There is a rectangular prism made of 1 in cubes that has been covered

Given volume of unfoiled prism \(=\,128\)

Let \(2x\) be the width of the un-foiled rectangular prism
Then
\(length\,*\,width\,*\,height\,=\,x\,*\,2x\,*\,x\,=\,128\)
\(2x^3\,=\,128\)
\(x\,=\,4\)
Width \(2x\,=\,8\)

Width of the foiled prism \(= 10\) \((\,2x\,+\,\)two foiled cubes on each side of the width\(\,=\,8\,+\,2)\)

Answer E

VERITAS PREP OFFICIAL SOLUTION

1) Don’t give up before you start!

I can feel your skin crawling as fear sets in, but believe in yourself! Let’s start working like we do with any other problem. This problem is at the end for a reason. You should have finished everything else already so all you need to focus on is this problem. Don’t worry about the time at this point. You will waste more time worrying than you have to spare. Start working! And start with what you know!

2) Start with what you know

We know: 1 in cubes is the same as inches cubed (volume in inches cubed is essentially a measure of how many cubes that are 1in x 1in 1in that can fit in a 3D shape). We also know that the 128 cubes are completely covered by other cubes so that none of their sides touch the outside. This means that there is essentially a bigger prism completely covering a smaller prism. Its like when you used to play with blocks (I know you did!) and you would completely enclose a block in other blocks to make an exact replica only bigger. So we have a rectangular prism that is 128 inches cubed. Finally, we also know that one side is twice as big as the other two.

3) Draw a picture (if possible)

Let’s try to draw what this might look like. If we had a cube that was 2 x 2 inches cubed, one side would look like this:

In order to cover it completely on all sides we would have to have a cube that had one more cube on each side. So the new face would look like this

So we essentially know that the final figure will have sides that are two greater than the sides of the interior structure, (again, this would enclose a smaller rectangular prism). Huzzah! We are getting somewhere.

4) Use general equations to get specific answers

Using the equation of a rectangular prism and the information that our smaller prism has a length that is two times the width and height we can start writing equations:

L x W x H = Area of a rectangular prism

Let’s make L = x That would mean that H = x also and W = 2x and we can plug that into our equation to get:

(x)(2x) (x) = 160 or (2x^3) = 128 —-> x = 4

5) ANSWER THE QUESTION

We are SO close to the end, but we need to answer the question being asked. Now the length and height of the smaller prism is 4 in (which is an answer choice) but that is NOT what the question asks. The width of our smaller prism is twice the length or 8 in. This is also NOT what the question asks! We know from our previous calculations that all the measurements of the final figure are two greater than the smaller so the dimensions of the larger prism would have to be L= (x+2), H= (x+2) and W = (2x+2). Thus our final answer is 2(4) + 2 = 10. WOOHOOO we DID IT!!

Solution:

Notice the width of the covered prism will be 2 more than the width of the figure created by the 128 cubes. Notice also that the length and height of the inner figure are equal and the width of the inner figure is twice the length/height. Using these facts, let’s test each answer choice:

A) If the width of the covered prism is 4, then the width of the inner figure is 4 - 2 = 2 in. Thus, the length and height of the inner figure are 1 in each. If there is one cube along the length, two cubes along the width and one cube along the height of the inner figure, there are 1 x 2 x 1 = 2 smaller cubes, which does not agree with the fact that the inner cube is made up of 128 smaller cubes.
B) If the width of the covered prism is 6, then the width of the inner figure is 6 - 2 = 4 in. Thus, the length and height of the inner figure are 2 in each. If there are two cubes along the length, four cubes along the width and two cubes along the height of the inner figure, there are 2 x 4 x 2 = 16 smaller cubes, which does not agree with the fact that the inner cube is made up of 128 smaller cubes.

C) If the width of the covered prism is 8, then the width of the inner figure is 8 - 2 = 6 in. Thus, the length and height of the inner figure are 3 in each. If there are three cubes along the length, six cubes along the width and three cubes along the height of the inner figure, there are 3 x 6 x 3 = 54 smaller cubes, which does not agree with the fact that the inner cube is made up of 128 smaller cubes.

D) The width of the covered prism cannot be 9 (or any other odd number), because then the width of the inner figure will be 7, which cannot be twice the number of smaller cubes along the length/height.

While we eliminated every answer choice besides E, let’s verify that W = 10 indeed satisfies the requirements of the question:

E) If the width of the covered prism is 10, then the width of the inner figure is 10 - 2 = 8 in. Thus, the length and height of the inner figure are 4 in each. If there are four cubes along the length, eight cubes along the width and four cubes along the height of the inner figure, there are 4 x 8 x 4 = 128 smaller cubes, which is consistent with the information given in the question.

Answer: E

When we remove the outer layer, reducing the width by 2, we get a smaller prism with integer dimensions and volume 128. The width is the largest dimension, so it must exceed the cube root of 128, and it definitely must be bigger than 4, and the width clearly needs to be a factor of 128 = 2^7. Glancing at the answer choices, only 10 gives us a factor of 128 larger than 4 if we subtract 2, so 10 must be the answer.

Can somebody explain me this?

If it says width is twice the length AND twice the height, doesn't that mean w = 2 l + 2 h = 4l, where l is the length of a side of cube?

Bunuel