Raven's Matrix

Anyone willing to help me solve this puzzle? I have a hunch that the answer is 5, but the method I used to come to that answer doesn't account for rectangles.

I would pick 5 because the number of horizontal and vertical rectangles are the same in each row

You're right, I didn't even notice that. I got 5 because the number of thin lines and thick curved lines are the same in every row.

it's gotta be 2,a pattern with a thick line can't appear twice on the same line

Not sure, I noticed the pattern between the number of horizontal and vertical lines in each row.

horizontal:
1 3 2
3 2 1
2 1 ?

vertical:
2 3 1
1 2 3
3 1 ?

so it must be 2 vertical lines with 3 horizontal, which is correct for 5. But what about the boxed and curved lines? Can't find a pattern yet..

It's 5. Each row contains a certain number of thin lines oriented either straight or sideways, a certain number of thick curved lines that are oriented either straight or sideways, and an equal number or horizontal and vertical rectangles. In row 3, we that one horizontal thin line is intersecting a vertical thick slanted one. We also see two thicc vertical slanted lines, and three vertical rectangles. This means that our answer must contain three horizontal rectangles, and two vertical thin horizontal lines. Pretty tough problem, not sure if I'd be able to solve that in a minute

Is this a troll thread? That is pretty much a medium level problem.

No, I'm not really sure why I struggled so much with it. I've solved much more complex matrices with relative ease. Maybe I overthought it, idk

5 is correct.

Your logic doesn't eliminate 4,5,8.

Your reasoning is good. You're explanation is poor. Can you further elaborate how you know what the pattern is? Your explanation leaves to misinterpretation.

its 5.

you got in the horizontal and vertical:
-1 and 2 rectangles
-1 an 3 lines
-1,2,3 curved lines

So you are missing the 2 line vertical and the 3 horizontal rectangles, and ta-da, its 5. It is that easy.

In each row: the number of vertical rectangles must equal the number of horizontal rectangles, the number of vertical thin lines must equal the number of horizontal thick curved lines, and the number the number horizontal thin lines must equal the number of vertical thick curved lines. By that logic, the answer can only be 5

>says it is 5
>states that he didn't eliminate 5

I don't think you can into logic very well in the first place.

Thanks. This is much more clear

In the second row you can see the pattern

Line -> rectangle -> thick curved

This holds for all rows and cycles through

In the third row, the second box's vertical is a thick curved so it cycles back to line, then the horizontal is a line so it turns into a rectangle. The number of these symbols on the vertical and horizontal is increasing on the vertical, then cycling back to 1 after it reaches 3, and decreasing on the horizontal, cycling back to 3 when it reaches 1.

It's 5

it is 5.
vertical in 1: thickA, lineB, rectangleC
ABC will always be successive in each line, so on the second line, we start with line (B), we know rectangle (C) will follow.
vertical N pattern: 1,2,3 (however it starts on 2 in this case, but will always be successive 123, relative to the number you start at, 3 will follow from 2 if you start with 2 and so on)
The last line will start on 3.
Horizontal: rectangleA, thickB, lineC
ABC will always follow successively, so if a line starts at B, C will follow, and so on.
same with the number.

How is this not just a matter of opinion?

Because the answer is unambiguously 5

It is 3.

For the figures with ( there is ( and (( and (((,
For the figures with ^ (upward arch) there is 1, 2 and 3 stacked arches.

That leaves 2 figures with horizontal lines, and we have those with 2 and 3, so we are looking for one with 1 or 4. And in Fig. 3 we find 4 stacked horizontal lines on vertical bars.

it's 5, but got it through a different reasoning. Basically it follows a clear pattern of 1,2,3 of each basic draw (rectangles vertical and horizontal, "parentheses", thin lines) in diagonals and triangles (think of a determinant), so it had to be 5.

sorry your a brainlet.

Sorry, you had no argument.

The fuck is this

Then you have three curved lines in a row.

What? Where do we have 4 stacked horizontal lines?

No it isn't. The number has to have a unique configuration of shapes and numbers. Two vertical bars and three horizontal lines were already in left and centre of the second row.

This riddle takes queues from both the horizontal and vertical axis. The horizontal decides the shape of the lines while the vertical decides the numbers.

So the proof goes likes this.
For the shapes, on a horizontal plane.
Bend/fat --> line --> rectangle --> bend/fat.

The number of lines is decided on the vertical plane, in a repeating sequence of 1, 2 or 3.

So the correct answer is 4

>4
My bad, i mend 5.

5. There are 6 total categories of line; thin vertical and horizontal, thick vertical and horizontal and curved vertical and horizontal. Each shape is composed of 2 of these 'elements'. Each category can have 1, 2 or 3 'elements' (so there must be 1, 2 and 3 vertical horizontal lines somewhere.) The only missing 'elements' are the double thin horizontal and the triple vertical thick.

>asking Veeky Forums to help with your IQ test
I got some bad news for you...

If you take each individual shape (regardless of the count) and put them by themselves on a 3x3 table, you can see that all of the shapes are going from 1 to 3 shapes diagonally or in a triangle.
The bottom right is part of the diagonal going from (1,1) to (3,3) and also of the triangle formed by (2,1), (1,2) and (3,3). If you then look at what individual lines these 2 bigger shapes have, you can see that there are 3 horizontal fat lines missing for the diagonal and 2 vertical thin lines missing for the triangle.

Therefore, it is indeed 5.

Raven's Progressive Matrices is just extracting random arbitrary patterns from random shit and hoping yours agrees with the tester's. Culture fair IQ test made by a brainlet psych major.

gg ez

Easiest and most competent answer!
Thank you.