Cube counting question

[Psa]

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How do we know that the little triangle that mouse is pointing to is another cube and not the right side of the visible cube?

 

[LazyP]

How do we know that the little triangle that mouse is pointing to is another cube and not the right side of the visible cube?

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Thats a good point... hope its just a coincidence and actually not something that we would see in the real deal since it can be argued either way in my opinion

 

[El Dentistador]

How do we know that the little triangle that mouse is pointing to is another cube and not the right side of the visible cube?

View attachment 24715

If I encountered a problem like that, then my instincts would assume the latter. It can be argued one way or another. On top of that the author of the question is human and could make a mistake, but that might not happen still if they implement some form of peer review for each question. Nonetheless, I would trust their oversight of the questions would not pose such duplicity.

One possible approach to solving such a problem would be to compare the answers given for the questions posed. You would need to compare the answers available for cubes with 2 exposed sides (2s) and cubes with 3 exposed sides (3s). If the possible cube is really present then the count for the 2s will be +2 and the count for the 3s will be -1. So say the question asking for the 2s cuts off at the high end at 5 or 6 and you have determined that the 2s could either be up to 5 or 7, then based on the question the count would have to be 5, which would also indicate what the actually count for the 3s would be. So basically, you are comparing the lower and upper extremes of each question. Unfortunately if the questions can accomodate all possibilities, then there is no way to definitively determine the count and you are left with your instincts.

 

[dent777]

I don't think you count that one. You have to see at least 2 complete sides, unless it is a supporting cube. In this case it is NOT a supporting cube, so it does not exist.

 

[Psa]

I don't think you count that one. You have to see at least 2 complete sides, unless it is a supporting cube. In this case it is NOT a supporting cube, so it does not exist.

That's what I thought while I was taking the test, but it turns out there actually IS a cube there. This question is from CDP test #5. I doubt they would give us something like this on the actual DAT.

If I encountered a problem like that, then my instincts would assume the latter. It can be argued one way or another. On top of that the author of the question is human and could make a mistake, but that might not happen still if they implement some form of peer review for each question. Nonetheless, I would trust their oversight of the questions would not pose such duplicity.

One possible approach to solving such a problem would be to compare the answers given for the questions posed. You would need to compare the answers available for cubes with 2 exposed sides (2s) and cubes with 3 exposed sides (3s). If the possible cube is really present then the count for the 2s will be +2 and the count for the 3s will be -1. So say the question asking for the 2s cuts off at the high end at 5 or 6 and you have determined that the 2s could either be up to 5 or 7, then based on the question the count would have to be 5, which would also indicate what the actually count for the 3s would be. So basically, you are comparing the lower and upper extremes of each question. Unfortunately if the questions can accomodate all possibilities, then there is no way to definitively determine the count and you are left with your instincts.

What you're saying makes sense now but during an actual exam I will be thinking about how bad I screwed up on angle ranking and how little time I have left on the clock. Therefore I would most likely just say "**** it" and move on without comparing anything.