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M23-06

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New post 16 Sep 2014, 01:18
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New post 16 Sep 2014, 01:18
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New post 26 Nov 2014, 17:12
Hello,

I am not sure I understand the answer.

I calculate the volume of both solids, then divide the rectangule's by the cube's. Why is this approach wrong?

Thanks !
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New post 27 Nov 2014, 03:31
sandrodab wrote:
Hello,

I am not sure I understand the answer.

I calculate the volume of both solids, then divide the rectangule's by the cube's. Why is this approach wrong?

Thanks !


Notice that we cannot cut the cubes. 12x16X10 box can accommodate only two layers of 4x4x4 cubes (so 2*12=25 cubes) on its 12X16 face. 12X16X2 spaces will be left unfilled because we cannot squeeze any 4x4x4 cubes there.
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New post 05 Dec 2014, 09:45
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I may be off base but could the way I found the answer was 4/10 =2.5 boxes rounded down to 2 , 4/12 = 3, 4/16= 4..... 2x3x4= 24 ..... I may have gotten lucky based on the numbers. Please tell me if my theory is flawed.
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New post 21 Jan 2015, 05:01
Bunuel wrote:
Official Solution:

What is the maximum number of 4x4x4 cubes that can fit in a rectangular box measuring 10x12x16 ?

A. 12
B. 18
C. 20
D. 24
E. 30

The 12x16 floor can be covered with 12 cubes. Another 12 cubes will form the second layer. The height of this construction will be 8; the box won't accommodate any more cubes.

Answer: D


How do we know that that the floor is 10x16 and it is not 12x16?

I am asking because since we are not told which value corresponds to l,w,h, then I would think that we need to use the volume, which does not require you to know which of the values is l,w or h.

In this sense, why isn't it volume of rectangular box 10x12x16 = 1920 divided by surface area of the cube 6*16= 96, which gives 20?

At first I thought of using the volume for the cube as well, but what we need to know is its outside area, right? So, the space it would assume inside the box.

Now, what I am not sure about is whether we also need to match each of the l,w,h of the square with those of the box. In this case though, shouldn't we be told specifically which value corresponds to l,w,h?
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New post 21 Jan 2015, 05:32
pacifist85 wrote:
Bunuel wrote:
Official Solution:

What is the maximum number of 4x4x4 cubes that can fit in a rectangular box measuring 10x12x16 ?

A. 12
B. 18
C. 20
D. 24
E. 30

The 12x16 floor can be covered with 12 cubes. Another 12 cubes will form the second layer. The height of this construction will be 8; the box won't accommodate any more cubes.

Answer: D


How do we know that that the floor is 10x16 and it is not 12x16?

I am asking because since we are not told which value corresponds to l,w,h, then I would think that we need to use the volume, which does not require you to know which of the values is l,w or h.

In this sense, why isn't it volume of rectangular box 10x12x16 = 1920 divided by surface area of the cube 6*16= 96, which gives 20?

At first I thought of using the volume for the cube as well, but what we need to know is its outside area, right? So, the space it would assume inside the box.

Now, what I am not sure about is whether we also need to match each of the l,w,h of the square with those of the box. In this case though, shouldn't we be told specifically which value corresponds to l,w,h?


Positioning of the box is not important. This is fairly simple problem from everyday life. You have 10x12x16 box and you want to know the maximum number of 4x4x4 cubes that can fit in it. No matter what dimensions the floor has the answer would be the same.
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New post 19 Mar 2015, 17:45
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As you see in the image I attached.
total 6 cubes (4*4*4 size) can get fit at level 1.
For H = 16 , there will be 4 level.
isn't it? :-D
so total cubes that can fit in the rectangular box will be 6 x 4 = 24.
I hope this helps. :-D
>> !!!

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New post 13 Jul 2015, 09:28
Bunuel wrote:
Official Solution:

What is the maximum number of 4x4x4 cubes that can fit in a rectangular box measuring 10x12x16 ?

