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A rectangular slab 2 inches depth, 20 length and 8 in width

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CEO
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A rectangular slab 2 inches depth, 20 length and 8 in width [#permalink] New post 09 Oct 2003, 19:11
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A
B
C
D
E

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A rectangular slab 2 inches depth, 20 length and 8 in width is completely
wrapped with paper. if a single sheet of paper is used without
patching the dimensions of paper, it would be of size (all in inches)?

a) 17 * 25
b) 21 * 24
c) 24 * 12
d) 24 * 14
e) 26 * 14

thanks
praetorian

Last edited by Praetorian on 14 Oct 2003, 22:57, edited 1 time in total.
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 [#permalink] New post 14 Oct 2003, 22:10
Not sure about the right answer but would pick C since is the only one that does not have a 7 or 17 factor in the multiplication. The surface area of the rectangular is:

2*(20*8) + 2*(20*2) + 2(8*2) = 432 or (2^4) * (3^3)
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 [#permalink] New post 18 Oct 2003, 11:49
B is the only answer that meets the requirements of the problem.

The 20" long edge of the slab will be placed along the 24" length of the paper so that we wil have 2" of extra lengths on both sides of the slab. The paper will be then wrapped around the slab (in a direction perpendicular to the length of the slab). This way, we will have 21" of paper to wrap in the direction perpendicular to the length of the slab. We will be able to wrap the paper with 1" overlap (8 x 2 + 2 + 2 is the length to be wrapped perpendicular to the 20" edge of the slab).

Afetr wrapping the paper as descibed above, we will have 2" left on both sides of the length of the slab. We can just flap those extra lenghths to cover the sides of the slab.

I hope I am correct in my understanding of the problem.
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 [#permalink] New post 20 Oct 2003, 09:33
am1974 wrote:
B is the only answer that meets the requirements of the problem.

The 20" long edge of the slab will be placed along the 24" length of the paper so that we wil have 2" of extra lengths on both sides of the slab. The paper will be then wrapped around the slab (in a direction perpendicular to the length of the slab). This way, we will have 21" of paper to wrap in the direction perpendicular to the length of the slab. We will be able to wrap the paper with 1" overlap (8 x 2 + 2 + 2 is the length to be wrapped perpendicular to the 20" edge of the slab).

Afetr wrapping the paper as descibed above, we will have 2" left on both sides of the length of the slab. We can just flap those extra lenghths to cover the sides of the slab.

I hope I am correct in my understanding of the problem.


I agree with B. I find the best way is to draw a picture here.
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  [#permalink] 20 Oct 2003, 09:33
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A rectangular slab 2 inches depth, 20 length and 8 in width

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