![]() |
| In the figure shown, point O is the center of |
|
GMAT 2007
Guest
|
We know AOD is a straight line -
Angle(AOB) + Angle(BOC) + Angle(COD) = 180---(1) We know Angle(AOB) = Angle(BAO)---(2) Because(AB = OB) Substitute in (1) Angle(BAO) + Angle(BOC) + Angle(COD) = 180 Also, Angle(BOC) = 180-2(Angle(OBC)) Because OB = OC (Radii of Circle) Substitute in (1) Angle(BAO) + 180 - 2(Angle(OBC)) + Angle(COD) = 180 Also, in Triangle BAO, Angle OBC is the exterior angle so Angle(OBC) = Angle(BAO) + Angle(BOA) (Exterior angle is equal to the sum of the opposite angles) So Angle(OBC) = 2(Angle(BAO)) ---from (2) substitute it back in (1) Angle(BAO) + 180 - 4Angle(BAO) + Angle(COD) = 180 We know Angle (COD) already so, (A) is also sufficient. Hope it helps GMAT 2007 |
||||||||||||
|
|
|||||||||||||
|
Stacey Koprince
MGMAT STAFF
|
This one's tough to show without the ability to draw. Try following through GMAT 2007's approach, above, or mine, below, while drawing at every step so you can see what's going on.
Re-draw just the straight lines - ignore the arc of the semi-circle. Label COD 60. Label BAO and BOA with a variable (I'll use "a") and label CBO and BCO with another variable (I'll use "b"). Use that info to label ABO. I know that AC = 180, so ABO = 180-b. That's part of a small triangle, which I can write: a+a+(180-b) = 180 which simplifies to 2a - b = 0 I know using the exterior angle rule that COD = BAO + BCO (look at the big triangle only). Given my labels, I can re-write that equation as: a+b = 60. I can now combine the two equations. 2a - b = 0 a+b = 60 Substitute however you want, but solve for a, not b (since that's what you are asked to find). 2a = b and b = 60-a, so 2a = 60-a. 3a=60. a = 20. |
||||||||||||
|
|
|||||||||||||
|
Luci
Guest
|
Thank you guys, great explanations. I think Stacey´s is easier, because it is more direct. Although I think GMAT 2007 is going to perform terrific in the quant part because he/she solves the toughest problems :-)
Thanks again. Luci |
||||||||||||
|
|
|||||||||||||
| Pls clarify |
|
rschunti
Guest
|
In this figure what is the reason of beleiving that "Line aegment "ABC" is a straight line?It could be possible that Line segments "AB" and "BC" may not be colinear as nothing is mentioned in this question that proves this.?Pls can you clarify the reason why we are assuming that ABC is a straight line?
|
||||||||||||
|
|
|||||||||||||
| Re: Pls clarify |
|
Ron Purewal
MGMAT STAFF
|
collinearity, and for that matter linearity in general, is one of the few things that you are allowed to assume on the gmat. the reason is pretty utilitarian: if you couldn't assume that things that look like straight lines are actually straight lines, then no diagram would ever be useful! (imagine if that cute little triangle in your diagram might turn out to be a heptagon... you get the picture) |
||||||||||||||
|
|
|||||||||||||||
|
mclaren7
Guest
|
Dear friends
AB = OC = OB Angle BAO = Angle BOA = X Angle ABO = 180 - 2X, therefore angle CBO = 2x. since OC = OB angle CBO = angle BCO = 2X, angle BOC = 180 - 4X Therefore angle COD = 180 - (180-4X) - X = 3X 1: angle COD = 3x = 60 , X = 20 angle BAO = 20 <== sufficient 2: angle BCO = 2x = 40 , X = 20 angle BAO = 20 <== sufficient KH |
||||||||||||
|
|
|||||||||||||
| In the figure shown, point O is the center of |
|
||
|
Powered by phpBB © phpBB Group
Content © Manhattan GMAT Forums
*GMAT and GMAT CAT are registered trademarks of the Graduate Management Admission Council,
which neither sponsors nor endorses this test preparation service.
Content © Manhattan GMAT Forums
*GMAT and GMAT CAT are registered trademarks of the Graduate Management Admission Council,
which neither sponsors nor endorses this test preparation service.



