Can someone solve this question?

[Danny X]

I can't solve it. Please explain how you solved it.

 

[Judge]

Set "A". Assuming of course this is about pattern recognition. Based on a diagonal pattern of each set/box of card suits where there are two of the same suit in each set with one other suit showing that are evenly distributed. Relative only to the top and bottom, left, right and center sets of card suits.

In the top left set, going diagonally I see two hearts and one club. In the center suit I see two diamonds and one heart. So in the bottom-right set of suits, it shows two clubs and one heart.

Conversely from the bottom-left there are two hearts and a club. In the center there are two diamonds and a club. And in the top-right set there are two clubs and one heart.

If there were other patterns I didn't catch them. But any time I see nine sets of something it usually means identifying something in a diagonal fashion.

 

[Danny X]

Last question I couldn't solve. Can you solve? @Judge

 

[Judge]

Last question I couldn't solve. Can you solve? @Judge

This one seems more difficult. Geez...lol. Pffffft...I can only guess at this one.

That there's two things to consider. The combinations of little black triangles that form an optimal number of triangles with different shapes. Patterns within patterns. However there's still one pattern that manages to repeat itself in the original set.

If so, then I chose "H". Because the combination of little black triangles forms a unique triangle not part of the existing eight sets. Plus it allows for an "O" diagonal win, to balance out the one "X" diagonal win.

LOL...naw...it's gotta be something else.

 

[FreeDiver]

Last question I couldn't solve. Can you solve? @Judge

I'm gonna guest "E" on this one.

 

[WildCat]

All I know for sure is that the pattern is rotated clockwise at the end of each row, but I can't figure out the rest of the logic right now. I took the same test with this same final question (iqtest.dk) and that's the one that stopped me in my tracks.

 

[Danny X]

I'm gonna guest "E" on this one.

No.

 

[xudo]

Last question I couldn't solve. Can you solve? @Judge

F?

 

[tlc]

This one seems more difficult. Geez...lol. Pffffft...I can only guess at this one.

That there's two things to consider. The combinations of little black triangles that form an optimal number of triangles with different shapes. Patterns within patterns. However there's still one pattern that manages to repeat itself in the original set.

If so, then I chose "H". Because the combination of little black triangles forms a unique triangle not part of the existing eight sets. Plus it allows for an "O" diagonal win, to balance out the one "X" diagonal win.

LOL...naw...it's gotta be something else.

The first one I'm still trying.

For the 2nd one, here's what I come up with (hope I looked at it correctly):

The only thing every set in the 1st row has in common, is X on the bottom left corner.

The only thing every set in the 2nd row has in common, is X O triangle on the top row.

The only thing every set in the 3rd row so far has in common, is X in the middle top row.

So to keep the pattern, the only choice that has that, is H. So I also choose H, for a different reason.

 

[Danny X]

The first one I'm still trying.

For the 2nd one, here's what I come up with (hope I looked at it correctly):

The only thing every set in the 1st row has in common, is X on the bottom left corner.

The only thing every set in the 2nd row has in common, is X O triangle on the top row.

The only thing every set in the 3rd row so far has in common, is X in the middle top row.

So to keep the pattern, the only choice that has that, is H. So I also choose H, for a different reason.

No.

 

[Danny X]

F?

No.

 

[tree]

No.

No.

It would be interesting to read your reasons
why those answers aren't correct.

 

[Danny X]

It would be interesting to read your reasons
why those answers aren't correct.