It is currently 22 Oct 2017, 05:22

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

In the figure above, point O is the center of the circle and

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 41884

Kudos [?]: 129248 [0], given: 12196

In the figure above, point O is the center of the circle and [#permalink]

Show Tags

16 Jan 2011, 16:56
Expert's post
100
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

60% (02:34) correct 40% (02:40) wrong based on 1114 sessions

HideShow timer Statistics

The Official Guide For GMAT® Quantitative Review, 2ND Edition

Attachment:

Untitled.png [ 13.16 KiB | Viewed 112918 times ]
In the figure above, point O is the center of the circle and OC = AC = AB. What is the value of x ?

(A) 40
(B) 36
(C) 34
(0) 32
(E) 30

Problem Solving
Question: 162
Category: Arithmetic Statistics
Page: 83
Difficulty: 600

GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

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:
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.

Thank you!
[Reveal] Spoiler: OA

_________________

Last edited by Bunuel on 16 Mar 2014, 05:42, edited 2 times in total.
Renamed the topic and edited the question.

Kudos [?]: 129248 [0], given: 12196

VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1285

Kudos [?]: 283 [16], given: 10

Show Tags

16 Jan 2011, 19:07
16
KUDOS
5
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

It goes like this -

In triangle OAB , Angle O + Angle A + Angle B = 180 -------1
OA = OB (Radius) => Angle A = Angle B

In triangle ACB, Angle C = Angle O + Angle OAC (Sum of interior opposite angles)
=> Angle ACB = 2x;
Also, AC = AB => Angle ACB = Angle ABC = 2x each

Thus Angle A = Angle B = 2x each.

So, substituting in 1

5x = 180 => x = 36 deg.

This will help.

Kudos [?]: 283 [16], given: 10

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7677

Kudos [?]: 17387 [40], given: 232

Location: Pune, India

Show Tags

18 Jan 2011, 20:51
40
KUDOS
Expert's post
16
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 95338 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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17387 [40], given: 232

Manager
Joined: 11 Jul 2010
Posts: 223

Kudos [?]: 105 [0], given: 20

Show Tags

19 Jan 2011, 23:57
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

Kudos [?]: 105 [0], given: 20

Senior Manager
Joined: 08 Nov 2010
Posts: 394

Kudos [?]: 128 [0], given: 161

Show Tags

20 Jan 2011, 00:03
I agree. i assumed that A and B are on the circle.
_________________

Kudos [?]: 128 [0], given: 161

Director
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 614

Kudos [?]: 1135 [1], given: 39

Show Tags

10 Mar 2011, 14:03
1
KUDOS
angle oac=x
angle acb=2x=angle b=2x=a (as oa and ob are radii)
so in triangle oab x+2x+3x=180
x=36
_________________

Collections:-
PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html
DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html
100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html
Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html
Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html

Kudos [?]: 1135 [1], given: 39

Manager
Joined: 14 Feb 2011
Posts: 185

Kudos [?]: 153 [0], given: 3

Show Tags

11 Mar 2011, 01:14
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.

Kudos [?]: 153 [0], given: 3

Senior Manager
Joined: 13 May 2013
Posts: 463

Kudos [?]: 198 [2], given: 134

Re: In the figure attached, point O is the center of the circle [#permalink]

Show Tags

14 Dec 2013, 17:41
2
KUDOS
1
This post was
BOOKMARKED
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.

b. 36

Kudos [?]: 198 [2], given: 134

Intern
Joined: 09 Feb 2014
Posts: 1

Kudos [?]: 1 [0], given: 21

Re: In the figure attached, point O is the center of the circle [#permalink]

Show Tags

18 Feb 2014, 02:13
1
This post was
BOOKMARKED
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.

Kudos [?]: 1 [0], given: 21

Math Expert
Joined: 02 Sep 2009
Posts: 41884

Kudos [?]: 129248 [0], given: 12196

Re: In the figure attached, point O is the center of the circle [#permalink]

Show Tags

18 Feb 2014, 02:28
Expert's post
1
This post was
BOOKMARKED
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:

Edited the original post.

Hope it helps.
_________________

Kudos [?]: 129248 [0], given: 12196

Senior Manager
Joined: 15 Aug 2013
Posts: 302

Kudos [?]: 82 [0], given: 23

Show Tags

25 May 2014, 14:31
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"?

Thanks!

