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Cylindrical tennis-ball cans are being packed into cartons.

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Cylindrical tennis-ball cans are being packed into cartons.  [#permalink] New post 22 Sep 2005, 05:09
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Cylindrical tennis-ball cans are being packed into cartons. If each can has a radius of 2 inches and a height of 12 inches, and the dimensions of each carton are 14 inches by 16 inches by 20 inches, what is the maximum number of tennis-ball cans that can fit in each carton?

(A) 12
(B) 15
(C) 20
(D) 24
(E) 4
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I got 20 [#permalink] New post 22 Sep 2005, 05:35
I got 20 (C). I don't know if this correct, but since the stem doesn't say that you have to stack them on top of each other, so stuff them in where the can fit.
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 [#permalink] New post 22 Sep 2005, 06:07
I made it 91 = (20/2 * 16/2) + 20/2 + 1

Have I missed something?
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 [#permalink] New post 22 Sep 2005, 06:24
I get 20...

Well you have to think of it logically...

so we know the diameter of the can is 4...

the height is 12...

so lets look at it, if we lay the cans on the side...say with the side measuring 14, then we can only put 3 such cans...

so in order to maximize, say we lay the cans in a fashion that we put em on the side measuring 20...that way we can have 5 cans laying on the side, then we can stack em along the side measuring 16, that way we can have 4 such cans...

so we can have 4*5=20 cans...
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 [#permalink] New post 22 Sep 2005, 07:03
Staggering the rows cans would allow more to fit than placing them side-by-each in neat rows, though.
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 [#permalink] New post 22 Sep 2005, 08:54
Does anyone knows standard way to solve this?
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 [#permalink] New post 22 Sep 2005, 11:17
duttsit wrote:
Does anyone knows standard way to solve this?


Let me share how I got 20.
First I figured the diameter is 4 and hight is 12

so in order to maximize the storage utilization 14 has to be used as hight of the box and 16X20 has to used as length and width. so we have an are of 16X20 to be filled up by cylinder with will be taking as much as 4X4 area.
so (16X20)/(4X4) = 20 cans.

still you have 2X14 free areas but you can not fit anything there.

So 20 is the right answer.
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 [#permalink] New post 22 Sep 2005, 18:40
Isn't there a proper way( :oops: ) to solve these sums. My sense of logic and math don't go well together!
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 [#permalink] New post 23 Sep 2005, 04:55
The OA is 20.

Fresinha and Nakib's logic are correct. You have to put the box in a way to maximize the space and then it's a simple calculation from there
  [#permalink] 23 Sep 2005, 04:55
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