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A rectangular carton is filled to capacity with k identical cylindrica

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A rectangular carton is filled to capacity with k identical cylindrica  [#permalink]

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New post 13 Sep 2018, 04:14
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46% (01:39) correct 54% (02:04) wrong based on 50 sessions

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A rectangular carton is filled to capacity with k identical cylindrical cans of fruit that stand upright in rows and columns. If the height if each can were 4 times greater but the volume was the same, how many cans would the carton fit?

(1) The volume of each can is 60π centimeters square.
(2) k is 200.

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A rectangular carton is filled to capacity with k identical cylindrica  [#permalink]

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New post 17 Sep 2018, 00:11
Bunuel wrote:
A rectangular carton is filled to capacity with k identical cylindrical cans of fruit that stand upright in rows and columns. If the height if each can were 4 times greater but the volume was the same, how many cans would the carton fit?

(1) The volume of each can is 60π centimeters square.
(2) k is 200.


There are a few interpretations to this question, in all of them our guiding approach should be Logical:
To calculate the difference in volume of something 'new' and 'old' we need to know the relation between their heights/widths/lengths/radii.
We'll look for statements that give us this information, a Logical approach.

If the carton has a lid, i.e. we care that the cans aren't 'above' the height of the carton, the answer is (E):
If the carton's height is less than 4 times the cans' height, it cannot hold even 1 can! (as the taller can is too tall!)
If the carton's height is more than 4 times the cans' height, at least 1 can will fit.
As 0 and (some positive number) are distinct options, none of the information given is sufficient.

If we don't care about height at all, and all of our rows/columns are 'in a grid' (i.e. one under the other, in square packing) the answer is (B):
As the volume of each can stays the same, increasing the cans' height by a factor of 4 implies reducing the radius by a factor of 2.
That is, every 'short and fat' can takes up 2r*2r space and every 'tall and thin' can takes r*r space.
This means that we can fit 4 'tall and thin cans' instead of every 1 'short and fat' can.
Meaning that no matter what the actual volume of the cans or of k, the number of cans increases by a factor of 4.
So if k = 200, the new box fits 4*200 = 800 cans. (1) does not give the value of k and is insufficient.

The last option is that we don't know if our rows/columns are 'in a grid' or not. In this case the answer is again (E):
Doing the actual math is a bit involved, but as opposed to square packing where the number of cans is exactly 4k, in hexagonal packing (see the above link) it is a bit less. You can figure this out by drawing a picture and SEEing that you can't fit 4 new cans instead of one old one.
So in this case the new number of cans could be 4k or could be a different number.
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A rectangular carton is filled to capacity with k identical cylindrica &nbs [#permalink] 17 Sep 2018, 00:11
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