IQ Test Question

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H.

And I can't explain why simply, but it fits with the ones to the left of the blank space,
and it fits with the ones above it, in a unique way. I don't recognize the whole pattern.

I determined this by using an ability that makes me always get the high score on picture matching games, but this ability doesn't interpret the information it only finds sameness.

If I am wrong then there is another answer, but only H "shines" for me.

edit:
I guess I guessed lucky. What I saw is that:
H had an X in the top middle like the two to the left of the space.
H has in common three same squares to the ones above as does the two to the left.
None of the others do.

edit2:
oops. My second criteria is incorrect so it is not the one I used.

Last edited by ValMikeSmith on 10 Dec 2008, 3:52 am, edited 2 times in total.

I managed to brute force the answer with several trial tests. It appears to be B.

Filling in the first twelve questions then guessing A to everything else gives me an IQ of 97 (! !) but if I guess B to the last question it is 98.

Now to figure out why

EDIT: Giving up for tonight. I should make an IQ test where your scores are adaptive based off of actual results people make.

I may be posting hastily again but B has one piece in common with what's on the left of the empty space, as does the one to the left of that, and B also has one piece in common with what's above it, as does the one above that. And it fits that binary pattern just like H. I have no other ideas.

Hard to explain but I think it is C. Each of the rows of blocks (top, center, bottom) are consistent within the row, but going from left to right, each block can be turned 90 degrees clockwise or counterclockwise. The 90 degree turned patterns aren't necessarily consistent as to the geometric shapes, but are consistent as to the pattern of shapes i.e. 2:1, 1:1:1, etc. The tricky part on the bottom row of squares is that it doesn't appear to be consistent but actually is because the rows of characters don't stay in the same order, but have consistent patterns within each row.

Given that the left column (a) is consistently 2:1 (pair of "1" and single "2"), turn square 01 counterclockwise to get 02 then turn 02 clockwise to get 03. Triangle=1; Circle=2; X=3.

01 02 03
abc acb abc
133 123 221
122 133 132
231 212 133

Z

If you look at which symbols are the 2 in the 2:1 property of columns, you will find the following pattern:



A and B fit the pattern. It is possible that this pattern is coincidence, but it does include only 2 patterns (one of which is accepted correct).

This is the second pattern I have found in columns of elements, spread across rows of squares. I have had no substantive findings in the rows of elements, so it might be best to try to discern what makes A and B distinct.

Another flimsy pattern that I just recognized is this:

In the 1:1:1 column of elements across rows of squares, the elements are ordered the same. Row one has {TXO} while rows two and three have {TOX}. B fits this pattern as well, but A does not.

Assuming that the 3 rules I put forward here are not flawed or coincidence, that would appear to prove B is the correct answer.

- - -



What did everyone get for this one? It was one of the ones that stumped me. What patterns I could find were not sufficiently decisive to elicit a singular answer.

I narrowed it down to C D F or G (based on the pattern of occurrence of "one of each non-white color in a row or column". It cyclically progresses as {Element-Row-3, Element-Column-3, Element-Column-1} across both rows and columns of squares).

G is also distinct from CDF, look at it in terms of white and non-white.

Some pretty simple rotation rules gives us a pattern of columns of heights 1, 2, and 3 in every square. Weird thing is the order of the columns seems to be stochastic.

The bottom-left-most 3 squares have 2 black, the upper-right-most 3 squares have 1 black. The 2 remaining diagonal squares have 3 black. 3 of the 4 possible answers fit this pattern, DFG. This seems to make D and F the most likely candidates.

Black always has light to its left, dark to its right, after being rotated so that each square forms ascending columns. This rules out D and permits F.

Ha! Talked myself through it. Correct answer is F. Confirmed through similar method as mentioned above.

Last edited by Death_of_Pathos on 10 Dec 2008, 2:49 pm, edited 5 times in total.

I don't think any more than 10 or 15 minutes tops is permitted for coming up with an answer.

So? You expect me to stop solving puzzles just because I don't get credit for them?

I had kind of eliminated B because none of the squares have three symbols that are alike, either horizontal or vertical. Only one square has a diagonal of 3 Xs.

Z

Just because a specific property did not occur in the first 8 iterations of a rule does not mean that the property can not occur in the ninth.

Three solid rules share only one possibility, B. That combined with the fact that the test-accepted correct answer is B (I brute forced it) makes me pretty certain that it is B.

What about this question?

There doesn't seem to be some sort of correlation here to solve this in my mind.
But if you say B, I'll just oblige. LOL.

I say it's H now DoP.

For problem 39:

0: There are three of each element in each square.

1: Across rows of squares, there is always exactly one column which has one of each element.

2: This column of elements has a predictable position. It is always positioned one column to the right of where it was on the previous square (going left-to-right across rows of squares)

3: This column of elements has a predictable ordering to its elements. It is always the same across rows of squares. The elements are not necessarily in the same position, though, but if a Triangle is positioned above a Circle in this column of elements in the first square of a row, a Circle will never be positioned above a Triangle in the other two columns of elements in that row of squares.

4: The other two columns in any given squares, have 2 of one type and 1 of another. By necessity the element represented only once per column is the same in both columns. Which element is present in both columns is dependent upon the square previous to it in that row of squares. Each of the three elements is represented once per row of squares in this manner. (or columns of squares)



Make sense now?

- - -



Thanks. I went ahead and 'looked up' the answer (it is H), but I haven't figured out the why yet.

All the dots move relatively one step in a linear fashion per phase of puzzle, notice the checkerboard adding itself everytime as well. Then notice that the dots change their colour every couple of steps; you could figure it out in your head.

So they do. Every time they occupy the same space as a square. I was making that way too hard, because I assumed that they would have the same color.

I enjoy these sort of puzzles everynow and then. Outta tests, I'd always score almost perfect on this sort of thing, that and trivial facts containing history, though math is not my strong point, eh eh.