Problem Solving

A cuboid of dimensions 80 cm * 100 cm *160 cm is painted on all faces. The cost of painting is Rs. 92. The cuboid is cut into several identical cubes, which are as large as possible. All the unpainted faces of all the small cubes were also painted. What is the cost, in rupees, of painting these unpainted faces of the small cubes.

A) 380
B) 388
C) 396
D) 404
E) 412


OA to follow soon....

I think the answer is B- 388.

First we find out how much it costs to paint per surface area of the cube, which will be price/total surface area.
Price = 92
Total Surface Area= 2(80*100) + 2(80*160) + 2(100*160)= 73,600
Therefore price per surface area = 92/73600

Next, we are told that it is cut into several identical cubes of the largest size, and so to determine the dimensions of this cube we just take the greatest common factor of 80,100 & 160, which is 20. Therefore each cube will have dimensions 20*20*20.

Given this information, we can re-label the cuboid as having dimensions 4*5*8, so we can say there are a total of 4*5*8=160 identical cubes.

Now, there are some cubes in the middle of the cuboid which haven't been painted at all. To find out this number, imagine you are 'peeling off' the outer layer of cubes leaving inside only the cubes in the middle.
This will be (4-2)*(5-2)*(8-2)=2*3*6= 36 cubes.
Therefore we have to paint each side of the cube for all these 36 cubes.
This gives us a surface area of painting of: 36*6*20*20= 86,400

Now we are able to only concentrate on the outer layer of cubes. Notice that the cubes on the edges of the cuboid will have 3 sides of them painted, so for these cubes, we only need to paint an additional 3 sides. Since there are 8 edges, this gives us:
8*3*20*20=9,600

Now we have 116 cubes left to be dealt (160-8-36)
Notice that on the sides of the cube, these cubes will have been painted on 2 sides already, so we only need to paint an additional 4 sides for these cubes but we must not include the cubes at the edges.
The number of these cubes are (2*4)+(3*4)+(6*4)=44.
So surface area of painting is 44*4*20*20=70,400

We now have 116-44=72 cubes left. These cubes have only been painted on one side. So we need to paint 5 sides of each of these cubes.
Surface area of painting for these cubes is 72*5*20*20=144,000

We are now finished. The total additional surface area of painting we have had to do is the sum of all of these, i.e.
86,400+9,600+70,400+144,000= 310,400

So the price we pay is = (price per surface area) * (surface area of painting)
Therefore, price = (92/73,600) * (310,400) = 388.

If someone knows of a much quicker way to solve this then please post it, also could you please confirm if this is the correct answer? Also out of interest, was this a GMAT question? I hope not, as there is no way I would have been able to do this in 2 mins lol... Thanks....

Hey I am overwhelmed to say u have got the answer correct.
I found this question in one of the Hard math problems file set.

But i am unable to comprehend how u re labelled the cuboid dimensions to 4, 5 and 8 and how u got to determine the number of cubes in the center etc. It would be great if you can help me with this.

The reason why I re-labelled the cube to 4x5x8 was to work with 'units' rather than length. To see what I mean look at the image that I uploaded. Let's say that the length of each side is 90 cm, and say that each individual small cube has a length of 30 cm. We can either describe the cube in terms of its total length i.e. 90x90x90, or we can describe it in terms of the number of cubes it contains i.e. 3x3x3. So the cube contains 27 little cubes.

In the example in the question, the cube has dimensions 80cmx100cmx160cm. We then determined that each individual cube has dimension 20cmx20cmx20cm. Therefore, one side of the cuboid will be able to fit a maximum of 4 cubes (since 80/20=4), one side will be able to fit 5 (100/20=5), and one side will fit 8 (160/20=8).

The reason why we decide to do this is that we no longer care about the total dimension of the original cuboid since we have cut it up into cubes, so essentially we just want to know how many cubes there are and how many sides of them we want to paint. So we are in a way 'dismantling' the cuboid to just leave over the individual cubes.

To determine the number of cubes in the center, it is helpful to work the other. Imagine you have a 2x2 cube, and that you want to add an extra 'layer' on top of it. You will have to place a cube on top of every cube, and this will increase the dimension of each side by 2. In the cube in the picture, if you add a cube over all the existing cubes, the resulting cube will have a dimension of 5x5x5. So whenever you want to work out the number of cubes inside the "outer layer" of cubes, then you just reduce the dimension of each side by 2, and then multiply it out. As a result, looking at the cube in the picture, if you remove all the cubes on the outer layer then you're only left with 1 cube, since (3-2)*(3-2)*(3-2)=1.

If you need me to explain anything else then just tell me because my explanations are sometimes badly worded and can be a bit confusing.

Hi,

There might be a short-cut right after figuring out there are 160 identical cubes.

The additional cost of painting = Unpainted area * Cost per unit area of painting

The additional cost of painting = (TSA of all the 160 smaller cubes - TSA of the huge cuboid) * (92 / TSA of the huge cuboid)

TSA of all the 160 smaller cubes = 6 * 40 * 40 * 160 (TSA = 6 * side * side) = 384,000

TSA of the huge cuboid = 2 * ( (80 * 100) + (100 * 160) + (160*80) ) = 73,600

The additional cost of painting = (384,000 - 73600) * (92 / 73600) = 388.

This way, we do not have to group the cubes based on the faces which have already been painted.

Hope this helps. Thanks.

Yes, this method is definitely quicker! I completely missed that. I think it's important to understand both methods to have a good understanding about cube questions in general. But thanks alot for pointing out the quicker- definitely a much better method!

4GMAT_Mumbai wrote:Hi,

There might be a short-cut right after figuring out there are 160 identical cubes.

The additional cost of painting = Unpainted area * Cost per unit area of painting

The additional cost of painting = (TSA of all the 160 smaller cubes - TSA of the huge cuboid) * (92 / TSA of the huge cuboid)

TSA of all the 160 smaller cubes = 6 * 40 * 40 * 160 (TSA = 6 * side * side) = 384,000

TSA of the huge cuboid = 2 * ( (80 * 100) + (100 * 160) + (160*80) ) = 73,600

The additional cost of painting = (384,000 - 73600) * (92 / 73600) = 388.

This way, we do not have to group the cubes based on the faces which have already been painted.

Hope this helps. Thanks.

Genius! thanks Naveenan.
guess this needs some level of quant grasping !!

Both the methods are very good in terms of understanding the cube problem..one is intense on theory and gets you the basics correct while the other is faster.

Thanks Naveen and SM2010.