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:

are you sure about \(\sqrt{3}\) as the answer. As per my calculations i got \(\sqrt{2}\) as the answer. If I am correct, then I will surely post my answer.
_________________

Step 1: Measure the radius. That will come out to be 2. Step 2: Observe the diagram. Since both OP and OQ are the radii, their measure will be 2 only. Step 3: Draw a line parallel to x axis, from P to Q. Step 4: Angle OQP and angle OPQ= 45 In the diagram attached, the angles marked X are 45 and 45 degrees. Step 5: Draw a perpendicular from Q to x axis. The point at which it meets the X axis, mark it as A. Step 5: The triangle OPA is a 45:45: 90 degree tiangle. This triangle has the property that the hypotenuese:side 1:side 2 will be in the ration \(\sqrt{2}\):1:1. Step 6: Since side 1(OA) and side 2(AQ) are equal, therefore let them be x. These will be equal to \(2/\sqrt{2}\). Step 7: The length of side 1 will come out to be \(2/\sqrt{2}\) or simply \(\sqrt{2}\).

Please let me know what my mistake, if my solution is wrong.

Step 1: Measure the radius. That will come out to be 2. Step 2: Observe the diagram. Since both OP and OQ are the radii, their measure will be 2 only. Step 3: Draw a line parallel to x axis, from P to Q. Step 4: Angle OQP and angle OPQ= 45 In the diagram attached, the angles marked X are 45 and 45 degrees. Step 5: Draw a perpendicular from Q to x axis. The point at which it meets the X axis, mark it as A. Step 5: The triangle OPA is a 45:45: 90 degree tiangle. This triangle has the property that the hypotenuese:side 1:side 2 will be in the ration \(\sqrt{2}\):1:1. Step 6: Since side 1(OA) and side 2(AQ) are equal, therefore let them be x. These will be equal to \(2/\sqrt{2}\). Step 7: The length of side 1 will come out to be \(2/\sqrt{2}\) or simply \(\sqrt{2}\).

Please let me know what my mistake, if my solution is wrong.

your solution is wrong in Step 3. PQ is parallel to x axis is an assumption in the solution and has not been stated in question. Figures are not drawn to scale.
_________________

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

Show Tags

10 Dec 2013, 05:34

Bunuel wrote:

zz0vlb wrote:

I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.

In the figure above, points P and Q lie on the circle with center O. What is the value of s?

Yes you are right.

Point Q in I quadrant \((1, \sqrt{3})\); Point P in II quadrant \((-\sqrt{3}, 1)\); Point T in III quadrant \((-1, -\sqrt{3})\) (OT perpendicular to OP); Point R in IV quadrant \((\sqrt{3},-1)\) (OR perpendicular to OQ).

Is it the case of reflection in Y axis ? I doubt that but still want to clarify..

I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.

In the figure above, points P and Q lie on the circle with center O. What is the value of s?

Yes you are right.

Point Q in I quadrant \((1, \sqrt{3})\); Point P in II quadrant \((-\sqrt{3}, 1)\); Point T in III quadrant \((-1, -\sqrt{3})\) (OT perpendicular to OP); Point R in IV quadrant \((\sqrt{3},-1)\) (OR perpendicular to OQ).

Is it the case of reflection in Y axis ? I doubt that but still want to clarify..

No. \((-\sqrt{3}, 1)\) is not a reflection of \((1, \sqrt{3})\) around the Y-axis. The reflection of \((1, \sqrt{3})\) around the Y-axis is \((-1, \sqrt{3})\).
_________________

Re: In the figure above, points P and Q lie on the circle with [#permalink]

Show Tags

11 Dec 2013, 11:51

In the figure above, points P and Q lie on the circle with center O. What is the value of s?

This problem took be a bit to conceptualize but I got it as soon as I realized that the graph I was looking at wasn't drawn completely accurately.

If you are to accurately graph the point -√3,1 on a piece of graph paper, you see that the angle it forms with regards to the origin (and the x plane) is less than 45 degrees. Because of this, and the fact that both points lie on the circle, and that the two radii form a 90 degree angle, Q must be higher on the circle than P. Upon looking at this graph, I thought that the x and y coordinates of Q would be the same (except positive) which is, I am sure, the intent of the test makers. The answer is simply the positive reciprocal of point P.

Re: In the figure above, points P and Q lie on the circle with [#permalink]

Show Tags

12 Dec 2014, 22:19

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.
_________________

Re: In the figure above, points P and Q lie on the circle with [#permalink]

Show Tags

24 Jul 2015, 05:28

Bunuel wrote:

zz0vlb wrote:

I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.

In the figure above, points P and Q lie on the circle with center O. What is the value of s?

Yes you are right.

