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A 5 cubic centimeter cube is painted on all its side. If it

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A 5 cubic centimeter cube is painted on all its side. If it [#permalink] New post 09 Jul 2007, 06:26
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A 5 cubic centimeter cube is painted on all its side. If it is sliced into 1 cubic centimer cubes, how many 1 cubic centimeter cubes will have exactly one of their sides painted?

A. 9
B. 61
C. 98
D. 54
E. 64
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Re: PS : Geometry [#permalink] New post 09 Jul 2007, 06:31
trahul4 wrote:
A 5 cubic centimeter cube is painted on all its side. If it is sliced into 1 cubic centimer cubes, how many 1 cubic centimeter cubes will have exactly one of their sides painted?

A. 9
B. 61
C. 98
D. 54
E. 64


D. 6*9=54. A cube has 6 sides, each of which has 9 cubes with one sides painted.
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Re: PS : Geometry [#permalink] New post 09 Jul 2007, 06:33
trahul4 wrote:
A 5 cubic centimeter cube is painted on all its side. If it is sliced into 1 cubic centimer cubes, how many 1 cubic centimeter cubes will have exactly one of their sides painted?

A. 9
B. 61
C. 98
D. 54
E. 64


9 such cubes for each of the 6 surfaces = 9 * 6 = 54 cubes. (D)
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 [#permalink] New post 09 Jul 2007, 07:26
Dear Wudy and Sumande

Please make the solution more clear.

First I don't understand how to get integer from 5 ^ 1/3 to get dimension of cube.

Also if you're talking about 9 cubes for every surface (3 x 3 x 3) and we're looknig for cubes with exactly one side painted that the solution is 6. One middle cube has only one painted side on every surface.
Or I didn't undertand the condition. Please explain.

Thank you in advance.
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Re: PS : Geometry [#permalink] New post 10 Jul 2007, 08:49
trahul4 wrote:
A 5 cubic centimeter cube is painted on all its side. If it is sliced into 1 cubic centimer cubes, how many 1 cubic centimeter cubes will have exactly one of their sides painted?

A. 9
B. 61
C. 98
D. 54
E. 64


are you sure it's a 5 cubic centimeter cube or is it a 5 cubic metre cube?? there should be five 1 cubic centimeter cubes when a 5 cubic centimeter cube is cut up
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 [#permalink] New post 10 Jul 2007, 09:38
Andrey2010 wrote:
Dear Wudy and Sumande

Please make the solution more clear.

First I don't understand how to get integer from 5 ^ 1/3 to get dimension of cube.

Also if you're talking about 9 cubes for every surface (3 x 3 x 3) and we're looknig for cubes with exactly one side painted that the solution is 6. One middle cube has only one painted side on every surface.
Or I didn't undertand the condition. Please explain.

Thank you in advance.


Well. I though the cube is 5*5*5 centimeter, or the question does not make sense-- how a 5 cubic centimeter cube is divided into 1 cubic centimeter cube?

The number listed above is how many faces painted for each 1cm*1cm cube on a surface.
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 [#permalink] New post 10 Jul 2007, 09:46
wudy wrote:
Andrey2010 wrote:
Dear Wudy and Sumande

Please make the solution more clear.

First I don't understand how to get integer from 5 ^ 1/3 to get dimension of cube.

Also if you're talking about 9 cubes for every surface (3 x 3 x 3) and we're looknig for cubes with exactly one side painted that the solution is 6. One middle cube has only one painted side on every surface.
Or I didn't undertand the condition. Please explain.

Thank you in advance.


Well. I though the cube is 5*5*5 centimeter, or the question does not make sense-- how a 5 cubic centimeter cube is divided into 1 cubic centimeter cube?

The number listed above is how many faces painted for each 1cm*1cm cube on a surface.
32223
21112
21112
21112
32223


My assumption was exactly the same. I didn't think that a cube of side 5^(1/3) cm being divided into 1 cubic centimeter cubes makes sense. So, I assumed that each of the sides of the cube was 5 cm (a 5 cm cube as opposed to 5 cubic centimeter).
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 [#permalink] New post 10 Jul 2007, 21:19
There is another way to do this problem. We know that all of the squares on each side have 2 sides (or 3 sides) painted. we want to remove them.

Make a new square which represents all of these squares. 5-2(one inch from each edge)=3

3.3=9 squares on each side.

9.6 sides is 54 squares
  [#permalink] 10 Jul 2007, 21:19
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