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.
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.
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.