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Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Re: Quant Rev #161 Advanced Triangles Problem [#permalink]
18 Jan 2011, 19:51

21

This post received KUDOS

Expert's post

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This post was BOOKMARKED

tonebeeze wrote:

In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)?

a. 40 b. 36 c. 34 d. 32 e. 30

A small diagram helps:

Attachment:

Ques1.jpg [ 9.23 KiB | Viewed 16751 times ]

In the figure, you see (180 - 2x) + y = 180 (straight angle) so 2x = y

Also, the moment you see the circle and its two radii, mark them equal and the corresponding angles equal. (the red angle = blue angle) x + 180 - 2y = y 5x = 180 (from above, y = 2x) x = 36 _________________

Re: Quant Rev #161 Advanced Triangles Problem [#permalink]
19 Jan 2011, 22:57

in your original diagram OA and OB don't appear to be radii - the end points are off the circle... without getting them to be equal I dont think you can answer this question

Re: Quant Rev #161 Advanced Triangles Problem [#permalink]
11 Mar 2011, 00:14

tonebeeze wrote:

In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)?

a. 40 b. 36 c. 34 d. 32 e. 30

The figure attached is ambiguous. A needs to be shown on the circle, or at least it should be mentioned that A is also a point on the circle. If A is any point inside the circle, you cant solve this question from given information.

Re: In the figure attached, point O is the center of the circle [#permalink]
14 Dec 2013, 02:35

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Re: In the figure attached, point O is the center of the circle [#permalink]
14 Dec 2013, 16:41

In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)?

OC = AC = AB tells us that triangle OAC and BAC are isosceles. Also, as with any triangle on a straight line with an exterior angle, the exterior angle (or in this case, angle ACB) can be found by subtracting the interior angle from 180.

We know that in triangle OAC (because it is an isosceles triangle) that angle o and angle a are equal to one another. Therefore, angle c = 180 - x - x --> angle c = 180-2x. Angle C (of the smaller isosceles triangle) also happens to share the same angle measurement as angle B. Furthermore, because OA and OB are radii, we know that they equal one another and that angle A = angle B.

We know that angle c in the smaller isosceles triangle = 180 - the measure of the obtuse angle C. As shown above the obtuse angle C = (180-2x) So, the small angle C = 180 - (180-2x) --> angle C = 2x. So, angle C = B = 2x. If angle A = B then A = 2x and becaause the obtuse triangle has two x measurements, we know that the measure of A in the small isosceles triangle = x. Therefore, in the small isosceles triangle we have 2x+2x+x = 180. 5x = 180. x = 36.

Re: In the figure attached, point O is the center of the circle [#permalink]
18 Feb 2014, 01:13

Is there another way to do it without assuming that OA and OB are radii? We are unable to state that OA and OB are radii based on the ambiguous diagram, and it doesn't mention OA and OB being the radii anywhere in the problem. It just so happened to work out in this problem, but maybe next time assuming it to be the radii won't result in the correct answer.

Re: In the figure attached, point O is the center of the circle [#permalink]
18 Feb 2014, 01:28

Expert's post

chuanchiehlo wrote:

Is there another way to do it without assuming that OA and OB are radii? We are unable to state that OA and OB are radii based on the ambiguous diagram, and it doesn't mention OA and OB being the radii anywhere in the problem. It just so happened to work out in this problem, but maybe next time assuming it to be the radii won't result in the correct answer.

Actually the diagram in the source (OG) is more accurate:

Re: Quant Rev #161 Advanced Triangles Problem [#permalink]
25 May 2014, 13:31

VeritasPrepKarishma wrote:

tonebeeze wrote:

In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)?

a. 40 b. 36 c. 34 d. 32 e. 30

A small diagram helps:

Attachment:

Ques1.jpg

In the figure, you see (180 - 2x) + y = 180 (straight angle) so 2x = y

Also, the moment you see the circle and its two radii, mark them equal and the corresponding angles equal. (the red angle = blue angle) x + 180 - 2y = y 5x = 180 (from above, y = 2x) x = 36

Hi Karishma,

Thanks for the detailed explanation but I still have a nagging issue. I was having a hard time setting this up with the variables. For example, <OAC, you have it broken up as 180-2y and x, which i see why you did. When I started working on this problem, I labeled <ACB and <ABC as "x" because of the congruent lines and I labeled OAC and COA as "x" too because all of those lines were same according to the question stem. Doing so, conflicted with the fact that <OAB and <ABO are equal and that's when my whole setup went kaboom Why is that wrong?

I guess I'm not following why you gave some variables "x" vs. some "y"?