Kudos [?]: 82 [0], given: 23

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7677

Kudos [?]: 17387 [2], given: 232

Location: Pune, India

Show Tags

25 May 2014, 19:42
2
KUDOS
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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17387 [2], given: 232

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1853

Kudos [?]: 2628 [27], given: 193

Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: In the figure above, point O is the center of the circle and [#permalink]

Show Tags

25 May 2014, 23:58
27
KUDOS
6
This post was
BOOKMARKED

Attachments

as.jpg [ 39.56 KiB | Viewed 86334 times ]

_________________

Kindly press "+1 Kudos" to appreciate

Kudos [?]: 2628 [27], given: 193

Senior Manager
Joined: 15 Aug 2013
Posts: 302

Kudos [?]: 82 [0], given: 23

Show Tags

27 May 2014, 16:15
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?

Hmm -- learn something new everyday! Thanks!

Kudos [?]: 82 [0], given: 23

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7677

Kudos [?]: 17387 [0], given: 232

Location: Pune, India

Show Tags

27 May 2014, 21:18
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 85949 times ]

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

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17387 [0], given: 232

Senior Manager
Joined: 15 Aug 2013
Posts: 302

Kudos [?]: 82 [0], given: 23

Show Tags

28 May 2014, 09:15
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.

Makes total sense. Thanks!

Kudos [?]: 82 [0], given: 23

Intern
Joined: 20 Dec 2014
Posts: 22

Kudos [?]: 3 [0], given: 31

Re: In the figure above, point O is the center of the circle and [#permalink]

Show Tags

29 Dec 2014, 14:43
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

Karishma - why is it that you did not have x+ x + (180 - 2y)+ y = 180 to include the entire triangle? I solved this equation by adding an additional line to form a diameter and then constructed another angle(titled z) and approached the solution that way but I am attempting to understand this method that you used to solve this problem.

Kudos [?]: 3 [0], given: 31

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7677

Kudos [?]: 17387 [0], given: 232

Location: Pune, India
Re: In the figure above, point O is the center of the circle and [#permalink]

Show Tags

29 Dec 2014, 20:07
GMAT01 wrote:
Karishma - why is it that you did not have x+ x + (180 - 2y)+ y = 180 to include the entire triangle? I solved this equation by adding an additional line to form a diameter and then constructed another angle(titled z) and approached the solution that way but I am attempting to understand this method that you used to solve this problem.

You need to find the relation between x and y. You can do it in any way you like; you will get the same result.

x+ x + (180 - 2y)+ y = 180
2x = y
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17387 [0], given: 232

Intern
Joined: 20 Dec 2014
Posts: 22

Kudos [?]: 3 [0], given: 31

Re: In the figure above, point O is the center of the circle and [#permalink]

Show Tags

29 Dec 2014, 20:38
VeritasPrepKarishma wrote:
GMAT01 wrote:
Karishma - why is it that you did not have x+ x + (180 - 2y)+ y = 180 to include the entire triangle? I solved this equation by adding an additional line to form a diameter and then constructed another angle(titled z) and approached the solution that way but I am attempting to understand this method that you used to solve this problem.

You need to find the relation between x and y. You can do it in any way you like; you will get the same result.

x+ x + (180 - 2y)+ y = 180
2x = y

Thank you.

I must have overlooked something because I get:

x+x+(180-2y)+y= 180
2x-y=0
2x-2x= 0

Kudos [?]: 3 [0], given: 31

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7677

Kudos [?]: 17387 [0], given: 232

Location: Pune, India
Re: In the figure above, point O is the center of the circle and [#permalink]

Show Tags

29 Dec 2014, 20:41
GMAT01 wrote:
VeritasPrepKarishma wrote:
GMAT01 wrote:
Karishma - why is it that you did not have x+ x + (180 - 2y)+ y = 180 to include the entire triangle? I solved this equation by adding an additional line to form a diameter and then constructed another angle(titled z) and approached the solution that way but I am attempting to understand this method that you used to solve this problem.

You need to find the relation between x and y. You can do it in any way you like; you will get the same result.

x+ x + (180 - 2y)+ y = 180
2x = y

Thank you.

I must have overlooked something because I get:

x+x+(180-2y)+y= 180
2x-y=0
2x-2x= 0

Ignore the highlighted step.
after 2x - y = 0, when you take y to the other side, you get
2x = y

Without doing this step, how did you substitute 2x for y in the highlighted step?
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17387 [0], given: 232

Re: In the figure above, point O is the center of the circle and   [#permalink] 29 Dec 2014, 20:41

Go to page    1   2   3    Next  [ 42 posts ]

Display posts from previous: Sort by