interesting. At 24,000, the front tires are .6 worn and the back .4 worn. The exact opposite proportion as the total km. That has to mean something, doesn't it?

[Unknown008]

Oh thanks! I found the ratio 2:3, I knew I had to make the front worn to some extent so that the ratio is inversed, but I somewhat messed up and came up with 2/3 worn at the front and 3/2 worn at the rear... I realise I had to use the total of ratios, 2/5 and 3/5 :o Thanks! :)

One I started, not yet done. The exams are taking my time...

Each face of a solid cube is divided into four as indicated in the diagram. Starting from vertex P , paths can be travelled to vertex Q along connected line segments. If each movement along the path takes one closer to Q, the number of possible paths from P to Q is
(A) 46
(B) 90
(C) 36
(D) 54
(E) 60

[morgaine300]

Yeah, I started it too... then got stuck. I got an answer that isn't a choice, realized what I screwed up, and now can't figure out how to fix it. But I will get it -- it shouldn't be this difficult, darn it.

[Unknown008]

I have to get some more time to think about it too :p Cya I have to go now :)

[galactus]

Oh thanks! I found the ratio 2:3, I knew I had to make the front worn to some extent so that the ratio is inversed, but I somewhat messed up and came up with 2/3 worn at the front and 3/2 worn at the rear... I realise I had to use the total of ratios, 2/5 and 3/5 :o Thanks! :)

One I started, not yet done. The exams are taking my time...

Each face of a solid cube is divided into four as indicated in the diagram. Starting from vertex P , paths can be travelled to vertex Q along connected line segments. If each movement along the path takes one closer to Q, the number of possible paths from P to Q is
(A) 46
(B) 90
(C) 36
(D) 54
(E) 60

Think of it as a 3-dimensional coordinate system. We have 2 in the x direction, 2 in the y direction, and 2 in the z direction. This is the number of ways we walk exactly 6 blocks from P to Q.

[Unknown008]

If I'm getting what you mean, this means that I have six vector directions:



All those twice. That makes 6! Now, each one is repeated twice, hence divided by 2!

Is that the "theory" behind it?

[galactus]

Yes, you can think of it that way. Think of x,y,z coordinates.

[Unknown008]

Ok, thanks! I'll try find more I'm stuck...