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

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Manager
Joined: 24 Jul 2009
Posts: 74
Location: United States
GMAT 1: 590 Q48 V24
Followers: 2

Kudos [?]: 59 [0], given: 124

The diagram shows circular sections of a cone. If x is 20 pe [#permalink]  15 Nov 2009, 08:49
00:00

Difficulty:

(N/A)

Question Stats:

33% (01:38) correct 67% (00:57) wrong based on 8 sessions
The diagram shows circular sections of a cone. If x is 20 percent
[Reveal] Spoiler: OA

Attachments

Cone&percentage_problem.jpg [ 31.8 KiB | Viewed 1172 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.
Math Expert
Joined: 02 Sep 2009
Posts: 28331
Followers: 4483

Kudos [?]: 45376 [3] , given: 6759

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: 74
Location: United States
GMAT 1: 590 Q48 V24
Followers: 2

Kudos [?]: 59 [0], given: 124

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: 203
Schools: Sloan '14 (S)
Followers: 3

Kudos [?]: 82 [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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