A rectangular box, with dimensions of 12 inches by 18 inches by 10 inc

Question

A rectangular box, with dimensions of 12 inches by 18 inches by 10 inches, contains soup cans. If each can is a cylinder with a radius of 3 inches and a height of 5 inches, what is the maximum number of soup cans that the box can contain?

A. 6
B. 12
C. 15
D. 30
E. 48

A. 6
B. 12
C. 15
D. 30
E. 48

BrentGMATPrepNow · Accepted Answer

The question comes down to determining how to orient the box. Do, we make the HEIGHT of the box 12 inches, 18 inches or 10 inches? 
Well, since the height of each CAN is 5 inches, we can see that making the HEIGHT of the box 10 inches is a great course of action. Otherwise, there will be empty space above the cans.

Also, if the radius of each can is 3 inches, then the DIAMETER = 6 inches
So, we want the base of the box to be such that we can fit as many 6-inch diameters into it. 
Since 12 inches and 18 inches are both multiples of 6 inches, we can maximize the number of cans by making the 12-inch by 18-inch part of the box the base.

Since with a 12-inch by 18-inch base, we can place 3 rows of 2 cans on the base. This is a total of 6 cans (so far)
Then, we can place a second level of 6 cans on top of the first 6 cans.

So, the most cans we can place in the box = 6 + 6 = 12

Answer: B

Cheers,
Brent

Michaelkalend13 · Answer

Is the answer C? Please advise.

Bunuel · Answer

You can check the Official Answer (OA) of a question under the spoiler in the first post. The OA for this question is B.

Michaelkalend13 · Answer

Hi Bunuel! Thanks for your prompt response. 
Can you please provide an explanation for the answer? Thanks in advance!

ashudhall · Answer

18/6=3 cans can be arranged along the length and12/6 = 2 along the breadth.so no of cans arranged in one layer would be 3*2=6 
since the height of box is 10, so two layers can be arrangedi.e 6*2=12
Hence answer should be B

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Abhishek009 · Answer

Maximum \ number \ of \ soup \ cans = \frac{Volume \ of \ Rectangular \ Box}{Volume \ of \ Cylinder}

Maximum \ number \ of \ soup \ cans = \frac{12*18*10}{π*3^2*5}

Maximum \ number \ of \ soup \ cans = \frac{12*18*10}{22/7*9*5}

Maximum \ number \ of \ soup \ cans = \frac{12*2*2*7}{22}

Maximum \ number \ of \ soup \ cans ~ 15.xx

Thus, the answer must be (C) 15

BrentGMATPrepNow · Answer

Hi Abhishek009,

This answer would be correct if we could squish the cans into any shape we wish, in which case we need only concern ourselves with the volume of each case. 
However, I believe the intent of this question is to keep the same cylindrical shape of each can, in which case there will be some space in the packed box that is just air.

Cheers,
Brent

ScottTargetTestPrep · Answer

Notice that each cylinder will take up an area of 6 x 6 = 36 square inches on the base of the box, even though the base area of each cylinder is less than that. Therefore, we can think of the soup cans as 6 x 6 x 5 rectangular boxes instead of cylinders. If the big box is laid on the 12 x 18 base, there will be 12/6 = 2 rows of cans and 18/6 = 3 columns of cans; totaling 2 x 3 = 6 cans on the base. Since the height is 10, 10/5 = 2 cans be stacked; therefore 6 x 2 = 12 cans is the maximum number to fit in the box.

Answer: B

A rectangular box, with dimensions of 12 inches by 18 inches by 10 inc

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A rectangular box, with dimensions of 12 inches by 18 inches by 10 inc

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