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# GmatPrep #2 - triangle + semicircle

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GmatPrep #2 - triangle + semicircle [#permalink]  19 Mar 2009, 18:41
Another problem that I thought was interesting.
I'll post the OA later.
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Re: GmatPrep #2 - triangle + semicircle [#permalink]  19 Mar 2009, 19:46
Given :

AB = OC BOA = ?

Stmt 1 :

COD = 60 BOC is an isoceles triangle

OBC = 0CB = 60

AB0 = 180 -60 = 120

ABO is an isoceles triangle.

BAO = 30 .Sufficient.

Stmt 2:

Similarly BCO = 40.

BOC is an isoceles triangle OBC = 40

ABO = 140.

BAO = 20.

Sufficient.

Ans D
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Re: GmatPrep #2 - triangle + semicircle [#permalink]  20 Mar 2009, 01:57
1
KUDOS
tkarthi4u wrote:
Given :

AB = OC BOA = ?

Stmt 1 :

COD = 60 BOC is an isoceles triangle

OBC = 0CB = 60

AB0 = 180 -60 = 120

ABO is an isoceles triangle.

BAO = 30 .Sufficient.

Stmt 2:

Similarly BCO = 40.

BOC is an isoceles triangle OBC = 40

ABO = 140.

BAO = 20.

Sufficient.

Ans D

I am sorry but I didn't understand why OBC = 0CB = 60 in the first statement. We don't know the angle BOC. Could pls explain that. Although I agree with the answer D

My expaination of the first statement takes more time:
Let's take OBC = 0CB=x and BAO=BOA=y
We have a system of two equations:
2x=180-(180-60-y)
x=180-(180-2y)

2x=60+y
x=2y

3y=60
y=20
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Re: GmatPrep #2 - triangle + semicircle [#permalink]  20 Mar 2009, 11:05
tkarthi4u, how did you come up with "OBC = 0CB = 60"?
Isoceles triangle yes but the degree could be 45 as well?

Senior Manager
Joined: 16 Jan 2009
Posts: 361
Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
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Re: GmatPrep #2 - triangle + semicircle [#permalink]  20 Mar 2009, 12:03
Thanks nlutsenko for the explaination.
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Lahoosaher

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Re: GmatPrep #2 - triangle + semicircle [#permalink]  20 Mar 2009, 20:34
Thanks Guys.

I was wrong but i was bit lucky.

i got ur explaination.
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Re: GmatPrep #2 - triangle + semicircle [#permalink]  22 Mar 2009, 12:09
The OA is D.
Thanks to everyone.
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Re: GmatPrep #2 - triangle + semicircle [#permalink]  23 Mar 2009, 17:42
1
KUDOS
nlutsenko wrote:
tkarthi4u wrote:
Given :

AB = OC BOA = ?

Stmt 1 :

COD = 60 BOC is an isoceles triangle

OBC = 0CB = 60

AB0 = 180 -60 = 120

ABO is an isoceles triangle.

BAO = 30 .Sufficient.

Stmt 2:

Similarly BCO = 40.

BOC is an isoceles triangle OBC = 40

ABO = 140.

BAO = 20.

Sufficient.

Ans D

I am sorry but I didn't understand why OBC = 0CB = 60 in the first statement. We don't know the angle BOC. Could pls explain that. Although I agree with the answer D

My expaination of the first statement takes more time:
Let's take OBC = 0CB=x and BAO=BOA=y
We have a system of two equations:
2x=180-(180-60-y)
x=180-(180-2y)

2x=60+y
x=2y

3y=60
y=20

How do you know that OCB = OBC ?
Re: GmatPrep #2 - triangle + semicircle   [#permalink] 23 Mar 2009, 17:42
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