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The diagram shows circular sections of a cone. If x is 20 pe

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Manager
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The diagram shows circular sections of a cone. If x is 20 pe [#permalink] New post 15 Nov 2009, 08:49
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A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

20% (02:17) correct 80% (00:57) wrong based on 7 sessions
The diagram shows circular sections of a cone. If x is 20 percent
[Reveal] Spoiler: OA

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Cone&percentage_problem.jpg
Cone&percentage_problem.jpg [ 31.8 KiB | Viewed 1041 times ]


Last edited by Bunuel on 07 Dec 2013, 05:14, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Cone & percentage problem [#permalink] New post 15 Nov 2009, 09:23
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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
Manager
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Joined: 24 Jul 2009
Posts: 76
Location: United States
GMAT 1: 590 Q48 V24
Followers: 2

Kudos [?]: 35 [0], given: 124

Re: Cone & percentage problem [#permalink] New post 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.
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Re: Cone & percentage problem [#permalink] New post 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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The diagram shows circular sections of a cone. If x is 20 pe

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