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DAT Cube Counting Help
Started by xatlasb
There is 1 solo cube in the back.
4 rows of cubes stacked 2 high = 8.
The tower of 3 in the front.
1 + 8 + 3 = 12
4 rows of cubes stacked 2 high = 8.
The tower of 3 in the front.
1 + 8 + 3 = 12

There are five rows of cubes in the back. Four of those rows have two cubes, and obviously, the row on the far left has just one cube.
There is a row in FRONT of the back row that is three cubes high. This 3-cube stack blocks a stack of 2 cubes behind it.
This is a tricky problem. It messes with your eyes.
I'm also lost here. I can see how it could be 12, but that would involve counting cubes that are completely hidden from view. I thought you weren't supposed to assume cubes exist if they aren't directly visible?
Last edited:
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I'm also lost here. I can see how it could be 12, but that would involve counting cubes that are completely hidden from view. I thought you weren't supposed to assume cubes exist if they aren't direectly visible?
Crack the PAT does stuff that is weird. I don't think they do illusions on the test.
I still can't see how there is 12. I mean I saw the answer in CDP but it looks like a different shape, especially when I rotated the answer to the angle it was shown, the lines didn't match up.
It seems like there are only 2 touching the ground, is that even possible for a cube-counting question?
It seems like there are only 2 touching the ground, is that even possible for a cube-counting question?
This is definitely a tricky question. Just thought of this idea and maybe this might be helpful to understand why it can't be 10. I think you have to realize that all of the cubes should be placed upon the same "ground" level. There wouldn't be a cube protruding from the lower end (or side for that matter, unless it is on the ground level) because when it is "painted" the bottoms should be untouched (from the paint). If you have cubes floating and protruding from the side, their bottoms will be painted. I'm not sure if this is the legit explanation, but I that's my best guess.
EDIT: I just saw your post^ and it seems you kind of got the idea of what I was going to say while I was coming up with the words for my post. In short, I believe all stacks of cubes must be touching the ground.
EDIT: I just saw your post^ and it seems you kind of got the idea of what I was going to say while I was coming up with the words for my post. In short, I believe all stacks of cubes must be touching the ground.
I'm also lost here. I can see how it could be 12, but that would involve counting cubes that are completely hidden from view. I thought you weren't supposed to assume cubes exist if they aren't directly visible?
This is a correct assumption. However, there also exists the rule that the figure has to be continuous (e.g., no breaks between cubes). Assuming that the non-visible tower of 2 cubes can't be counted because of the "can't see it, can't count it" rule immediately violates the "must be continuous" rule.
Therefore, the trick here is to know that in order for the visible solo cube + two towers of 2 cubes on the left of and behind the 3-cube tower to be continuous with the cubes on the right, there must be another tower of 2 hidden behind the 3-cube tower.
This is definitely a tricky question. Just thought of this idea and maybe this might be helpful to understand why it can't be 10. I think you have to realize that all of the cubes should be placed upon the same "ground" level. There wouldn't be a cube protruding from the lower end (or side for that matter, unless it is on the ground level) because when it is "painted" the bottoms should be untouched (from the paint). If you have cubes floating and protruding from the side, their bottoms will be painted. I'm not sure if this is the legit explanation, but I that's my best guess.
EDIT: I just saw your post^ and it seems you kind of got the idea of what I was going to say while I was coming up with the words for my post. In short, I believe all stacks of cubes must be touching the ground.
Using the "floating" perception of this graphic and not counting the bottom sides still gives you the same "how many sides are x-shaded" answers as the "grounded" perception. The website Predds dot net has a sample test with this image, and he talks about using the "floating" method of perception and not counting the bottom sides as if they were grounded.
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