"Top five percent of the world's population"? That's top 0.5 percent.  I am suspicious that maybe there is no right answer to this question, and that it's only there to throw you off. I looked at it for some time and couldn't come up with anything.

Maybe a new reader can give this a try. I haven't found anyone who can figure it out (although I have to admit, I haven't found anyone who was interested in trying to figure it out after looking at it for about 10 seconds).  Is the answer C?  I think the answer is E. I think the rules are as follows:

- No circle can be chosen more than once;
- The next circle should be the one with the least inner circles;
- If the choice of no. inner circles is the same, then the next circle should be the one linked to the fewest lines.


So, starting at J, we have the option of either K, E or I. E is immediatedly discounted as it has an inner circle. That leaves K and I, neither of which have an inner circle. K is lined to 5 lines, I to 2, so we select I.

By the time we get to A, there are five choices : F, B, C, D, E. We've already been to F, so we can't choose that one again. B and D both have two inner circles, so we discount them. That leaves C and E. C is linked to 7 lines, E to 6, so the final choice is E.

Of course, I could be wrong, but that's how I see it. Let me know what you think!  ... but not to tricky. Got 166 on this one, though I have to admit that I do perform worse on timed tests.  I think the answer is C. The answer E is almost reasonable to me, but I think that there is to much chance involved. I think that there would have to be a pattern. And there is a pattern in the row of letters at the bottom and the next in the pattern would be C. That is just my geuss. I'm not saying that that this is definitely correct, or that anyone else is wrong. This is simply my answer.  I think the answer may be B, because for the chain to complete the rest of the circles, without having to go back over a previous circle; so, the chain goes from A to B to C to D then E and finally back to the J . It's the only sequence whereby the path can be completed , without having to digress.  From the given decision at J, we determine that o-o is preferred to 0- and -.
From I, - > o- & outweighs number of circles.
From H, O > 2*O
From G, O > 3*O
From K, (- + -o) > -o
We need to assume a letter cannot be revisited, otherwise G>F
From F, (o- + o- + o-) > - or -o, so a reasonable consideration is that the value of lines is additive.
Then C & D are ruled out since (- + - + -) > (- + -) and lines outweigh circles.
E has fewer circles, therefore is preferred to B.  lets start with "J", some of the lines that connects a letter with another have a circle at the end of the lines, as you can see, the one linked from "J" to "I" is the only one that has a circle on both ends, so we're going to use that as our first step.
From "I" to "H" the line that connects the letters do not have a circle on it and it had add an inner circle to the letter "H"
the following it's from "H" to "G" which has add a circle to one of the ends facing the opposite way we're going,
from "G" to "K" it has add a second line with no circle on it and to the right of line added before.
on the next step the lines that connect "K" and "F" are two as well but change positions and direction and add an inner circle to "F" also
then we have "A" which is the center and final but the beginning of our headache, it is connected with "F" as a final connection on the question, a new line has been added to the sequence which now makes it three that have change directions again...

From here if we were to finish not only to the next letter from "A" but all the sequence, we need to see back to what we have already done and think. We have ( B,C,D & E) we need to follow the same sequence we did in the beginning, with the circles and lines but backwards.

ANSWER:
From "A" to "B" it has 3 plain lines, that are followed by two plain lines that connects "C" with "B"
From "C" to "D" it has two lines which one them has a circle going the opposite way we're going,
From "D" to "E" it has a single line with a circle on the end, but it had change directions, same as "K" and "F" the one sequence before finishing to "A"

but Like Terry said, I COULD BE WRONG...

I know it's not the best explanation of the procedure, but I'm just Mexican... jejeje, if you want a better explanation I could do it but in Spanish LOL...  1) I think that there is a formula, calculating the cost traveling from X to Y:
1 * the count of circles around the letter at Y
+
1/3 * the count of lines connecting X with Y that don't have dots at the Y's end

2) We start from J and choose cheapest connected letter.
3) We don't use letters we have already been.

Based on these 3 simple rules:

We are at J.
J - K: 0.33
J - I: 0
J - E: 1.33

From I:
I - D: 2.33
I - H: 1.33

From H:
H - C: 1.33
H - G: 0.33

From G:
G - B: 2.66
G - K: 0.66

From K:
K - F: 1.33

From F:
F - B: 2
F - A: 1
F - E: 1.33

From A:
A - B: 3
A - C: 1.66
A - D: 2.66
A - E: 2

We finally move to C

contacts: FireSt@Gmail.com please poke me with the URL on replays.