The dimensions of a ream of paper are 8 1/2 inches by 11 inches by 2 1

Question

The dimensions of a ream of paper are  8 \frac{1}{2}  inches by 11 inches by  2 \frac{1}{2}  inches. The inside dimensions of a carton that will hold exactly 12 reams of paper could be

A.  8 \frac{1}{2}  in by 11 in by 12 in
B. 17 in by 11 in by 15 in
C. 17 in by 22 in by 3 in
D. 51 in by 66 in by 15 in
E. 102 in by 132 in by 30 in

PS25602.01
Quantitative Review 2020 NEW QUESTION

TornDilemma · Accepted Answer

Volume of a ream of paper= \frac{17}{2}*11*\frac{5}{2} 
Volume of 12 reams of paper= \frac{17}{2}*11*\frac{5}{2}*12=17*11*5*3=17*11*15.

Imo, Option B.

ScottTargetTestPrep · Answer

Let’s analyze each answer choice.

A) Since 12/2.5 = 4.8, the dimensions in choice A can only hold 4.8 reams of paper.

B) Since 17/8.5 = 2, 15/2.5 = 6 and 2 x 6 = 12, the dimensions in choice B can hold 12 reams of paper. To fill the carton, then, two reams will be placed side-by-side in the bottom and then 5 reams will be stacked on top of each.of the two reams, making 2 stacks of 6 reams each.

Answer: B

Snezanelle · Answer

I approached the problem in a different way: I thought that 2,5 in is the width of the ream of paper so we multiply it by 12 to get the needed space to hold all paper. I do understand that there is no information that 2,5 in is the width but it's a common logic to assume that other numbers (8,5 in and 11 in) are too big to be the width of a realm. We get 30 in as the 3rd dimension and our number in E.
What's wrong with my reasoning?

globaldesi · Answer

given dimensions are 17/2,11 and 5/2
total number of boxes to be filed are 12.
thus 12 can be written as (1,1,12), (2,2,3),(1,2,6) .. in any form
where each of these in (x,y,z) represent the total number divisible by corresponding side of carton
for quick solution
lets start looking at options
a)  8(1/2) in by 11 in by 12 in divisible by corresponding l,b,h in 1 , 1 and not divisible(the divisor must be an integer)
B. 17 in by 11 in by 15 in divisible by l, b and h in 2,1, 6 
Hence answer B

sakshamchhabra · Answer

One important point to note in the questions stem is that it mentions the word  "exactly" , which makes the second most selected option (E) wrong.

The dimensions of the box in option E are sufficient enough to fit in 12 reams of paper but we need a box with   exact   dimensions.

Now, the given dimensions are--> 8.5, 11, 2.5 , lets consider l=8.5 , b=11, and h=2.5 ( the order would not matter here )

now all we need to do is- 8.5 * 11 * 2.5 * 12, the problem is that in options the "12" has been spread out

i.e.in option (B) 8.5*2 * 11*1 * 2.5*6 
basically 2 colums of 6 reams each which makes a total of 12 reams

Considering all the options

A. 8.5 in by 11 in by 12 in

wrong because only the third dimension has been changed and it should've been 30 (2.5*12) i.e all the boxes are stacked in 1 single column on each other

B. 17 in by 11 in by 15 in

C. 17 in by 22 in by 3 in

same as option A the 3rd dimension should've been 7.5 (2.5*3)

D. 51 in by 66 in by 15 in

A box with these dimensions can sure accommodate 12 reams but we need a box with exact dimensions

E. 102 in by 132 in by 30 i

A box with these dimensions can sure accommodate 12 reams but we need a box with exact dimensions

energetics · Answer

You don't need to do any calculations, just match choices:

(L=17/2 * W=11 * H=5/2) * 12 reams 
(17*11*5)/4 * 12 
( 17*11*5 ) * 3 = Volume necessary

If you notice, it's 4 prime numbers, the answer must have ONLY these numbers, i.e. since it says "exactly" the volume must be the same, and the prime factorization of any n>1 is unique.

A.  8 \frac{1}{2}  in by 11 in by 12 in --> 17/2 * 11 * 3*4 --> 17 * 11 * 3 --> missing 5
B. 17 in by 11 in by 15 in --> 17*11*(5*3) --> this is correct
C. 17 in by 22 in by 3 in --> 17*(11*2) * 3 --> extra 2, missing 5
D. 51 in by 66 in by 15 in --> (17*3)*(11*6)*(5*3) --> extra 3^2 and 2
E. 102 in by 132 in by 30 in --> (17*6)(11*12)*(5*6) --> extra 3^2 and 2^4

Snezanelle 
It actually doesn't matter what you take as length, width, height because we're only multiplying. So if you have a L=2, W=4, H=6 rectangular solid, the volume would be 48. If you are stacking L=1, W=1, H=1 boxes, you could stack the boxes in different ways (for example 2*4*6 or 2*2*12, both Vol = 48). The idea is that you can take different L,W,H as long as they equate to the same volume. It's the same in this problem, other possible correct choices could be ( 51*11*5 ) or (17*33*5 ). The issue with answer E) is that it also multiplies the other dimensions, so in fact the volume is 72 times too large.

