# What's the solution to this IQ Test?

My brother showed me an IQ test from a magazine - one of those “what’s the next in this sequence?” deals. He and his friends had solved 14 of the questions but the fifteenth one was a stumper. I found the first fourteen questions prettty easy, but number 15 was like running into a brick wall.

Here’s the question. The answer is supposed to be3 - the trianglebut none of us have any idea why.

Note that the fuzzy/wavy line thing is an artifact of the image process; all lines are non-fuzzy and of equal thickness in the original.

The only thing that jumps out at me is that on the bottom row of designs, the two parts (circle and line segments) appear to change in a pattern. Row 1 has 2 circles and one line segment, Row 2 has 1 circle and 2 line segments…if we keep that pattern, you get row 3 has 0 circles and 3 line segments. The triangle fits the pattern; it has 0 circles and 3 line segments (attached at the ends)

Going of Flander’s idea:

The bottom row starts with two intersections, then one intersection, then zero intersections (just ends).

The rightmost column starts with five segments, then four segments, then three segments.

I think it’s because #3 doesn’t look like any of the other 8 images.

#1 looks like the top right minus the straight line.
#2 looks like the bottom center with the two elements shifted.
#4 looks like the top left with the two elements shifted and one warped.
#5 is the top left done in squares.

#3 is unique, as are the other 8 images.

This sounds like the only relevant images for deciding on what fits in the ? space are the ones in the same column and the ones in the same row, and that the four images in the upper left are irrelevant. Am I understanding your hypothesis correctly?
Roddy

Just as I suspected. Being intelligent is just too damn much work.

I haven’t taken a test like this in a while, but aren’t all of the figures usually relevant? Aren’t there progressions (horizontal, vertical or diagonal) that hold true for all nine positions?

If we count discrete enclosed spaces using triangle as last answer

Top row Left to right 4 - 2 - 1 or not inc. exterior cube space 3-1-0
Middle row Left to right 8 - 4 - 2 or not inc. exterior cube space 7-3-1
Bottom rom Left to right 3 - 2 - 2 or not inc. exterior cube space 2-1 -1

Hmmm… not so good.

What’s the magazine & the issue?

Another possibility:

Count the number of straight line segments in each square:
0 1 1
0 4 4
1 2 3

Note that we can add columns one and two to get column three:

0+1=1
0+4=4
1+2=3

So far I think that everyone’s looking at the wrong things. The answer has to do with the amount of symmetry in the figures. The 3 figures in the upper left corner have only one axis of symmetry. The level of symmetry increases as you move to the lower right. You can eliminate choices 1, 2, and 4 because they only have 1 axis of symmetry. Between 3 and 5, choice 3 has more symmetry and its symmetry is in line with the figures to the left and above the ? cell.

Something like this might be right, but it’s still a little confusing exactly what they were going for:

Number of axes of symmetry:
1 1 0
1 2 4
2 4 3

That’s not a terribly convincing pattern. Let’s look at the following:

1/minimum fraction of a turn before re-attaining original figure:
1 1 2
1 2 4
2 4 3

That’s a much more “patterny” looking thing, although it’s still not clear why one would be expected to infer 3 and not any other number for the bottom-right.

Hm, although, I was wrong; looking more closely, the actual table is:

1/minimum fraction of a turn before re-attaining original figure:
1 1 1
1 2 4
2 4 3

Also not terribly convincing…

I think everyone is over-analyzing the problem. Since Priceguy said the rest of the problems were easy, the solution to this last one shouldn’t be overly complicated. I think KneadToKnow is right, in that the triangle is the only one that doesn’t share any components with the remaining figures.

The solution is:

14 K of G in a F P D

Obviously.

I don’t particularly like that solution, because, if you’re using the logic to eliminate answers 1, 2, 4, and 5, that KneadToKnow is enumerating in his answer, you can make arguments that certain pairs of figures in the square itself are not unique figures.
The logic is too vague and can be used to justify a number of answers. If answer one is eliminated because it’s just the top right corner, rearranged, with the line segment removed, you can just as well say that the top right and bottom left are the same, with the arcs rearranged into circles. Or you could say the middle top or bottom left are the same figure, with an extra circle. There’s just too much leeway to what manipulation is permitted to prevent an answer from being labeled as “unique.”

I don’t like this question, because there seems to be too many possible ways of arriving at a reasonable answer, although I think there may be something to the symmetry hypothesis, although I’m not satisfied with any of the explanations yet.