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

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Bunuel
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If a brick which is .2 inches by .05 inches by .05 inches is placed in [#permalink] New post   13 Jun 2018, 11:24
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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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B
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EgmatQuantExpert
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Re: If a brick which is .2 inches by .05 inches by .05 inches is placed in [#permalink] New post   13 Jun 2018, 12:22
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Solution



Given:
    • Dimension of a brick = 0.2 * 0.05 * 0.05
    • Dimension of a cylinder = radius 0.05 and height 0.2
    • Brick is placed inside the cylinder

To find:
    • What portion of the cylinder is not taken up by the brick

Approach and Working:
    • Volume of the cylinder = \(π * 0.05^2 * 0.2\) = 0.0005π
    • Volume of brick = 0.2 * 0.05 * 0.05 = 0.0005

Therefore, the portion of cylinder not taken up by the brick = 0.0005 π – 0.0005

Hence, the correct answer is option B.

Answer: B
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Re: If a brick which is .2 inches by .05 inches by .05 inches is placed in [#permalink] New post   17 Jun 2018, 19:49
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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. \(0\)


The volume of the brick is .2 x .05 x .05 = .0005

The volume of the cylinder is π x .05^2 x .2 = .0005π

Thus, .0005π - .0005 cubic inches of the cylinder is not taken up by the brick.

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