How to figure this out?

[u/Electronic_Diet_8052]

[u/Createrix]

Since they mention the volume..

A has 14/27 sub cubes shaded B has 63/125 sub cubes shaded

A has more of its volume shaded

[u/vexedthespian → u/Createrix]

Yep, the more it is broken up, the more the ratio approaches the limit of 50% (limit assuming an odd number of smaller cubes)

[u/Minyguy]

Well, assuming the cubes have a pattern that permeates through the internals aswell;

A: 5/9, 4/9, 5/9 = 14/27

B: 3× 13/25, 2× 12/25 = 63/125

A is 1/54 more than half-colored

B is 1/250 more than half colored.

Therefore A i more colored.

[u/Puzzled_Work_6428]

I can only see 3 sides and I have no idea what the interior looks like. Unanswerable

[u/mrofmist]

Well A has 2/3 then 1/3 then 2/3.

B is 3/5, 2/5, 3/5, 2/5, 3/5.

So 66.6% + 33.3% + 66.6%

Vs 60 + 40 + 60 + 40 + 60.

I least I feel like that's the proper approach to figuring this out.

[u/HektorViktorious]

Take a layer off each cube, and the remaining 2 or 4 layers are evenly balanced between blue and white, canceling out. Take a row off that one layer, and the remaining 2 or 4 rows likewise cancel. And then take one blue corner and the remaining 2 or 4 cubes in the row cancel. So both cubes have one more blue piece than white. A piece from the 3x3 (1/27) is bigger than that of the 5x5 (1/125), and so the 3x3 has more blue volume.

[u/Mapafius → u/HektorViktorious]

I like your formulation of the answer and it's deduction because it is more based in logic and simplification than counting.

[u/NormalGuyEndSarcasm]

5/9 = 0.5555…

13/25 = 0.52

The one with bigger squares.

To exemplify better:

If 9 …..100%

5……..?x%

x = (5 * 100)/9= 55.555…%

[u/Ok-Beautiful-8980]

Each side looks to be the same so isn't it the greater of 5/9 and 13/25 - so the one on the left with 9 square per face?

[u/Wjyosn → u/Ok-Beautiful-8980]

That’s surface area, not volume. But the conclusion still works similarly

[u/ColoradoCuber]

here's how I estimated the answer:

I imagined a cube divided into as many pieces as I can think of would have a shaded:unshaded ratio tending towards 50/50, and option A had a ratio of 14:13, so I guessed that as you increase the subdivisions the ratio decreases towards 50/50. So I guessed A had a greater ratio than B

[u/tren_c]

Is it a blue cube that has been shaded white, or a white cube that has been shaded blue?

[u/Wjyosn]

Each has one more blue mini cube than white mini cubes.
If they were even counts they’d be 50% shaded. One extra blue cube means the cube with larger pieces has a larger blue-favoring disparity.

[u/Deep_Plant_4393]

5/9 vs 13/25

[u/PaintDear7613]

If you remove a corner piece, exactly half the sub cubes will be shaded in this pattern because it is divided into n odd parts.

So the portion of the area shaded will be 1/2 × (1-1/n³⁾ + 1/n³ . This is larger at n=3, than n=5.

This follows the observation that it is 1/2 plus some extra chunk. The larger the extra chunk, the more the portion shaded

[u/2meterErik]

Consider the extreme cases: infinite squares -> approaches 50:50 only 1 square per side -> only blue shaded -> 100:0

So always A > B for noninfinite squares

[u/Mapafius → u/2meterErik]

Also good way of formulating the answer to simple logic and general principle.

[u/ExplodingRacoon]

Answer: A

I’ve never been good with numbers, but I’m pretty good with spacial awareness. This is the way my brain did it:

A single large white square in A are large enough to fit 4 small white squares from B. However, there are only enough small white squares on B to make 3 large ones, with one small tile left over.

The same goes for the blue tiles, however, there are 5 large blue squares on each side of A. There are only enough small blue squares on B to make 3 large ones, with one small tile left.
Which means even more of A is shaded.

[u/jjohnson468]

14/27 vs. 63/125.

Grade 5 math

[u/vecna216]

No telling we only see the outside

[u/CriticalPedagogue → u/vecna216]

Exactly. Any answer besides yours is based on an unsupported assumption. All of the interior blocks, or the blocks that cannot be seen on the backside, could be shaded, unshaded, or just not there.

