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# Cone & percentage problem

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Manager
Joined: 24 Jul 2009
Posts: 79
Location: United States
GMAT 1: 590 Q48 V24
Followers: 2

Kudos [?]: 30 [0], given: 123

Cone & percentage problem [#permalink]  15 Nov 2009, 08:49
00:00

Difficulty:

5% (low)

Question Stats:

33% (02:17) correct 66% (00:54) wrong based on 3 sessions
source 800.
D

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Cone&percentage_problem.jpg [ 31.8 KiB | Viewed 873 times ]

Math Expert
Joined: 02 Sep 2009
Posts: 15073
Followers: 2519

Kudos [?]: 15480 [3] , given: 1552

Re: Cone & percentage problem [#permalink]  15 Nov 2009, 09:23
3
KUDOS
Expert's post
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%
_________________
Manager
Joined: 24 Jul 2009
Posts: 79
Location: United States
GMAT 1: 590 Q48 V24
Followers: 2

Kudos [?]: 30 [0], given: 123

Re: Cone & percentage problem [#permalink]  15 Nov 2009, 09:56
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
Posts: 204
Schools: Sloan '14 (S)
Followers: 3

Kudos [?]: 66 [0], given: 16

Re: Cone & percentage problem [#permalink]  15 Nov 2009, 10:04
Nice catch Bunuel... I was going to say E for this one as well.
Re: Cone & percentage problem   [#permalink] 15 Nov 2009, 10:04
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