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Senior Manager
Joined: 24 Mar 2011
Posts: 450
Location: Texas
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Kudos [?]: 184 [0], given: 20

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03 May 2011, 10:22
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there is no figure..
Senior Manager
Joined: 21 Jul 2010
Posts: 272
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Kudos [?]: 29 [0], given: 17

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03 May 2011, 14:02
You might want to repost in GMAT quant section, as this is the share your GMAT experience section... Might get more help that way...

Posted from my mobile device
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Kudos [?]: 254 [0], given: 10

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04 May 2011, 21:32
let the angle be x,

(x/360)* pi * 4^2 = 2 pi

x = 2 * 360 / 16 = 45

Thus C.
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Senior Manager
Joined: 21 Jul 2010
Posts: 272
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Kudos [?]: 29 [0], given: 17

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04 May 2011, 23:37
amit2k9 wrote:
let the angle be x,

(x/360)* pi * 4^2 = 2 pi

x = 2 * 360 / 16 = 45

Thus C.

Either you are clairvoyant, or know what problem # this is... care to share what # out of what problem set it is?
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1334
Followers: 17

Kudos [?]: 254 [0], given: 10

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04 May 2011, 23:42
humphy wrote:
amit2k9 wrote:
let the angle be x,

(x/360)* pi * 4^2 = 2 pi

x = 2 * 360 / 16 = 45

Thus C.

Either you are clairvoyant, or know what problem # this is... care to share what # out of what problem set it is?

I just used the whatever I know,

area of a sector = (unknown angle / 360 deg) * pi * (radius)^2
since radius is given and area of the shaded region is given as 2 * pi.

I substituted the values and got 45 deg.
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Re: in the figure   [#permalink] 04 May 2011, 23:42
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