Cone & percentage problem

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Manager

Joined: 24 Jul 2009

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GMAT 1: 590 Q48 V24

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Cone & percentage problem [#permalink]

15 Nov 2009, 09:49

00:00

Question Stats:

0% (00:00) correct

100% (00:54) wrong

based on 0 sessions

source 800.

[Reveal] Spoiler: Answer

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Re: Cone & percentage problem [#permalink]

15 Nov 2009, 10:23

This post received
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Area of the bigger circle C: \pi*R^2
Area of the smaller circle B: \pi*(0.2*R)^2=0.04*\pi*R^2

\frac{C}{B}=\frac{\pi*R^2}{0.04*\pi*R^2}=25

C is 25 times bigger then B, which means 2400% more (like twice more is 100% more)

OR to get this directly:

\frac{Difference}{Area of the smaller circle B}=\frac{\pi*R^2-0.04*\pi*R^2}{0.04*\pi*R^2}=\frac{0.96}{0.04}=24

Transform 24 into the percentage --> 24*100=2400%
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Manager

Joined: 24 Jul 2009

Posts: 84

Location: United States

GMAT 1: 590 Q48 V24

Followers: 2

Kudos [?]: 29 [0], given: 123

Re: Cone & percentage problem [#permalink]

15 Nov 2009, 10:56

Ohh i missed that last step, i got confused with the OA, hence posted it here.
Thank you again n again Bunuel.

Manager

Joined: 13 Aug 2009

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Schools: Sloan '14 (S)

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Re: Cone & percentage problem [#permalink]

15 Nov 2009, 11:04

Nice catch Bunuel... I was going to say E for this one as well.

Re: Cone & percentage problem [#permalink] 15 Nov 2009, 11:04

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