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

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Manager
Joined: 29 Jul 2007
Posts: 182
In the figure below, points P and Q lie on the circle with [#permalink]

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25 Oct 2007, 11:13
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

a)1/2

b)1

c) 2^(1/2)

d) 3^(1/2)

e) (sq rt (2))/2
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DS_SemiCircle.JPG [ 3.78 KiB | Viewed 1315 times ]

Manager
Joined: 02 Aug 2007
Posts: 145

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25 Oct 2007, 12:44
Young gun, can you explain further? I can't seem to get that without an assumption.
Current Student
Joined: 31 Aug 2007
Posts: 369

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25 Oct 2007, 12:52
1
KUDOS
yuefei wrote:
Young gun, can you explain further? I can't seem to get that without an assumption.

sure, you know from the coords of point P, the left triangle is a 30-60-90...using that, you can figure out that the radius of the circle is 2. then using the same method you can solve the Q1 triangle, with hypotenuse of 2, using the 30-60-90 features, since you know the angle closest to origin in Q1 is 30.
makes sense?
VP
Joined: 08 Jun 2005
Posts: 1145

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25 Oct 2007, 13:01
See attachment:
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xy1.JPG [ 20.05 KiB | Viewed 1269 times ]

Director
Joined: 11 Jun 2007
Posts: 914

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25 Oct 2007, 13:12
i also got 1.. created another picture - hopefully makes more sense!
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geom.JPG [ 14.96 KiB | Viewed 1261 times ]

VP
Joined: 09 Jul 2007
Posts: 1100
Location: London

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25 Oct 2007, 13:23
what i could find from the stem is S^2+T^2=4, and the other part is just assumption as young_gun described.
hmm, is it just the assumption that we can reach the answer using?
yeh. thanks beckee.
Manager
Joined: 29 Jul 2007
Posts: 182

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25 Oct 2007, 13:31
Thx guys. My main issue was w/ the diagram. I was under the impression that you can use the diagram to make any assumptions as long as it didn't say "Not drawn to scale". As the 90 angle is bisected by the Y axis, i thought it was a 45 degree angle. Clearly, i was wrong in making this assumption.

So, what's the offical "rule" for diagrams then?

Last edited by Skewed on 25 Oct 2007, 13:35, edited 1 time in total.
VP
Joined: 09 Jul 2007
Posts: 1100
Location: London

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25 Oct 2007, 13:33
Skewed wrote:
Thx guys. My main issue was w/ the diagram. I was under the impression that you can use the diagram to make any assumptions as long as it didn't say "Not drawn to scale". As the 90 angle is bisected by the Y axis, i thought it was a 45 degree angle. Clearly, i was wrong in making this assumption.

So, what's the offical "rule" about diagrams then?

never assume anything about any diagram unless otherwise stated in the statement. all the diagramms in gmat PS are not for scale.
Manager
Joined: 21 Feb 2007
Posts: 80

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25 Oct 2007, 13:55
beckee529 just for my understanding...as per your diagram shouldn't 2sqrt3 and not 2?
Senior Manager
Joined: 28 Jun 2007
Posts: 315

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25 Oct 2007, 14:22
beckee529 wrote:
i also got 1.. created another picture - hopefully makes more sense!

Ok this diagram is freaking me out.

(1) Here OP and OQ are the radius of the circle with center at O. If OP = OQ = 2 then angle OPQ = angle OQP = 45, correct?

(2) From Beckee's diagram I can see that angle OPQ = 30 and angle OQP = 60 must be right. But, if OPQ is a 30 - 60 - 90 triangle then how can the sides facing the 30 degree and 60 degree angle be both of the same length?

So what is going wrong with my understanding?
Attachments

geom_129.jpg [ 14.96 KiB | Viewed 1230 times ]

Manager
Joined: 19 Aug 2007
Posts: 169

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25 Oct 2007, 18:41
ok guys so i understand now that the left triangle can be a 30-60-90 triangle and thus the radius will have to be 2. but then how do we assume that the triangle to the right is also a 30-60-90 triangle?
Senior Manager
Joined: 28 Jun 2007
Posts: 315

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25 Oct 2007, 20:57
gluon wrote:
beckee529 wrote:
i also got 1.. created another picture - hopefully makes more sense!

Ok this diagram is freaking me out.

(1) Here OP and OQ are the radius of the circle with center at O. If OP = OQ = 2 then angle OPQ = angle OQP = 45, correct?

(2) From Beckee's diagram I can see that angle OPQ = 30 and angle OQP = 60 must be right. But, if OPQ is a 30 - 60 - 90 triangle then how can the sides facing the 30 degree and 60 degree angle be both of the same length?

So what is going wrong with my understanding?

Killersqurirrel, bkk145, anyone? How can a 30 - 60 - 90 triangle OPQ have two sides OP = OQ = 2?
Manager
Joined: 01 Oct 2007
Posts: 138

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26 Oct 2007, 04:00
To my calculation the answer is D means square root of 3, means 3 raise to the power 1/2 and its very quick to calculate.

Since OP and OQ both are radius of same circle, their points on circle are also same but with different quadrant. So for point Q the co-ordinates will be (+ sq. root of 3, 1)

So value of s is + sq. root of 3.

Whats the OA?
Director
Joined: 23 Sep 2007
Posts: 783

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12 Dec 2007, 03:30
the OA is 1

but I encountered this question several times before and I got it wrong again.

Beckee illustration makes sense.

But from the official diagram, IF we are NOT allowed to assume that the "Y-line" bisects the right 90 degree angle, then why are we asked to assume that the Y-line is the Y-axis, and the x-line is the x-axis?

For all that I know, the arrow lines show/define the directions of x and y ONLY. By convention, an x/y coordinate drawing must include tick marks on the actual x/y axis(s). Without the actual definition of the axis(s), the center O could be of coordinates (.5,-10) or whatever.
Intern
Joined: 13 Jun 2007
Posts: 48

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12 Dec 2007, 05:58
You have 3 triangles. Left, Center and Right and all are right angled

Left triangle we know the sides x=sqrt(3) and y=1
We know that 30-60-90 is 1-sqrt(3)-2
So the angle of the left triangle, the angle that is part of the center, is 30.

Then 180-30-90=60

So we know that the Right triangle is a 30-60-90 Right angled triangle and we know the hypotenuse which is 2 which is the radius
So we find that s=1 and t=sqrt(3)

I don't agree with beckee529's picture. You can draw this only when the Left and Right triangles are 45 degrees right angled
Director
Joined: 23 Sep 2007
Posts: 783

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15 Dec 2007, 02:06
I just want to add some additional info to further my contention that this question is "bogus".

According to the instruction from the gmatprep software and the official gmac materials, to paraphrase: "diagrams and figures are drawn as accurately as possible".

And from the princeton review 2007, page 155. It states that: unless a diagram is marked "not drawn to scale", it is drawn "as accurately as possible".

This question does not have the "not drawn to scale" disclaimer. Thus, it would completely reasonable and within the GMAC rule to safely "assume" that a equilateral triangle would form if you draw a line from point P to point Q, and that line PQ is parallel to the x-axis. Thus D could be the correct answer.
15 Dec 2007, 02:06
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