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805+ (Hard)|   Geometry|      
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rajarshee
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A cube with each side = 1 inch is half filled with water 30 Oct 2015, 16:42
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B
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95% (hard)

Question Stats:

30% (01:57) correct 70% (02:12) wrong based on 77 sessions

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A cube with each side = 1 is half filled with water. Now it is held in such a manner that only one of vertices is touching the floor and the diagonal of the cube is perpendicular to the floor.

What is the surface area of the water?

(1) √2
(2) 2
(3) √3
(4) 2√2
(5) 1
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B
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kunal555
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A cube with each side = 1 inch is half filled with water 30 Oct 2015, 18:30
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rajarshee
A cube with each side = 1 is half filled with water. Now it is held in such a manner that only one of vertices is touching the floor and the diagonal of the cube is perpendicular to the floor.

What is the surface area of the water?

(1) √2
(2) 2
(3) √3
(4) 2√2
(5) 1
Show more

This question can be better understood with diagram. I hope someone provides solution with the diagram.
When the cube is rested on the vertex such that the diagonal of cube is perpendicular to the floor, a pyramid with three right triangular faces and a square base is obtained within the cube.
Each triangle of this pyramid has hypotenuse \(\frac{\sqrt{5}}{2}\) and, sides 1 and 1/2.
So area of each triangle will be,
1/2*Perpendicular*base = 1/2*1*1/2 = 1/4
Total area of all 3 triangles will be 3*1/4= 3/4
Each side of the base of this pyramid is formed by joining vertex closest to the floor(not the vertex resting on the floor) to the midpoint of a side of the cube. This side is same as the hypotenuse of the right triangle i.e \(\frac{\sqrt{5}}{2}\).
So the area of the square base of the pyramid will be 5/4
So total surface area of the water or the total surface area of this pyramid is
5/4+3/4 = 8/4 = 2

Answer:- B
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davedekoos
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A cube with each side = 1 inch is half filled with water 08 Jan 2016, 15:28
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rajarshee
A cube with each side = 1 is half filled with water. Now it is held in such a manner that only one of vertices is touching the floor and the diagonal of the cube is perpendicular to the floor.

What is the surface area of the water?

(1) √2
(2) 2
(3) √3
(4) 2√2
(5) 1
Show more


This problems takes some mental geometrical gymnastics of Olympic proportions. And this is almost certainly not a real GMAT question... But it's a really interesting fun problem, so here goes.

When you orient a cube so that one vertex is touching the ground and the diagonal is perpendicular to the ground, you'll basically get two triangular pyramids (one pointing up on top and the other pointing down on the bottom), with a triangular hexagonal hybrid shape in the middle. (What do you mean you have no idea what I'm saying?? :wink: )

I've drawn some different perspectives of a cube here to help

Attachment:
Wireframe cube 2.png
Wireframe cube 2.png [ 36.94 KiB | Viewed 3981 times ]

#1 and #2 show the cube oriented the way we want with a slight perspective to show where the water line would be.
#3 shows the cube directly from the side, with the water line exactly halfway.
#4 and #5 show a more common view of a cube in isometric view. This time the water line is shown at an angle (it's the same water line from #2 and #3), and at this angle we can see more clearly the shape that the surface of the water will have.
#6 and #7 show the cube from the top, looking down towards the ground, so we can see the full shape of the surface of the water.

As we can see it is a hexagon.

The hexagon is formed by joining the midpoint of the edges that do not connect to the vertices at the top or at the bottom. Since each side of the cube is 1, the length of each side of the hexagon is \(1/\sqrt{2}\). From there the area of the hexagon is \(\frac{3\sqrt{3}(side^2)}{2}\)

So the area of the hexagon, which is the surface area of the water is \(\frac{3\sqrt{3}}{4}\)

This is not one of the answer choices, but I feel pretty good about it. And seeing as this isn't a real GMAT question anyway, who knows where it and the answer choices came from?

Anyone want to have a go and see if I made an error?

At least some kudos for the drawing though... :-D

Cheers
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Beixi88
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A cube with each side = 1 inch is half filled with water 10 Jan 2016, 04:04
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davedekoos
rajarshee
A cube with each side = 1 is half filled with water. Now it is held in such a manner that only one of vertices is touching the floor and the diagonal of the cube is perpendicular to the floor.

What is the surface area of the water?

(1) √2
(2) 2
(3) √3
(4) 2√2
(5) 1


This problems takes some mental geometrical gymnastics of Olympic proportions. And this is almost certainly not a real GMAT question... But it's a really interesting fun problem, so here goes.

