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DS Geometry

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DS Geometry [#permalink] New post 18 Jun 2006, 02:50
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A
B
C
D
E

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Hi

Some thoughts
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 [#permalink] New post 18 Jun 2006, 04:26
I assume that the figure in statement 1 is 3690 degrees

3690 = (360)*10 +90

Statement 1

Insuff as r is not known

Statement 2

Insuff as no info on motion of beam

Combine

The beam will be making 90-40 = 50 degrees with the positive hand of x axis. Hence will lie in q4
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 [#permalink] New post 18 Jun 2006, 04:35
I`d say statement 1 is sufficient. After 10 sweeps, the remainder of 90 degrees must put the line somewhere in the II quandrant. Even if r = 1~89 degrees, it doesn`t matter.

(A)
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 [#permalink] New post 18 Jun 2006, 08:28
GMATT73 wrote:
I`d say statement 1 is sufficient. After 10 sweeps, the remainder of 90 degrees must put the line somewhere in the II quandrant. Even if r = 1~89 degrees, it doesn`t matter.

(A)


GMATT73,

I'm not entirely sure I agree with you (you can correct me if I'm wrong!), you are making your calculation purely by looking at the diagram and disregarding the instructions. If I were to have a circle and was not allowed to infer from it, then I would certainly need the position of the sweep, or the angle (r) which would tell me which quadrant I am in. Let me put it this way - what if the diagram of the circle that is shown, is rotated w.r.t it's frame of reference? you have no way of knowing it, right?

So my question is - can you make a conclusion of a DS question by "inferring" from the diagram and ignoring an option? or am I overanalyzing this whole thing? I think the answer should be (c), you need both.

Any thoughts?
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 [#permalink] New post 22 Jun 2006, 13:36
I agree with GMAT analysis. The radar has to be in the bottom right quuarant after 3,690 degrees. So A is my choice.
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 [#permalink] New post 22 Jun 2006, 13:48
I go on (C) as well.

We need to know r. r could equal to 100° or 90°.
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 [#permalink] New post 22 Jun 2006, 13:48
I am going with C on this.
You cannot look at the figure and assume a certain position for the beam i.e. the value of r.
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 [#permalink] New post 22 Jun 2006, 14:40
I remember having read somewhere that the figure can be taken as accurate, unless specifically mentioned otherwise... :roll:

A, in that case.
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 [#permalink] New post 22 Jun 2006, 15:03
(A)

90degree clockwise rotation will put it in bottom right quadrant irrespective of value of r ..
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Re: DS Geometry [#permalink] New post 22 Jun 2006, 18:11
agree with A, however it can be disputed.

it makes sense that we donot know the value of r as it could be anything from 0< to <90. but i believe it is not 0 or 90 since it is clear that the line isnot either on x axis or on y axis. if we assume r is 0, we can also assume that r is on any quadrant.

therefore after a run of 3690 degree, it should be on Qd iv.
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 [#permalink] New post 23 Jun 2006, 07:42
+1 for A
in 30 secs the ray will make 10 full swips plus 1/4 - i.e. it will be in the next clockwise quadrant, no matter the angle
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 [#permalink] New post 24 Jun 2006, 09:20
OA: A

But it is a controvesrsial one........
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 [#permalink] New post 24 Jun 2006, 09:28
I believe it is not drawn to scale in PS questions and in DS it is to scale unless otherwise stated. I am not 100% on this but it would agree with A being the OA.

I was leaning towards C since you dont know what R is and it could be anything, thinking its not drawn to scale.
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 [#permalink] New post 25 Jun 2006, 14:55
I also went for C but have to agree for A as OA is A.
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 [#permalink] New post 26 Jun 2006, 08:48
I believe that we could assume that it lies in the first quandrant but the angle could be any betweeb 0 and 90.
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 [#permalink] New post 26 Jun 2006, 09:22
I don't get this question. The answer A is based on the assumption that the diagram is accurate. We are not given the current position, if we were the ans would clearly be A.
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 [#permalink] New post 26 Jun 2006, 17:13
The stem says that the figure above shows the current position of .... That means to me that we can be sure that it lies in the first quandrant. However, without information about r we cannot measure r in the figure and assume it is "accurate".
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 [#permalink] New post 26 Jun 2006, 17:33
HongHu wrote:
I believe that we could assume that it lies in the first quandrant but the angle could be any betweeb 0 and 90.


Honghu, do you mean that 0 and 90 both fall in/on/under 1st quadrant?

1st qdt = 0 - 90.
2nd qdt = 90< but <180.
3rd qdt = 180< but <270.
4th qdt = 270< but <360 (or = 360?). correct me, if any.

i guess yes, that makes sense. Thanx..
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 [#permalink] New post 26 Jun 2006, 17:53
Hey MA, how's everything going with you?

Yes you are right. Though I'm not sure how the axises are attibuted. Are they counted as being in one of the quandrants? In other words, I'm not sure if r could be exactly 0 or 90. Although this does not really matter, for whatever the rule is, it will be consistent. In other words if it lies in first quandrant, 90 degrees later it will be in the second quandrant, no matter what exact r is.
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 [#permalink] New post 26 Jun 2006, 18:08
HongHu wrote:
Hey MA, how's everything going with you?

Yes you are right. Though I'm not sure how the axises are attibuted. Are they counted as being in one of the quandrants? In other words, I'm not sure if r could be exactly 0 or 90. Although this does not really matter, for whatever the rule is, it will be consistent. In other words if it lies in first quandrant, 90 degrees later it will be in the second quandrant, no matter what exact r is.


thank you honghu.

so far so good. was busy with lots of works and out of internet. therefore, i have been out of touch with you wonderful guys. however when i had time, i had always been reading you guys. now i am moved to the capital and so i will be regular in the forum. i took once but could not able to cross the nervus 7xx. i want to give a shot one more time. lets see what happens?

how about you? are you in simon? i believe, yes. am i right? :wink:
  [#permalink] 26 Jun 2006, 18:08
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