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Bunuel
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chetan2u

Can you further explain these types of questions where there is 'excess' room left over? I have run into these a few times now and don't know what to do with that leftover space and how to think about it.

So r = 3, d = 6, h = 6

18/6 = 3
12/6 = 2

3 x 2 = 6 <-- # of cylinders that can fit on the face defined by the height and width

Next, since the height is 6...20/6 = 3

3 x 6 = 18 cylinders can fit.

We have left over

18 x 12 x 2

I concluded that you cannot fit anything with a height > 2

Hence, C is the answer.

But what if we had a larger box that could fit more?
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Please let me know if this is a correct approach.
1. Since radius is 3 so diameter is 6. Height is also 6. So I assumed the cylinders as cubes with dimensions 6*6*6. Vol is 216 unit^3.
2. Vol of the box is (12*18*20)=4320 unit^3.
3. 4320/216=20.
4. Since it is a cylinder, some amount of unused space will persist. I looked for a value in the options nearer to 20. Option C matched.
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