Re: Quant Rev #161 Advanced Triangles Problem [#permalink]
25 May 2014, 18:42

Expert's post

russ9 wrote:

VeritasPrepKarishma wrote:

tonebeeze wrote:

In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)?

a. 40 b. 36 c. 34 d. 32 e. 30

A small diagram helps:

Attachment:

Ques1.jpg

In the figure, you see (180 - 2x) + y = 180 (straight angle) so 2x = y

Also, the moment you see the circle and its two radii, mark them equal and the corresponding angles equal. (the red angle = blue angle) x + 180 - 2y = y 5x = 180 (from above, y = 2x) x = 36

Hi Karishma,

Thanks for the detailed explanation but I still have a nagging issue. I was having a hard time setting this up with the variables. For example, <OAC, you have it broken up as 180-2y and x, which i see why you did. When I started working on this problem, I labeled <ACB and <ABC as "x" because of the congruent lines and I labeled OAC and COA as "x" too because all of those lines were same according to the question stem. Doing so, conflicted with the fact that <OAB and <ABO are equal and that's when my whole setup went kaboom Why is that wrong?

I guess I'm not following why you gave some variables "x" vs. some "y"?

Thanks!

When two sides of a triangle are equal, the two opposite angles are equal. But can you say what the two angles are? No. Say a triangle has two sides of length 5 cm each. Do we know the measure of equal angles? No. They could be 40-40 or 50-50 or 80-80 etc. So if you have two different triangles with 2 sides of length 5 cm each, the equal angle could have different measures - in one triangle the equal angles could be 50-50, in the other triangle, the equal angles could be 70-70.

In triangle OAC, since OC = AC, you have two equal angles as x each. The third angle here is 180 - 2x.

In triangle ACB, since AC = AB, angle ACB = angle ABC but what makes you say that they must be x each too? This is a different triangle. Even if the sides have the same length as the sides of triangle OAC, there is no reason to believe that the equal angles need to be x each. So you call the angles y. The third angle here is 180 - 2y. _________________

Re: Quant Rev #161 Advanced Triangles Problem [#permalink]
27 May 2014, 15:15

VeritasPrepKarishma wrote:

russ9 wrote:

Hi Karishma,

Thanks for the detailed explanation but I still have a nagging issue. I was having a hard time setting this up with the variables. For example, <OAC, you have it broken up as 180-2y and x, which i see why you did. When I started working on this problem, I labeled <ACB and <ABC as "x" because of the congruent lines and I labeled OAC and COA as "x" too because all of those lines were same according to the question stem. Doing so, conflicted with the fact that <OAB and <ABO are equal and that's when my whole setup went kaboom Why is that wrong?

I guess I'm not following why you gave some variables "x" vs. some "y"?

Thanks!

When two sides of a triangle are equal, the two opposite angles are equal. But can you say what the two angles are? No. Say a triangle has two sides of length 5 cm each. Do we know the measure of equal angles? No. They could be 40-40 or 50-50 or 80-80 etc. So if you have two different triangles with 2 sides of length 5 cm each, the equal angle could have different measures - in one triangle the equal angles could be 50-50, in the other triangle, the equal angles could be 70-70.

In triangle OAC, since OC = AC, you have two equal angles as x each. The third angle here is 180 - 2x.

In triangle ACB, since AC = AB, angle ACB = angle ABC but what makes you say that they must be x each too? This is a different triangle. Even if the sides have the same length as the sides of triangle OAC, there is no reason to believe that the equal angles need to be x each. So you call the angles y. The third angle here is 180 - 2y.

Hi Karishma,

This is news to me -- you can't assume that because length AC is shared, that implies that the corresponding angle in Triangle ACO will equal the corresponding angle in ACB?

Re: Quant Rev #161 Advanced Triangles Problem [#permalink]
27 May 2014, 20:18

Expert's post

russ9 wrote:

Hi Karishma,

This is news to me -- you can't assume that because length AC is shared, that implies that the corresponding angle in Triangle ACO will equal the corresponding angle in ACB?

Hmm -- learn something new everyday! Thanks!

Corresponding angles are equal only when you have parallel lines with a common transversal. Make two triangles with a common side. Can you make the angles made on the common side such that the angles have very different measures? Sure!

Attachment:

Ques3.jpg [ 6.81 KiB | Viewed 7874 times ]

Here one angle is 90 degrees and the other is acute. _________________

Re: Quant Rev #161 Advanced Triangles Problem [#permalink]
28 May 2014, 08:15

VeritasPrepKarishma wrote:

russ9 wrote:

Hi Karishma,

This is news to me -- you can't assume that because length AC is shared, that implies that the corresponding angle in Triangle ACO will equal the corresponding angle in ACB?

Hmm -- learn something new everyday! Thanks!

Corresponding angles are equal only when you have parallel lines with a common transversal. Make two triangles with a common side. Can you make the angles made on the common side such that the angles have very different measures? Sure!

Attachment:

Ques3.jpg

Here one angle is 90 degrees and the other is acute.

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