# Distances on line

• Sep 26th 2012, 10:42 AM
algorithm
Distances on line
Hi,

Then multiple choice answers for the attached question are \$\displaystyle 3, 5, 7, 9, 11\$. I get different answers depending on how I arrange the letters. I must be missing a trick!

Thanks.
• Sep 26th 2012, 11:23 AM
johnsomeone
Re: Distances on line
1) Triangle inequality, which here is just an absolute value game.
2) Notice the word "could". That's different than "is always"/"necessarily"/"implies"/etc.. It could be a special cases, but even so, it *could* be a valid solution.
3) Draw things to extremes, and not that only two possible extremes exist in these scenarios. Extremes often clarify things I think - and when you're hunting for what *could* be the case, you need to allow for things to be as extreme as possible.
• Sep 26th 2012, 12:08 PM
algorithm
Re: Distances on line
Thanks, but I've still been unable to get the right answer consistently. Could you please post a solution? At the moment I'm drawing lines to find an answer.
• Sep 26th 2012, 12:16 PM
Plato
Re: Distances on line
Quote:

Originally Posted by algorithm
Then multiple choice answers for the attached question are \$\displaystyle 3, 5, 7, 9, 11\$. I get different answers depending on how I arrange the letters. I must be missing a trick!

Consider the order \$\displaystyle z-x-w-y\$.
• Sep 26th 2012, 05:04 PM
johnsomeone
Re: Distances on line
The fact that you get different answers depending on how you arrange them is the whole point of this problem.

First Hint: Ignore point W. They tell you its distance from X, and then nothing about is used again in the problem. It's completely extraneous.

Second Hint: Put the points on the number line, with X = 0. You could've made Y = 0, or X = 15.1 or several other choices, but this makes things the easiest, for several reasons:
0 is an easy number to compute with, one of your givens is a distance from X (so easier to compute when it's a "distance from 0"), and finally, the thing you're trying to find is a distance from X.

Third Hint: You now have a very few choices - in fact, exactly 4 choices.
Since X = 0, and the distance from X to Y is 4, means that on the number line, Y is either at 4 or Y is at -4. I'm going to "cheat" and put both points on my number line, except I'll call them Y1 and Y2. Thus Y1 = -4 and Y2 = 4. Thus Y could be either Y1 or Y2.
Now we need to do Z. Where to put Z? That depends on where Y is, since all we know about Z is that the distance from Y to Z is 9. But Y could be at Y1 or Y2. Thus you must do it for both cases: Y = Y1 and Y = Y2.
Suppose Y = Y2. Then Y = 4. Since the distance from Y to Z is 9, Z is either 9 to the right of Y, i.e. Z is at 4+9 = 13 on the number line, OR Z is 9 to the left of Y, i.e. Z is at 4-9 = -5 on the number line. Playing the same trick, let's call Z1 = 13 and Z2 = -5 be two points on the number line.
Then Z is one of those two points WHEN Y = Y2 (i.e. when Y = 4). However, Y could also be Y1 (i.e Y = -4).
When Y = Y1, you'll again have two possible Z's, call them Z3 and Z4, found the same way - one is by moving 9 units to the right of Y2, and the other is by moving 9 units to the left of Y2.
These Z1, Z2, Z3, Z4 that you get are the only possibilities given that you chose X = 0. If you had choosen a different X, you'd get different values for those 4 Z's, BUT, the distance between your new X and your new Zs (which it's our goal to find) would be the same as the distance current X (= 0 on the number line) and the 4 current Z's (two of which are Z1 = 13, Z2 = -5). In effect, the structure of those points is rigidly defined by the distances, but where you put that structure on the number line gives you a lot of freedom to slide the whole structure up and down the number line.
Now
Distance X to Z = 13 when Z = Z1 = 13.
Distance X to Z = 5 when Z = Z2 = -5.
Distance X to Z = ??? when Z = Z3 = ???
Distance X to Z = ??? when Z = Z4 = ???

I've left it to you to find Z3 and Z4, and so to fill out the rest of the possibilities. So far the answer 5 on your list is possible.
• Oct 13th 2012, 08:16 AM
algorithm
Re: Distances on line