This is the real difference between an official question and one from a prep company: even if an official question is time-consuming there will be ALWAYS at least ine shortcut to solve the problem.

Hope this helps.

Regards

Can someone please explain this? @carcass?

Question:
A cube shaped cake is frosted on all sides except the bottom and then cut into 27 equally sized cube shaped pieces
Quantity A: The number of pieces of cake that have exactly two sides frosted
Quantity B: One more than the number of pieces of cake that have fewer (than) two sides frosted


Solution:

There are 27 pieces = 3 x 3 x 3
Thus, each edge has 3 small cubes and each face has 9 small cubes

Quantity A: Pieces of cake that have exactly two sides frosted

Small cubes with 2 faces frosted will be found along the edges of the cube
There are 12 edges with 1 center small cube having 2 faces frosted (the corners have 3 faces frosted)
Thus, there are 12 small cubes.

However, if we look at the base:
There are 4 edges along the base face. The center small cube on each such edge has only 1 face frosted (not 2 - since the bottom face is not frosted). Thus, these 4 small cubes are ignored.
However, in the 4 corners of the bottom face, the 4 small cubes have 2 faces frosted (not 3 - since the bottom face is not frosted). Thus, these 4 will be added.

Hence, the total small cubes with 2 faces frosted remain 12.
Quantity A = 12

Quantity B: One more than the number of pieces of cake that have fewer (than) two sides frosted

Number of pieces of cake that have one side frosted:
Along the 5 faces, there is one center small cube with 1 face frosted i.e. 5 small cubes
Along the bottom face (base) edges, the cube in the center of each edge has 1 face frosted i.e. 4 more small cubes

Number of pieces of cake that have no side frosted:
Along the bottom face, the center small cube has no face frosted i.e. 1 small cube
Also, in the 3 x 3 x 3 cube, the center piece (obtained by stripping off the top, bottom, left, right, front and back layers) will also have no face frosted i.e. 1 more small cube

Thus, total such small cubes = 5 + 4 + 1 + 1 = 11
Thus, 1 more than the above = 12
Quantity B = 12


Hence, the quantities are equal

Answer C

You preceded me Sir.

Thank you.

Great explanation

Stunning

is this a GRE level question? I find it really hard.

Thank you!

For me it goes too far even for a GRE 5 level question.

Regards

I think it's within the score of the GRE

If the lead quant mod says this. Turns out I am wrong.

However, I think it is slightly over. But maybe, that is me

It's in the 169.7 to 170.1 range

what prep company is mgre?

Hey jantje

It's Manhattan GRE

A total of 5*9 = 45 sides are shared by 27 cubes.
Recognize that the 4 top corners have 3 sides each, 2 cubes in the middle has 0 side.
Rest are cubes with 2 or 1 sides. Therefore, solve the system:
2a + b = 45 - 3*4
a + b = 27 - 4 - 2
cubes with 2 sides: a = 12
cubes with 1 side: b = 9
cubes with sides < 2 will be 9 + 2 = 11

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