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 Your Progress
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Stoolfi ur explanation for statement (2) is good. I dont know why I miss out those things. Probably due to time constrains. Anyways from (1) we are given that PR= sqrt 3, we know that PR=OR (Bisects), so OP will be 2sqrt3 and as both OP and OS are radius, OP=OS. So OS= 2sqrt3
Re: In the figure above, O is the center of the circle. If OP an [#permalink]
08 Sep 2013, 00:38
1
This post received KUDOS
1
This post was BOOKMARKED
fozzzy wrote:
In the figure above, O is the center of the circle. If OP and QS are perpendicular and bisect each other. What is the length of OS?
1) The length of PR is \(\sqrt{3}\) 2) The length of QS is 6
Please provide detailed explanations!
In the given figure,OS=OP=radius "r" and PR=RO=r/2 QR=RS,angleORS=90 degrees We need find length of OS, which is radius of the circle
From stmt 1)PR= [square_root]3 and PR=r/2 hence radius=2*[square_root]3, hence 1 alone is sufficient From stmt 2)QS=6 => RS=3 (OP is bisector of QS,R is the midpoint of QS) OS=radius=r and OR=r/2 (R is the midpoint of OP, OP being the radius of the circle) r^2=(r/2)^2+3^2 r=2*[square_root]3 Hence stmt 2 alone is sufficient
Re: In the figure above, O is the center of the circle. If OP [#permalink]
10 Sep 2013, 02:19
1
This post received KUDOS
Expert's post
fozzzy wrote:
Is there any property tested here? I'm really clueless when it comes to geometry...
In the figure above, O is the center of the circle. If OP and QS are perpendicular and bisect each other. What is the length of OS?
Notice that OS is the radius of the circle.
(1) The length of PR is \(\sqrt{3}\). PR is half of the radius, so the radius is twice of that. Sufficient.
(2) The length of QS is 6. This one tests pythagorean theorem. Triangle ORS is a right triangle. We know that RS=QS/2=3 and we know that the other two sides are r (OS) and r/2 (OR) --> r^2=(r/2)^2+3^2 --> \(r=2\sqrt{3}\). Sufficient.
Re: In the figure above, O is the center of the circle. If OP [#permalink]
10 Sep 2013, 04:31
Expert's post
fozzzy wrote:
So for such questions never rely on the figure. In this particular question, the figure is misleading.
Why it's misleading?
OG13, page 150: Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.
OG13, page 272: A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2). Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight. You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees. All figures lie in a plane unless otherwise indicated.
Re: In the figure above, O is the center of the circle. If OP [#permalink]
28 Nov 2013, 16:56
Bunuel wrote:
fozzzy wrote:
So for such questions never rely on the figure. In this particular question, the figure is misleading.
Why it's misleading?
OG13, page 150: Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.
OG13, page 272: A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2). Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight. You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees. All figures lie in a plane unless otherwise indicated.
Hope it helps.
Hi Bunuel, question for you How do we know that PR = OR from the fact that the chord is perpendicular to the radius? I mean what's the property here if you will Thanks Cheers J
Re: In the figure above, O is the center of the circle. If OP [#permalink]
29 Nov 2013, 01:20
Expert's post
jlgdr wrote:
Bunuel wrote:
fozzzy wrote:
So for such questions never rely on the figure. In this particular question, the figure is misleading.
Why it's misleading?
OG13, page 150: Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.
OG13, page 272: A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2). Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight. You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees. All figures lie in a plane unless otherwise indicated.
Hope it helps.
Hi Bunuel, question for you How do we know that PR = OR from the fact that the chord is perpendicular to the radius? I mean what's the property here if you will Thanks Cheers J
We are told that OP and QS are perpendicular and bisect each other, which means that PR = OR and QR = RS.
Re: In the figure above, O is the center of the circle. If OP [#permalink]
04 Dec 2013, 09:52
From statement 2.) it seems like we know that triangle ORS is a 30:60:90. How do we know this?
I think I got it but I want to be sure:
We know OP = OS because they are both radiuses. QS bisects OP so OR (where R is the midpoint of OP) is equal to 1/2 OS. We can say that OS = 2x and OR = x. From statement 2 we know that QS = 6 and RS therefore must = 3 because it is bisected by OP. We can now set up Pythagorean equation a^2 + b^2 = c^2 --> x^2 + 3^2 = 2x^2 --> x^2 + 9 = 2x^2 --> Subtract x^2 from both sides --> x^2 = 9 --> x = 3. If x = 3, then 2x (i.e. line OS) = 6.
Re: In the figure above, O is the center of the circle. If OP [#permalink]
04 Aug 2015, 04:02
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...