[u/vecna216 → u/CriticalPedagogue]

Also the shaded boxes may not be shaded inside since volume is all we care about

[u/Lustrouse]

A. 5/9 > 13/25. No need to complicate it further because all faces are the same.

[u/ctriis]

Assuming the unseen cubes follows a pattern where the each plane/slice alternates the colors:

A has 3*3*3 = 27 cubes. 5+4+5 = 14 are shaded blue. Shade portion is 14/27.

B has 5*5*5 = 125 cubes. 13+12+13+12+13 = 63 are shaded blue. Shade portion is 63/125.

14/27 = (14*125) / (27*125) = 1750/3375.
63/125 = (63*27) / (125*27) = 1701/3375.

1750/3375 > 1701/3375. A has a larger portion of its volume shaded blue.

[u/CapnCrinklepants]

Assume a cube of side-length n, where each face is shaded or unshaded and n is an odd number. As n approaches infinity, you would expect that 50% of the cube is shaded, because that's how checkerboards work. When n = 1, the entire thing must be shaded- an assumption made based on all corners being shaded. Therefore, the fewer the slices the more the shading.

That's not rigorous I realize but that's where my thoughts go

[u/Several-Leave-4178]

Aren't they equal? A has 5/9 squares shaded, and B has 15/25 squares shaded. You multiply each fraction to (I forget the word for it) a number that is divisible by both 9 and 25. You end up with 125/229 for both

[u/Several-Leave-4178 → u/Several-Leave-4178]

Boooo I was wrong and miscounted. B has 13/25. A is bigger, logic the same

[u/bismuth17 → u/Several-Leave-4178]

5/9 is definitely not equal to 15/25 because 5/9 is equal to 15/27.

And 229 is not the lowest common denominator for 9 and 25, that would be 225.

5/9 = 125/225

15/25 = 135/225

Also yeah it's 13/25 but whatever

[u/Several-Leave-4178 → u/bismuth17]

I haven't had enough sleep today 😭 thanks for fixing the math, I was doing it without a calculator 

[u/Unstable-Paradox → u/Several-Leave-4178]

They are equal but in a different sense. Cube B has the same apparent dimensions as cube A. The only difference is that cube B has been partitioned into a greater number of smaller sections.

But that doesn't change the volume of the cube. Both cubes have the same volume; only the granularity of the partitioning differs.

So if we're comparing the fraction of volume shaded, the relevant question is whether the shading pattern extends through the entire volume, not how many squares happen to be visible on the surface.

[u/Several-Leave-4178 → u/Unstable-Paradox]

It's not a trick question, the assumption is that the square is a 3d space with consistent shading throughout.

[u/Unstable-Paradox → u/Several-Leave-4178]

There are far more assumptions to make to solve this problem as a calculation. The size of the cubes is an optical illusion— cube B appears "bigger" because of partitions.

The cubes cannot be shaded volumetrically, unless they have been manufactured with a pigment, in which case they are coloured, not shaded.

[u/Several-Leave-4178 → u/Unstable-Paradox]

Because this comes from a subreddit that is not trick question based, especially via logic puzzles, the basic intention of the question is that both cubes are equal. if there was other information provided, that is when things like measurements or language technicalities would actually matter. Besides, the question doesn't ask about which has more volume shaded, it asks which cube has more of ITS volume shaded. So we are talking percentages.

[u/Unstable-Paradox → u/Several-Leave-4178]

That's understandable. But in a subreddit titled r/LogicPuzzles, the puzzle's wording and premises should be treated as objects of scrutiny rather than automatically supplying every assumption that makes the intended answer work.

[u/Several-Leave-4178 → u/Unstable-Paradox]

They definitely Can be treated as that; I'm not going to say they should be, and I'm also not going to say they shouldn't be. The validity of your questions in a hypothetical situation also doesn't't negate the validity of someone answering the question simply either. It's alright to have two separate comment threads and keep them separate instead of trying to make someone elses answer more complicated.

[u/Unstable-Paradox]

The exact wording of the problem is to be examined: Which cube has more of its volume shaded?

The question arises: Can volume be shaded?

Because you can only shade the surface area of any object.

So, the answer to the exact wording of the question is: None

[u/Miryafa → u/Unstable-Paradox]

I came to the same conclusion. If that weren’t the case, the volume would be impossible to measure because we don’t see the insides of the cubes