When you orient a cube so that one vertex is touching the ground and the diagonal is perpendicular to the ground, you'll basically get two triangular pyramids (one pointing up on top and the other pointing down on the bottom), with a triangular hexagonal hybrid shape in the middle. (What do you mean you have no idea what I'm saying?? :wink: )

I've drawn some different perspectives of a cube here to help

Attachment:
Wireframe cube 2.png

#1 and #2 show the cube oriented the way we want with a slight perspective to show where the water line would be.
#3 shows the cube directly from the side, with the water line exactly halfway.
#4 and #5 show a more common view of a cube in isometric view. This time the water line is shown at an angle (it's the same water line from #2 and #3), and at this angle we can see more clearly the shape that the surface of the water will have.
#6 and #7 show the cube from the top, looking down towards the ground, so we can see the full shape of the surface of the water.

As we can see it is a hexagon.

The hexagon is formed by joining the midpoint of the edges that do not connect to the vertices at the top or at the bottom. Since each side of the cube is 1, the length of each side of the hexagon is \(1/\sqrt{2}\). From there the area of the hexagon is \(\frac{3\sqrt{3}(side^2)}{2}\)

So the area of the hexagon, which is the surface area of the water is \(\frac{3\sqrt{3}}{4}\)

This is not one of the answer choices, but I feel pretty good about it. And seeing as this isn't a real GMAT question anyway, who knows where it and the answer choices came from?

Anyone want to have a go and see if I made an error?

At least some kudos for the drawing though... :-D

Cheers
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Do you know what the volume of the water would be in the shape you're working with? I bet the problem is that your solution has a different water volume than 0.5 units^3.

Anyways, for the answer, I assumed the water level would be less than half of the cube when the cube is tipped. If this is wrong, I'm sorry, it's 3 AM and I'm very tired. Either way, if the water level is less than half, if the water level is just low enough, you can use pyramidal formulas. The volume of a pyramid is always 1/3(a*a*h). If you take 1/2 as your water volume, you get 1/2 = 1/3(a*a*h), which becomes (3/2) = a*a*h. From here, I engage luck mode and hope that because we've seen nothing but clean numbers so far, our height and our final answer will be clean as well. That is, no roots. From this, we see that only B and E are applicable. E sets off red flags due to the 1, so I opted for B. B ended up being the answer.

Please note that my methodology and assumptions are pretty bad if you're sitting the gmat. I just wanted to see what happens if I use a line of assumptions to reach an answer.

If anyone has a real method to solve this question in a reasonable amount of time, please let me know. I'm curious!
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A cube with each side = 1 inch is half filled with water 10 Jan 2016, 18:48
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Beixi88


Do you know what the volume of the water would be in the shape you're working with? I bet the problem is that your solution has a different water volume than 0.5 units^3.

Anyways, for the answer, I assumed the water level would be less than half of the cube when the cube is tipped.
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Hi Beixi,

If the cube is filled halfway, then no matter how you orient it, it will always be filled halfway. It can't be any other way, since the shape is symmetrical about the center point.

Your question about the volume is an interesting problem in and of itself though, and something I didn't check (because I'm pretty confident that half a cube is always half a cube). For fun, I did verify that the volume of water in my solution is still 0.5\(u^3\) though.

I was going to try to illustrate it myself, but instead, I found some beautiful explanations from people who have better graphical tools than I do. If I had found this before I worked it all out myself, it would have made my life much easier, but in any case, here is an excellent explanation and illustration from Mathworld.Wolfram.com. https://mathworld.wolfram.com/Cube.html The part that interests us is about halfway down the page starting with this image:


Attachment:
CubeCutByPlanes_1100.gif
CubeCutByPlanes_1100.gif [ 10.03 KiB | Viewed 3851 times ]

As we can see from this image of a unit cube cut by various planes, the surface of the shape in our problem will indeed be a hexagon with sides \(\frac{\sqrt{2}}{2}\) (aka \(\frac{1}{\sqrt{2}}\)), and from the table below the image on the webpage, the volume of the section is 1/2 and the surface area of the hexagon is \(\frac{3\sqrt{3}}{4}\), as in my solution.

I think this leaves little doubt that the real answer is in fact \(\frac{3\sqrt{3}}{4}\) and is does not appear in the answer choices.

N.B. I did have a thought that maybe the question was asking for the total surface area of the volume of water, not just the "exposed" area inside the cube. But in that case, the answer would be \(3+\frac{3\sqrt{3}}{4}\), which is quite a bit larger than any of the answer choices given, so I ruled that possibility out.


Cheers,
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A cube with each side = 1 inch is half filled with water 04 Jan 2024, 13:57
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