Point Q in I quadrant \((1, \sqrt{3})\); Point P in II quadrant \((-\sqrt{3}, 1)\); Point T in III quadrant \((-1, -\sqrt{3})\) (OT perpendicular to OP); Point R in IV quadrant \((\sqrt{3},-1)\) (OR perpendicular to OQ).

Hi Bunuel

Cant we do this by applying 45-45-90 getting distance between O and Q as 1/sqroot 3? As OP=OQ (radii)? but with this I get t/s=1/sqroot3. Pls advise

I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.

In the figure above, points P and Q lie on the circle with center O. What is the value of s?

Yes you are right.

Point Q in I quadrant \((1, \sqrt{3})\); Point P in II quadrant \((-\sqrt{3}, 1)\); Point T in III quadrant \((-1, -\sqrt{3})\) (OT perpendicular to OP); Point R in IV quadrant \((\sqrt{3},-1)\) (OR perpendicular to OQ).

Hi Bunuel

Cant we do this by applying 45-45-90 getting distance between O and Q as 1/sqroot 3? As OP=OQ (radii)? but with this I get t/s=1/sqroot3. Pls advise

Let me try to answer this.

\(\angle{POA} = 30\) by following steps:

In triangle POA, OA = \(\sqrt{3}\), AP = 1, thus this triangle is a 30-60-90 (as the sides are in the ratio \(1:\sqrt{3}:2\)) triangle (either remember this or use trigonometry to figure it out!).

Now, per your question, triangle OPQ is isosceles with 45-45-90 triangle and sides in the ratio \(1:1:\sqrt{2}\)

Thus, PQ should be = \(\sqrt{2}\) and not 1/sqroot 3

Re: In the figure above, points P and Q lie on the circle with [#permalink]

Show Tags

16 May 2016, 12:37

1

This post received KUDOS

Hello all, I just got this question in my GMATprep practice test, and came looking here for an explanation afterwards.

I understand the math in the comments above, and why the answer is what it is. What concerns me, is that the image is not drawn to scale. It did not say the image was not drawn to scale, and this is an official question, so why could I not assume the image was drawn to scale?

(When one has time to look at the question properly it is obvious that it is not drawn to scale - but I have up to now just been assuming that geometry questions are drawn to scale unless clearly indicated that they are not.)

Hello all, I just got this question in my GMATprep practice test, and came looking here for an explanation afterwards.

I understand the math in the comments above, and why the answer is what it is. What concerns me, is that the image is not drawn to scale. It did not say the image was not drawn to scale, and this is an official question, so why could I not assume the image was drawn to scale?

(When one has time to look at the question properly it is obvious that it is not drawn to scale - but I have up to now just been assuming that geometry questions are drawn to scale unless clearly indicated that they are not.)

You cannot solve the questions by assuming that the diagram is drawn to scale. Two angles which look equal may not actually be equal. You can just assume the basics - straight lines that look straight are straight; points are in the order in which they are drawn. But don't take calls based on relative length of lines, relative measure of angles according to the diagram drawn etc.
_________________

Hello all, I just got this question in my GMATprep practice test, and came looking here for an explanation afterwards.

I understand the math in the comments above, and why the answer is what it is. What concerns me, is that the image is not drawn to scale. It did not say the image was not drawn to scale, and this is an official question, so why could I not assume the image was drawn to scale?

(When one has time to look at the question properly it is obvious that it is not drawn to scale - but I have up to now just been assuming that geometry questions are drawn to scale unless clearly indicated that they are not.)

OFFICIAL GUIDE:

Problem Solving Figures: All figures accompanying problem solving questions are intended to provide information useful in solving the problems. 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.

Data Sufficiency: Figures: • Figures 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 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.
_________________

In the figure above, points P and Q lie on the circle with [#permalink]

Show Tags

17 May 2016, 01:32

Quote:

Problem Solving Figures: All figures accompanying problem solving questions are intended to provide information useful in solving the problems. 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.

Thanks for the extra info, to me the bold/underlined part above is not met in this question though. I will be more careful in the test, but I still feel this figure can be drawn much more accurately or should state that it is not accurate. Clearly an image which makes √3 seem equal to 1 in length is not "as accurate as possible".

Re: In the figure above, points P and Q lie on the circle with [#permalink]

Show Tags

04 Aug 2016, 13:31

etienneg wrote:

Quote:

Problem Solving Figures: All figures accompanying problem solving questions are intended to provide information useful in solving the problems. 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.

Thanks for the extra info, to me the bold/underlined part above is not met in this question though. I will be more careful in the test, but I still feel this figure can be drawn much more accurately or should state that it is not accurate. Clearly an image which makes √3 seem equal to 1 in length is not "as accurate as possible".

I agree with you, I have always worked that way until now. Time for a change!
_________________

Send some kudos this way if I was helpful! !

gmatclubot

Re: In the figure above, points P and Q lie on the circle with
[#permalink]
04 Aug 2016, 13:31

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...