A. 12
B. 18
C. 20
D. 24
E. 30

The 12x16 floor can be covered with 12 cubes. Another 12 cubes will form the second layer. The height of this construction will be 8; the box won't accommodate any more cubes.

Answer: D


One question, cannot the floor also be 10*16? Because this would mean that the first layer would fit 160/16 = 10 cubes, up to the second layer 20, and up to the third layer 30. The height of this construction would be the side of the cube times the number of the layers, so 3*4 = 12.

In this case then the solid can fit 30 cubes measuring 4*4*4...or not...? :?

OK I just realised my mistake. When I take two sides, one of them being 10, this leaves out 2, because the same side of the cube would be 4. So, it would fit 2 cubes and a half. So, I missed the amount of space each side occupies.
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New post 15 Aug 2015, 02:39
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. why shouldnt i use the area formulae as in 10*12*16/64
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New post 15 Feb 2016, 15:08
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Alternate approach to this question :

Find the highest multiple less or equal to the edges of the cuboid. You have a cube with an edge of 4, to be fit in 10X12X16. So that would be 08*12*16 cube space that can be occupied. Now find the cubes -> 2*3*4 = 24.
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New post 21 Feb 2018, 09:59
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Cube size is=4x4x4
Rectangular box size=10x12x16
We have to find no of cubes in rectangular box.

Available size for cubes to put in rectangular box=8x12x16
Because we can put 2 cubes in 10th side, 3 cubes in 12th side and 4 cubes in 16th side.

So, 8x12x16/4x4x4 = 1536/64 = 24

Answer=D
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New post 17 Sep 2018, 07:50
Bunuel wrote:
What is the maximum number of 4x4x4 cubes that can fit in a rectangular box measuring 10x12x16 ?

A. 12
B. 18
C. 20
D. 24
E. 30



buneul if the question was to number of 5*5*5 cubes then it is 12...am i right ??

10*12*16 can take only 10*10*15 /5*5*5 =12?


thanks
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New post 16 Nov 2018, 11:17
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Please elaborate.
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New post 21 Jun 2019, 09:43
Bunuel

What is the problem with the following solution:

We have a cuboid with dimensions : 10*12*16, now since these are not marked I can take L, B and H randomly.
If 16 is the length 10 be the breadth and 12 be the height - The floor area is 160 so I can have maximum of two parallel rows of 4*4*4 cubes, as the breadth is only 10.
Total number cubes = 10*2 =20 so far.

Now available height = 12-4 = 8, Area of side face ( B*H) = 8*10 = 80, so total number 4*4*4 that we can fit in here = 80/16= 5

Maximum number of cubes= 20+5= 25

And 25 is not even the option :(
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New post 23 Jun 2019, 02:22
Manku wrote:
Bunuel

What is the problem with the following solution:

We have a cuboid with dimensions : 10*12*16, now since these are not marked I can take L, B and H randomly.
If 16 is the length 10 be the breadth and 12 be the height - The floor area is 160 so I can have maximum of two parallel rows of 4*4*4 cubes, as the breadth is only 10.
Total number cubes = 10*2 =20 so far.

Now available height = 12-4 = 8, Area of side face ( B*H) = 8*10 = 80, so total number 4*4*4 that we can fit in here = 80/16= 5

Maximum number of cubes= 20+5= 25

And 25 is not even the option :(


If the base is 16 by 10, then you can place 4 cubes along 16 and only 3 along 10, so on the base you get 8 cubes only and since the height is 12, then you'll have 3 layers: 8*3 =24.
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New post 01 Aug 2019, 11:37
I saw the official explanation but don't completely understand why is the approach I describe below wrong:

Given that the dimension of the rectangular box is 10*12*16, and the dimension of the cube to be fit in is 4*4*4.
So I calculated the volume of the rectangular box using the formula lxbxh = 10x12x16= 1920.
The surface area of the small cube can be found using the formula 6a^2 = 96.

So the number of boxes that can be fit should be 1920/96 = 20.

Could someone please help me with this?
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