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

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If a brick which is .2 inches by .05 inches by .05 inches is [#permalink] New post 30 Mar 2018, 09:50
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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. \(0\)
[Reveal] Spoiler: OA

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Re: If a brick which is .2 inches by .05 inches by .05 inches is [#permalink] New post 31 Mar 2018, 17:22
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This is basically like a shaded area problem, except with volume. If we figure out the volume of the cylinder and subtract that of the brick, we'll get the answer. It's always good practice to eliminate wrong answers that we know can't be correct.

In this case, E must be out. How can we fit a box-shaped brick into a cylinder and have it take up all of the volume exactly? Less obviously, D is out. We're going to have the volume of the cylinder minus the volume of the box, and the former must have a π in it somewhere. Answer choice D has something with a π but it doesn't subtract anything. So we're left with A, B, or C.

Another key in this problem is that we don't actually need both volumes. Since the cylinder volumes are different in A, B, and C, as are the brick volumes, we only need to figure out one of them in order to eliminate the others. Let's choose the easier, which I'd say is the brick.

This brick is .05x.05x.2 inches. Seems pretty small for a brick. Even "pebble" seems a bit too grandiose a term for this thing. Anyway, the easiest way to multiply a bunch of decimals is to just ignore the decimals, multiply the resulting numbers, and then simply add back in the decimals we ignored later. So 5x5x2 = 50. Since we ignored a total of 5 decimals, we'll put them back in now, giving us .0005. Since B is the only option subtracting that amount, B must be the answer.

EXTRA WORK: In case you're nervous about ignoring the cylinder, we can do that too. The volume of a cylinder is πr^2h, or the area of the circle times the height. Since we've been told that the radius is .05 and the height is .2, we're going to multiply the exact same numbers we did a moment ago, and π: πx.05x.05x.2. This should give us an answer of .0005π, which you'll not matches the first term in answer choice B.
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Re: If a brick which is .2 inches by .05 inches by .05 inches is [#permalink] New post 31 Mar 2018, 20:57
portion of the cylinder is not taken up by the brick = volume of cylinder(VC) - volume of brick(VB)

VC= pi * r*r*h =pi* 0.05*0.05*0.2 =0.0005pi
VB= l*b*h= 0.2*0.05*0.05 = 0.0005

ans= 0.0005pi-0.0005
Re: If a brick which is .2 inches by .05 inches by .05 inches is   [#permalink] 31 Mar 2018, 20:57
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