What is the maximum number of rectangular blocks, each with

What is the maximum number of retangular blocks, each with dimensions 12 centimeters by 6 centimeters by 4 centimeters, that will fit inside rectangular box X?

1. When box X is filled with the blocks and rests on a certain side, there are 25 blocks in the bottom layer.
2. the inside dimensions of Box X are 60 centimeters by 30 centimeters by 20 centimeters.

25 = 25*1 or 5*5 so the only possible arrangement of the boxes is 25 of one row or 5*5 rows with each arrangement we can have differnt dimension for the big box........insuff

from 2

thinking volume 12*6*4 = 288 for each samll box , 36000 = volume of big one thus max number is 36000/288 = 125 boxes..suff

B is my answer

I didn't get this explanation. Can someone explain? Thanks.

What is the maximum number of rectangular blocks, each with dimensions 12cms by 6cms by 4cms, that will fit inside rectangular box X?

1. When box X is filled with the blocks and rests on a certain side, there are 25 blocks in the bottom layer.
2. The inside dimensions of box X are 60cms by 30cms by 20 cms.

Clearly (B)

Statement 1: This will give us two dimensions of the larger box, but since we do not know the height of the larger box, this is insufficient.
Statement 2: We know the dimensions of the larger box so we can calculate. Sufficient.

B it is

Merging similar topics.



Please DO NOT reword or shorten the questions.

Bumping for review and further discussion.

statement (1): Now here are 25 blocks in bottom layer but we don't know which face is on the lower layer so we can't calculate the max no of block.

so statement is insufficient.

statement (2): dimension of rectangular box X = 60x30x20 cm3
volume = 60x30x20 = 36000 cm3
so dividing by volume of blocks , no. of max blocks that can be accommodated = 36000/288 = 125.

also area of faces of rectangular box = 60x30, 30x20, 60x20 = 1800,600, 1200 cm3.

now 1800/72 = 25 so 12x6 face exactly fits on 60x30 face. 20/4=5 thus overall 125 blocks sets fully inside it ..which was max capacity.
so it is sufficient
also if we check for other dimensions like 600/24 = 25, and 60/12=5 ..so it same
also if we check 1200/48 = 25, and 30/6 = 5..so it is same.

if we take any other face ... u can't put max. blocks.. Hence B

Most of you are making a BIG mistake. The correct answer is B but many of you are incorrectly justifying your answer choice.

In this scenario, you can't multiply the dimensions of the box to attain the total volume and then divide that number by the volume of the rectangular box to determine the total number of blocks that could fit. That's wrong.

For example, I could ask what is the total number of rectangular blocks (10x2x1) that fit in a rectangular box of (5x4x1).

Do you see the problem? With the method most of you are using, your answer to this question would be 1 block. However, while both volumes equal 20 cubic units, you can't fit a rectangular block with side length 10 into a box with side length 5 no matter how you try.

The correct way of approaching this problem is demonstrated by Brunel:



Do NOT make this mistake on test day.

Hi All,

This DS question is more about logic than about math. We're given the dimensions of the blocks (12x6x4) and we're asked how many of these blocks would fit in a certain box. Since the dimensions have common multiples (12, 24, 36, etc.), we'll need to know a lot of specific info to answer this question.

1): 25 blocks fit in the bottom layer of the box.

Fact 1 gives us no information on the dimensions of the bottom layer, if the blocks are in 1 row or 5 rows, the height of the box, etc.
Fact 1 is INSUFFICIENT

2) The dimensions of the box are 60x30x20

With the dimensions of the box, we can absolutely figure out how many blocks can fit.
Fact 2 is SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

I'm going to explain how can we solve this question in detail.

Let's call A is rectangular that need to be filled in to block X
=> We have to determine how many A will fit in the X.

First, we must choose 2 edges that are rectangular A and X are resting on and the edges of the rest are the height.

Then the 1st layer (or bottom layer) we can put = square of the bottom face of X/ square of the bottom face of A
However, u should choose 2 sides resting that each side of X is divisible by each side of A.

