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Intern  Joined: 31 Jul 2003
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Location: CT
In the figure above, O is the center of the circle. If OP  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 54% (01:54) correct 46% (01:56) wrong based on 480 sessions

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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
Manager  Joined: 26 Dec 2003
Posts: 202
Location: India

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A? We need to find OS which is equal to OP as both are radius. from 1 we can find the OS
Intern  Joined: 31 Jul 2003
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the answer is D actually...anyone have any ideas on why D?
Senior Manager  Joined: 28 Oct 2003
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1
Rakesh explains why 1 is sufficient. Here's why 2 is sufficient:

The length of QS is 6.

Hence, QR and RS are each 3.

Call OR and PR BOTH "x".

(Since OR and PR combine to form a radius, just like OS)

Call OS "2x."

Since we have a right triangle, we can establish that X^2 +3^2=(2x)^2.

Solve this and you see that x= sqrt 3, and 2x (in this case line OS) is 2*sqrt3.

I think
Manager  Joined: 26 Dec 2003
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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
Director  Joined: 29 Nov 2012
Posts: 703
In the figure above, O is the center of the circle. If OP an  [#permalink]

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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!
Attachments circle.jpg [ 10.67 KiB | Viewed 6556 times ]

Intern  Joined: 14 Aug 2013
Posts: 31
Location: United States
Concentration: Finance, Strategy
GMAT Date: 10-31-2013
GPA: 3.2
WE: Consulting (Consumer Electronics)
Re: In the figure above, O is the center of the circle. If OP an  [#permalink]

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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)
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

Ans is D
Math Expert V
Joined: 02 Sep 2009
Posts: 58381
Re: In the figure above, O is the center of the circle. If OP an  [#permalink]

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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!

Merging similar topics.
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Director  Joined: 25 Apr 2012
Posts: 660
Location: India
GPA: 3.21
WE: Business Development (Other)
Re: In the figure above, O is the center of the circle. If OP  [#permalink]

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rc197906 wrote:
Attachment:
Untitled.png
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

Hi,

St 1 is obvious so lets not dwell on that.

From St 2 we get QS =6 -----> QR=RS=3

Let OR=x since it is being bisected by QS then PR is also x
Also OQ= 2x since OP=OS as radius of circle
Now in triangle OQR we have 3^2+x^2= 4X^2

9=3 x^2
x=\sqrt{3}

Hence OQ= 2\sqrt{3}

Ans is D
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Director  Joined: 29 Nov 2012
Posts: 703
Re: In the figure above, O is the center of the circle. If OP  [#permalink]

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Is there any property tested here? I'm really clueless when it comes to geometry...
Math Expert V
Joined: 02 Sep 2009
Posts: 58381
Re: In the figure above, O is the center of the circle. If OP  [#permalink]

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1
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.

Hope this helps.
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Director  Joined: 29 Nov 2012
Posts: 703
Re: In the figure above, O is the center of the circle. If OP  [#permalink]

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So for such questions never rely on the figure. In this particular question, the figure is misleading.
Math Expert V
Joined: 02 Sep 2009
Posts: 58381
Re: In the figure above, O is the center of the circle. If OP  [#permalink]

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fozzzy wrote:
So for such questions never rely on the figure. In this particular question, the figure is 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.
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Re: In the figure above, O is the center of the circle. If OP  [#permalink]

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Bunuel wrote:
fozzzy wrote:
So for such questions never rely on the figure. In this particular question, the figure is 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 Math Expert V
Joined: 02 Sep 2009
Posts: 58381
Re: In the figure above, O is the center of the circle. If OP  [#permalink]

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jlgdr wrote:
Bunuel wrote:
fozzzy wrote:
So for such questions never rely on the figure. In this particular question, the figure is 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.

Hope it's clear.
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Senior Manager  Joined: 13 May 2013
Posts: 405
Re: In the figure above, O is the center of the circle. If OP  [#permalink]

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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.
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Re: In the figure above, O is the center of the circle. If OP  [#permalink]

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_________________ Re: In the figure above, O is the center of the circle. If OP   [#permalink] 21 Oct 2018, 17:04
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