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# If a brick which is .2 inches by .05 inches by .05 inches is placed in

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Math Expert
Joined: 02 Sep 2009
Posts: 58464
If a brick which is .2 inches by .05 inches by .05 inches is placed in  [#permalink]

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27 Feb 2019, 23:07
00:00

Difficulty:

55% (hard)

Question Stats:

44% (01:26) correct 56% (02:04) wrong based on 16 sessions

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If a brick which is .2 inches by .05 inches by .05 inches is placed inside a right circular cylinder with a radius of .05 and a height of .2, what portion of the cylinder is not taken up by the brick?

A. $$.005 \pi - .005$$

B. $$.0005 \pi - .0005$$

C. $$05 \pi - .05$$

D. $$.0005 \pi$$

E. $$0$$

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Re: If a brick which is .2 inches by .05 inches by .05 inches is placed in  [#permalink]

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28 Feb 2019, 03:29
Bunuel wrote:
If a brick which is .2 inches by .05 inches by .05 inches is placed inside a right circular cylinder with a radius of .05 and a height of .2, what portion of the cylinder is not taken up by the brick?

A. $$.005 \pi - .005$$

B. $$.0005 \pi - .0005$$

C. $$05 \pi - .05$$

D. $$.0005 \pi$$

E. $$0$$

vol of the cyld ; 0.05^2 * .2*pi = 5*10^-4 pi----1
and vol of brick = .2*.05*.05 ; 5*10^-4---2
so volume not taken by brick 1-2
i.e $$.0005 \pi - .0005$$
IMO B
Re: If a brick which is .2 inches by .05 inches by .05 inches is placed in   [#permalink] 28 Feb 2019, 03:29
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