any one feel confident on Geometry?

IMO E.

1)the ratio of OP = 1x, ratio of OQ = 2x

r = 3x.

Insufficient. x could be any number.

2) doesn't give any information about radius other than AP = BP.

together we can draw a right triangle by making r as the hypotenuse: 3x.

OP = 1x

and figure out AP as 2*sqrt(2)x

But this is not sufficient either as we have no information on x.

For problem number 2, you are missing the second requirement.

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brick2009: Master | Next Rank: 500 Posts; Posts: 171; Joined: Mon Jun 01, 2009 8:59 pm; Thanked: 8 times

2nd prob

by brick2009 » Sat Jul 11, 2009 7:34 am

The answer C

1.) L(arc) = ((r deg)360) x 2*pi*radius

L(arcc) = 18, but we have 2 variables, angle R and radius ---> insuf

2.) Same equation... L(arc) cant be calculated as radius is unknow..--> insuf

When you combine the two.. r=s=> its an equilateral triangle..as bottom two angles are S...

So angle is 60..L(arc) = 18...only Radius unknow... hence u can find the circumference

So C..

[/list]

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kc_raj: Master | Next Rank: 500 Posts; Posts: 117; Joined: Wed May 13, 2009 7:41 am; Thanked: 5 times

IMO

by kc_raj » Sat Jul 11, 2009 7:49 am

IMO both E,

brick why angle is 60 why not both r=s=45,

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gmat740: MBA Student; Posts: 1194; Joined: Sat Aug 16, 2008 9:42 pm; Location: Paris, France; Thanked: 71 times; Followed by:17 members; GMAT Score:710

by gmat740 » Sat Jul 11, 2009 9:33 am

brick why angle is 60 why not both r=s=45,

dude,it is an equilateral triangle.

all angles are equal = 60

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crackthetest: Senior | Next Rank: 100 Posts; Posts: 40; Joined: Mon Feb 16, 2009 8:50 pm; Thanked: 3 times

by crackthetest » Sat Jul 11, 2009 9:51 am

IMO,

Q1: E

(ssmiles - the length of AP I calculated to be sqrt(2)x, ( b^2 = 3x^2 - x^2) )

Q2: C (same method as brick)

You can calculate without working for XOY, but yes angle(XOY) = 2* angle(XYZ).

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Naruto: Master | Next Rank: 500 Posts; Posts: 121; Joined: Thu Jun 11, 2009 1:16 am; Thanked: 4 times; GMAT Score:720

by Naruto » Sun Jul 12, 2009 1:30 am

OA please

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kc_raj: Master | Next Rank: 500 Posts; Posts: 117; Joined: Wed May 13, 2009 7:41 am; Thanked: 5 times

IMO

by kc_raj » Sun Jul 12, 2009 7:14 am

My bad,

IMO 1) E,
2) C

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tohellandback: Legendary Member; Posts: 752; Joined: Sun May 17, 2009 11:04 pm; Location: Tokyo; Thanked: 81 times; GMAT Score:680

by tohellandback » Sun Jul 12, 2009 7:17 am

IMO E,C
for qstn1) neither of the statements are sifficient. and combining both, you can draw any number of circles with that ratio.

for qstn2: draw a chord parallel to the side containing two angles s.
the triangle formed is an equilateral triangle. Just by symmetry you can tell that the circumference will be 27

The powers of two are bloody impolite!!

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real2008: Legendary Member; Posts: 527; Joined: Thu May 01, 2008 12:06 am; Thanked: 7 times

by real2008 » Sun Jul 12, 2009 12:13 pm

both answers are E

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mgshorrGMAT: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Tue Jul 14, 2009 6:15 am; Location: DC; Thanked: 1 times; GMAT Score:710

2nd Question...

by mgshorrGMAT » Thu Jul 16, 2009 2:08 am

Using Statement 2, You can determine that it is a 60-60-60 Triangle...Therefore the arc XZ is 120 degrees, or 1/3 of the circle...Using Statement 1, you are told that the remaining portion of the circle, the other 2/3 is = to 18...

2/3=18 so 3/3=27...Answer should be C....