Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 15 Jul 2019, 23:22

### 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

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 07 Apr 2012
Posts: 95
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
GMAT 1: 590 Q48 V23
GPA: 3.9
WE: Operations (Manufacturing)
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

10 Dec 2013, 06:34
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..
Math Expert
Joined: 02 Sep 2009
Posts: 56234
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

10 Dec 2013, 06:40
ygdrasil24 wrote:
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..

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 $$(-1, \sqrt{3})$$.
_________________
Manager
Joined: 27 Aug 2014
Posts: 71
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

24 Jul 2015, 06:28
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
CEO
Joined: 20 Mar 2014
Posts: 2620
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

24 Jul 2015, 07:10
sinhap07 wrote:
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

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
Attachments

GPrep.jpg [ 6.59 KiB | Viewed 2903 times ]

Intern
Joined: 06 May 2016
Posts: 15
WE: Education (Education)
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

16 May 2016, 13:37
1
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.)
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9428
Location: Pune, India
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

17 May 2016, 01:39
1
etienneg wrote:
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.
_________________
Karishma
Veritas Prep GMAT Instructor

Math Expert
Joined: 02 Sep 2009
Posts: 56234
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

17 May 2016, 02:12
etienneg wrote:
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.)

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.
_________________
Intern
Joined: 15 Aug 2017
Posts: 5
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

25 Mar 2018, 08:00
can you pls explain the answer to this question in detail.
thankyou!

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).
Senior PS Moderator
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 751
GMAT 1: 740 Q50 V40
GMAT 2: 770 Q51 V42
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

25 Mar 2018, 08:27
1
pranavpal wrote:
can you pls explain the answer to this question in detail.
thankyou!

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 pranavpal,

Many people have already solved the Q in prev posts, but I will attempt it as you pinged on main chat Hope I can help.

Let PO make an angle of x (angle POA) with the negative x axis.. this angle comes out to 30 degrees using the tangent rule... ( i.e the slope of line PO is 1/sq root 3)

This angle is complementary to the angle made by QO with the positive X axis ... hence the angle QOB is 90 - 30 = 60 degrees.

In the two traingles POA and QOB... only the hypotenuse is common and we can call this to be r (as it is also the radius of the circle)

Applying the cosine rule on both right angled triangles..

1) $$\sqrt{3}/r = \sqrt{3}/2$$ ... applying the cosine rule on POB triangle.

This gives us r = 2.

2) $$s / r = 1/2$$ ... cosine rule on right angled traingle QOA. ( we know r = 2 from above)

Hence s = 1.

Option (B).

Hope it makes sense.

Best,
Attachments

IMG_20180325_205501.jpg [ 387.31 KiB | Viewed 1789 times ]

_________________
Regards,

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)
Intern
Joined: 31 Mar 2018
Posts: 2
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

31 Mar 2018, 18:33
I think I understand how to solve the problem now, but I am having trouble understanding one assumption that people are making in many of the solutions: that O is at the origin. When I tried to solve this problem in the practice test I got to the point where I knew the radius was 2 and that line QO had a slope that was the negative reciprocal of line PO (since they were perpendicular bisectors), but I couldn't solve for slope m of line QO because I was stuck with 1=-sqrt(3)*m+b.

Can we assume that O is at the origin even though it isn't stated in the question, just because it looks like it is? Why can we assume that and not that the y-axis is equidistant from points P and Q (the diagram shows the y-axis passing directly in the middle of the 90 degree angle shown)?
Math Expert
Joined: 02 Sep 2009
Posts: 56234
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

01 Apr 2018, 02:31
1
ktjorge wrote:
I think I understand how to solve the problem now, but I am having trouble understanding one assumption that people are making in many of the solutions: that O is at the origin. When I tried to solve this problem in the practice test I got to the point where I knew the radius was 2 and that line QO had a slope that was the negative reciprocal of line PO (since they were perpendicular bisectors), but I couldn't solve for slope m of line QO because I was stuck with 1=-sqrt(3)*m+b.

Can we assume that O is at the origin even though it isn't stated in the question, just because it looks like it is? Why can we assume that and not that the y-axis is equidistant from points P and Q (the diagram shows the y-axis passing directly in the middle of the 90 degree angle shown)?

O is the origin because it's on the intersection of x and y axis. Also, if we did not know that, we would not be able to solve the question and since it's a PS questions, we should be able to.

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.
_________________
CEO
Joined: 12 Sep 2015
Posts: 3846
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

19 Apr 2018, 14:39
Top Contributor
christoph wrote:

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

A. $$\frac{1}{2}$$

B. $$1$$

C. $$\sqrt{2}$$

D. $$\sqrt{3}$$

E. $$\frac{\sqrt{2}}{2}$$

Attachment:
image.JPG

Here's one approach:

So, s = 1

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
Intern
Joined: 23 Jul 2016
Posts: 6
Location: India
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

22 Jul 2018, 11:40
Here if we consider the left angle as x.
And consider the right angle as y
Then we will have the condition as x+y+90=180 --> y=90-x ---------------(1)

Now we can find out the x using pythagorean
sinx = 1/2
x=30

y =60 ---- from ( 1)

so , cos 60 = 1/2 =s/2
therefore s=1
Intern
Joined: 14 May 2018
Posts: 3
Location: United States (GA)
Concentration: Strategy, General Management
WE: Consulting (Energy and Utilities)
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

23 Nov 2018, 14:49
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).

Bunuel,

Can you please go over this logic in detail? I'm having a hard time wrapping my head around it.
Manager
Joined: 29 Jun 2018
Posts: 52
Location: India
GPA: 4
Re: In the figure above, points P and Q lie on the circle with  [#permalink]

### Show Tags

18 Jun 2019, 11:59
muralikrishnag wrote:
Not sure if this is simpler or even different approach from the others(I maybe just formulizing qpoo's approach!)u.....

Lets say slope of OP =m1 and OQ =m2
Since OP perpendicular to OQ => m1m2=-1
Since m1= -1/sqrt(3)
=>m2=sqrt3
=>s=1,t=sqrt(3)

you have mentioned m2 = Sqrt3 and hence t= sqrt(3).
But it could have been that s: 1/2 and t=Sqrt3/2 then also you would have got the same m2 as Sqrt3.
_________________
Re: In the figure above, points P and Q lie on the circle with   [#permalink] 18 Jun 2019, 11:59

Go to page   Previous    1   2   [ 35 posts ]

Display posts from previous: Sort by