Side note: It becomes trickier when the dimensions of the object being stacked don't conform to the ratio of box (this problem says "exactly" so it's not an issue). In that case you have left over space or go over the max of at least 1 dimension (there's min-max problems that test this, where knowing Vol is not enough, you need to have the specific dimensions to say how many objects you can place in the box).

P.S. If experts can confirm my analysis it would be appreciated.

Sarjaria84 · Answer

A ream measures = 8.5 x 11 x 2.5, and we need to fit in exactly 12 reams in the box.

Therefore, of the answer choices whichever option will give 12 will be the answer, as in,

Option A = 8.5 x 11 x 12

This means that we can fit,

8.5/8.5 x 11/11 x 12/2.5

= 1 x 1 x 4 = 4 reams, not our answer

(2.5 x 5 = 12.5, thus for the given size (anything, height , length or width) of 12 we cannot completely fit 5 reams and can only fit 4. Therefore, 1 x 1 x  4 )

Option B = 17 x 11 x 15

We can fit,

17/8.5 x 11/11 x 15/2.5

= 2 x 1 x 6 = 12 reams

Thus answer is option 'B'.

ayushkumar22941 · Answer

i thought that if we put ream in random way that would destroy its physical structure so did not choose b.

MHIKER · Answer

Volume of one ream =  \frac{17}{2}*11*\frac{5}{2}

Volume of 12 reams =  \frac{17}{2}*11*\frac{5}{2}*12=17*11*5*3=17*11*15.

The answer is B

CEdward · Answer

Question about terminology:

1. What is a 'ream'? and more importantly,

2. I often see such questions which state  "inside dimension" . What is meant by that?

Chitroll · Answer

This approach would've worked assuming that the rest of the dimensions did not change. Moreover, you can attack this problem by simply multiplying the given dimensions by 12 and then checking for that in the answer choices.

ccontesse · Answer

Please someone explain why my reasoning is wrong!

My reasoning for this question was entirely different. I was more restrictive with the calculation of the volume.  A ream of paper is not a liquid that you can fill in any figure that has the same volume. It is a solid rectangular figure , if you don't want to bend the papers you need to keep 2 dimension and the other one multiply it by 12 to fill the 12 rectangular figures.

If you are going to pack 12 reams of paper one on top from the other, you only need to multiply by 12 the height of the ream. You have 3 different options of height: 8.5, 11 and 2.5. So there are 3 different possible answers:

1) Assuming 8.5 is the height:  102in by 11in by 2.5in 
1) Assuming 11 is the height:  8.5in by 132in by 2.5in 
1) Assuming 2.5 is the height:  8.5in by 11in by 30in

The correct answer (B) for example has a dimension of 17in by 11in by 15in. With my reasoning (paper is not a liquid and you cant or dont want to bend it) you can only pack  2  reams  if 17 is the height ,  1  ream  if 11 is the height  and  6  if  15 is the height

Edit: I figured out my problem, I assumed that you have to pack the reams one on top from the other. You can put some reams on top or on the side.

Abhishek009 · Answer

Volume of each ream is  \frac{17}{2} * 11 * \frac{5}{2}

Volume of box that can hold 12 reams is  \frac{17}{2} * 11 * \frac{5}{2}*12 = 17*11*5*3 > 17*11*15 , Answer must be (B)

tkorzhan1995 · Answer

GMATNinja , Bunuel, could you please explain why we should not multiply each dimension by 12?

GMATNinja · Answer

Imagine what would happen if we stacked all 12 reams on top of each other. (If you're uncomfortable with the term "ream", just picture a rectangle.) Would all three dimensions increase by a factor of 12? Or just one or two of them?

Once you get your head around that, the question becomes much easier. Obviously, stacking 12 reams vertically isn't the only way to arrange 12 reams inside a carton. But hopefully, the image of a vertical stack will get you to some useful logic about how the dimensions change when we start stacking things up.

I hope that helps a bit!

jaydavido · Answer

the boxes might be stocked 6 and 6. E is wrong because the dimensions are wrong, if stacked on top of each other the two other widths and lengths would stay the same.

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The dimensions of a ream of paper are 8 1/2 inches by 11 inches by 2 1

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The dimensions of a ream of paper are 8 1/2 inches by 11 inches by 2 1

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