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 Post subject: GMAT ProbabilityPosted: Mon Sep 20, 2010 9:06 am

Joined: Sun May 30, 2010 3:15 am
Posts: 424
A wooden cube whose edge length is 10 inches is composed of smaller cubes with edge lengths of one inch. The outside surface of the large cube is painted red and then it is split up into its smaller cubes. If one cube is randomly selected from the small cubes, what is the probability that the cube will have AT LEAST one red face?
A. 36.0%
B. 48.8%
C. 50.0%
D. 52.5%
E. 60%

(B) The large cube is composed of 10 small cubes in each dimension, so it is composed of 1000 small cubes in all because the volume of a cube is:
Volume = length × width × height = 10 × 10 × 10 = 1000.

The cubes on the surface will have at least one face painted, but the inside cubes will not. It is difficult to directly solve for the number of small cubes along the surface, so we will calculate the number of cubes on the inside (not painted), and subtract this number from the total number of cubes, to get the number of cubes that are painted.

The number of small cubes that are on the inside (not painted) is:
8 × 8 × 8 = 512.
This is because the first and last cube in each dimension will be painted, meaning that there are rows of 8 unpainted cubes in each dimension.

Then, the number of painted cubes is:
1000 – 512 = 488.
Therefore, the probability of selecting a cube with at least one face painted is:
488/1000 = 48.8%.

The correct answer is choice (B).

Note: If you chose choice (A), you were probably working with a 2-dimensional figure instead of a cube (3-dimensional).
---------
I can't imagine this problem in 3D. Could you help, please?

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 Post subject: Re: math: 3D geometry, probability.Posted: Tue Sep 21, 2010 5:42 am

Joined: Sun May 30, 2010 2:23 am
Posts: 498
At the beginning we have a 10 × 10 cube which sides are painted red:

We want to count number of unit cubes that are NOT painted (it's easier). For that we take off outer layer of painted unit cubes:

As the result we get inner 8 × 8 cube which is made of unit cubes that are NOT painted:

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 Post subject: Re: math: 3D geometry, probability.Posted: Mon Dec 06, 2010 1:26 pm

Joined: Sun May 30, 2010 3:15 am
Posts: 424
Why will the number of cubes painted at least on one side ie the cubes on the outside not be 600, 10 × 10 on each side × 6 sides?

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 Post subject: Re: math: 3D geometry, probability.Posted: Mon Dec 06, 2010 1:41 pm

Joined: Sun May 30, 2010 2:23 am
Posts: 498
questioner wrote:
Why will the number of cubes painted at least on one side ie the cubes on the outside not be 600, 10 × 10 on each side × 6 sides?

In this case the corner cubes (marked green) will be counted 3 times (because each one belongs to three sides) and the edge cubes (marked blue) will be counted 2 times (because each one belongs to two sides).

Considering that there are 8 corner cubes and 8 × 12 = 96 edge cubes, the correct counting of the painted cubes is:

(100 × 6) – (8 × 2) – (96 × 1) = 488

It's really easy to make a mistake counting the number of painted cubes, while counting the number of unpainted cubes is much simpler: 8 × 8 × 8 = 512.

 Attachments: corner_cubes.gif [17.67 KiB] Not downloaded yet
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 Post subject: Re: GMAT ProbabilityPosted: Sun Dec 09, 2012 5:24 am

Joined: Mon Nov 26, 2012 5:39 pm
Posts: 11
I spent a lot of time on this problem and I question your total number of cubes that comprise the 10 inch cube. There may be a 1000 faces on the outside of the cube, but the cubes at the corners account for 3 of those faces each and the cubes at the edges account for 2 of those faces each. Using similar logic to that used to find the number of cubes at the interior of the block. I calculated that the total number of cubes that make up the block is 760 not 1000. I would appreciate it if you would get back to me with a response. I have been struggling with the quantitative section and would appreciate knowing if my calculations were faulty.

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 Post subject: Re: GMAT ProbabilityPosted: Sun Dec 09, 2012 5:42 am

Joined: Sun May 30, 2010 2:23 am
Posts: 498
Quote:
There may be a 1000 faces on the outside of the cube, …
1000 are not just small cubes on the outside layer, they are ALL small cubes that make the wooden cube.

When you study math, start with simple objects and move on to more complicated ones. Imagine that we have just 1 row of 10 small cubes lying on a table.

How many cubes are there? 1 × 10 = 10

Then imagine we place 10 rows and 10 columns of small cubes on a table.

How many cubes are there? 10 × 10 = 100

Now imagine we add another level on top the previous one.

How many cubes are there? 2 × 10 × 10 = 200

Now you should fully understand why there are 10 × 10 × 10 small cubes in the original wooden cube:

And also why there are 8 × 8 × 8 = 512 not-painted small cubes:

Quote:
…, but the cubes at the corners account for 3 of those faces each and the cubes at the edges account for 2 of those faces each.
This is a good observation. But we use it not when counting the whole number of small cubes that compose the wooden cube, but if we count the number of small cubes that make the surface.

In this case the corner cubes (marked green) will be counted 3 times (because each one belongs to three sides) and the edge cubes (marked blue) will be counted 2 times (because each one belongs to two sides). While we need to count all the surface cubes only once!

There are 10 × 10 = 100 small cubes on each side and there are 6 sides. Considering that there are 8 corner cubes and 8 × 12 = 96 edge cubes, the correct counting of the painted cubes is:

(100 × 6) – (8 × 2) – (96 × 1) = 488

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