For example:
A: 12 6 4
X: 60 30 20
opt1: we should choose:
- the bottom face of A with 2 sides: 6 and 4
- the bottom face of X with 2 sides: 30 and 20
so that
1. we can put (30:6)x(20:4)= 5x5 = 25 blocks to X - the bottom layer
The block X is 60 tall => a maximum of 60:12=5 => a maximum of 5 layers will fit inside the box
=> 25x5= 125 blocks would fit inside the box

opt 2: if we choose
- the bottom face of A with 2 sides: 6 and 4
- the bottom face of X with 2 sides: 60 and 20
so that we can put (60:6)x(20:4)= 10x5=50 blocks - the bottom layers

The block X is 30 tall => a maximum of 30:12=2,... => a maximum of 2 layers will fit inside the box

=> 50x2= 100 bocks would fit inside the box

So we will choose option 1 to have a maximum of blocks A that fit inside the block X
=> Statement 2 is sufficient

---------------------
Statement 1. Since there are 25 blocks in the bottom layer
1. If block A is resting on the side that is 6 and 4.
The side on which block X is resting could be B: 30 and 20
=> there would be (30:6)x(20:4)= 25 blocks (as the statement 1 said)

If the box X is 12 tall => a maximum of 25 blocks A would fit inside the box X
If the box X is 36 tall => a maximum of 25x(36:12)=75 blocks A would fit inside the box X

=> Not sufficient

Hi Bunuel, KarishmaB for concept clarity, say we have to find the maximum number that can fit. Am I wrong to think that that would require all 9 cases to be calculated or is there a smart way to select the l*b pair for area covered and h for layers made to maximize the fit?

3 of the 9 cases: (where we assume the smaller boxes are kept 12*6 with height 4)

Bigger box: 30*60 with height 20
Number of boxes per layer = (30*60)/(12*6)= 25
Number of layers = 20/4=5
Total fit = 25*5 = 125

Bigger box: 20*60 with height 30
Number of boxes per layer = (20*60)/(12*6)= 100/6 (fraction so some space is wasted, nearest lower integer = 16)
Number of layers = 30/4
Total fit = 16*30/4 =120

Bigger box: 20*30 with height 60
Number of boxes per layer = (20*30)/(12*6)= 25/3 (fraction so some space is wasted, nearest lower integer = 8)
Number of layers = 60/4=15
Total fit = 8*15 =120

Now here the maximum fit 125 but can the other orientation of smaller box (say 6*4 with height 12) give us a better fit?. For example in a similar question(https://gmatclub.com/forum/what-is-the-maximum-number-of-cubic-blocks-of-wood-that-will-fit-into-247028.html#p1907074) i see that all 3 cases gave maximum fit, in that case was there a way to foresee that by looking at the dimensions itself and avoiding the calculation of all 3 but doing just 1 .

Thank you for your time.

Whether we will need to calculate the values in various orientations depends on the sizes given . We need to maximise the volume used of Box X. So I will look for orientations in which the rectangular blocks fit perfectly. Dimensions of the box are 60, 30 and 20 and those of the blocks are 12, 6 and 4. Since 30 is divisible by 6 only (out of 12, 6 and 4), side of 6 should correspond to side of 30 to ensure that no volume is leftover. Also 20 is divisible by 4 so side of 4 should correspond to 20. Since 60 is divisible by all 12, 6 and 4, we have no problem there.

This is how we will place the blocks inside the box such that all volume of the box is utilised.
Consider the length of box X as 60 so the rectangular block can be placed inside such that its length side is 12 which means 5 of these will be placed next to each other. The width of box X is 30 cm so 6 cm should be the width of the rectangular blocks. So 5 of these will be placed width-wise too. Hence, we will have 25 boxes in the bottom layer. Now, height of box X is 20 cm and height of each rectangular block is 4 cm so 5 layers can be stacked one of top of another.

Since all the volume is utilised, there is no way we can place anymore blocks inside the box. Had the dimensions not matched perfectly, we would need to consider a few orientations.
Say if the box dimensions were 60, 30 and 18, how many maximum blocks can be fit in?

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