##### This topic has expert replies
Junior | Next Rank: 30 Posts
Posts: 21
Joined: 04 May 2010
Thanked: 4 times

### MGMAT

by yuliawati » Mon Nov 01, 2010 11:00 pm

A certain cube is composed of 1000 smaller cubes, arranged 10 by 10 by 10. The top layer of cubes is removed from a face, then from the adjacent face above it, then from the adjacent face to the right of the first. The process is repeated on the same three faces in reverse order. Finally, a last layer is taken from the first face. How many smaller cubes have been removed from the larger cube?

(A) 488
(B) 552
(C) 612
(D) 722
(E) 900

OA later

GMAT Instructor
Posts: 905
Joined: 12 Sep 2010
Thanked: 378 times
Followed by:123 members
GMAT Score:760
by [email protected] » Tue Nov 02, 2010 12:28 am
I would probably just find what cube I have remaining, and subtract the new number of cubes from the 1000.

Remove a top layer from the front, so the cube is only 9 cubes "wide": 10*9*10
shave a layer from the top face ("above it"), so the cube is only 9 cubes "high": 10*9*9
Now shave a layer from the side ("to the right of the first") so the cube is only 9 cubes "long": 9*9*9

Every time you shave a layer, you reduce 1 from one dimension, leaving the others as they are. Thus, if we repeat the process for all three dimensions (even in reverse order), we get 8*8*8. If we take another layer from the first face, we take another 1 from the width to get 8*7*8
How we call the dimensions "length, width height" does not really matter: the dynamic is the same, and the result will be two 8s and a 7.

Now the question asked how many have been removed, so the final answer is {original number - new number} = 10*10*10 - 8*8*7 = 1000 - 64*7 = 1000 - 448 = 552. Answer should be B.
Geva
Senior Instructor
Master GMAT
1-888-780-GMAT
https://www.mastergmat.com

Junior | Next Rank: 30 Posts
Posts: 21
Joined: 04 May 2010
Thanked: 4 times
by yuliawati » Tue Nov 02, 2010 1:30 am
Great!! thank Geva!

Master | Next Rank: 500 Posts
Posts: 235
Joined: 26 Oct 2016
Thanked: 3 times
Followed by:5 members
by Anaira Mitch » Sat Jan 14, 2017 4:17 pm
We can answer this by keeping track of how many cubes are lopped off of each side as the cube is trimmed (10 x 10 + 10 x 9 + 9 x 9 + ...), but this approach is tedious and error prone. A more efficient method is to determine the final dimensions of the trimmed cube, then find the difference between the dimensions of the trimmed and original cubes.

Let's call the first face A, second face B, and third face C. By the end of the operation, we will have removed 2 layers each from faces B and C, and 3 layers from face A. So B now is 8 cubes long, C is 8 cubes long, and A is 7 cubes long. The resulting solid has dimensions 8 x 8 x 7 cubes or 448 cubes. We began with 1000 cubes, so 1000 - 448 = 552. Thus, 552 cubes have been removed.

GMAT Instructor
Posts: 2630
Joined: 12 Sep 2012
Location: East Bay all the way
Thanked: 625 times
Followed by:118 members
GMAT Score:780
by [email protected] » Wed Jan 18, 2017 7:19 pm
Here's a quick way:

We're removing three layers from the length, two layers from the width, and two layers from the height.

We originally had 10 * 10 * 10, but now we've got (10 - 3) * (10 - 2) * (10 - 2), or 448.

The difference between 1000 and 448 is the number that we've removed, so 1000 - 448, or 552.

• Page 1 of 1