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# Circle

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Manager
Joined: 29 May 2007
Posts: 97
Followers: 2

Kudos [?]: 2 [0], given: 0

Circle [#permalink]  17 Aug 2007, 04:54
Folks.See attachment for the question
Attachments

Score.png [ 11.91 KiB | Viewed 729 times ]

VP
Joined: 10 Jun 2007
Posts: 1465
Followers: 6

Kudos [?]: 133 [0], given: 0

Re: Circle [#permalink]  17 Aug 2007, 05:21
r0m3416 wrote:
Folks.See attachment for the question

E.

Solving what it is asking for...
Set R = radius of circle
r = radius of semi circle
P*(R^2) < P*(r^2) / 2
Same as if
(R/r)^2 < 1/2
R/r < 1/1.4
R/r < 5/7

(1) R/r < 1
Can't tell, INSUFFICIENT

(2) 2*P*R < P*r + 2r
=> 2*P*R < r*(P+2)
=> R/r < (P+2) / (2*P)
=> R/r < 5.14 / 6.28
=> R/r <~ 5/6
Can't tell, INSUFFICIENT

Together, still get R/r < 5/6, INSUFFICIENT.
Manager
Joined: 15 Aug 2007
Posts: 70
Followers: 2

Kudos [?]: 5 [0], given: 0

Correct approach...

I ll also go for E....
Director
Joined: 12 Jul 2007
Posts: 865
Followers: 12

Kudos [?]: 218 [0], given: 0

good explanation. E it is.
Senior Manager
Joined: 13 May 2007
Posts: 252
Followers: 2

Kudos [?]: 7 [0], given: 0

Re: Circle [#permalink]  19 Aug 2007, 01:04
that is one hell of a way to solve this

bkk145 wrote:
r0m3416 wrote:
Folks.See attachment for the question

E.

Solving what it is asking for...
Set R = radius of circle
r = radius of semi circle
P*(R^2) < P*(r^2) / 2
Same as if
(R/r)^2 < 1/2
R/r < 1/1.4
R/r < 5/7

(1) R/r < 1
Can't tell, INSUFFICIENT

(2) 2*P*R <P> 2*P*R <r> R/r <P> R/r <5> R/r <~ 5/6
Can't tell, INSUFFICIENT

Together, still get R/r < 5/6, INSUFFICIENT.
Re: Circle   [#permalink] 19 Aug 2007, 01:04
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