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The figure above shows the present position on a radar [#permalink]
16 May 2008, 16:37

00:00

A

B

C

D

E

Difficulty:

25% (low)

Question Stats:

54% (01:49) correct
45% (00:43) wrong based on 90 sessions

Attachment:

Radar.jpg [ 2.9 KiB | Viewed 2907 times ]

The figure above shows the present position on a radar screen of a sweeping beam that is rotating at a constant rate in a clockwise direction. In which of the four quadrants will the beam lie 30 seconds from now ?

(1) In each 30-second period, the beam sweeps through 3690° (2) r = 40

9. The figure above shows the present position on a radar screen of a sweeping beam that is rotating at a constant rate in a clockwise direction. In which of the four quadrants will the beam lie 30 seconds from now ? (1) In each 30-second period, the beam sweeps through 3.690° (2) r = 40

Please see attached document for teh diagram.

From what I see, the answer would be A. The current position of the Radar beam is somewhere in the middle of the first quadrant and it moves only 3.690degrees, which would still be well within 90degrees. Statement 2 by itself can't offer much since it doesn't tell you the rate of movement.

To answer this question, we need to know the current position of the beam (i.e r) and we need to know the rate of rotation

1) gives us rate of rotation only, insuff 2) gives us r=40 degrees, insuff

1)2). in 30 seconds, the beam will be in q1, with r = 40-3.69

The answer is C

I think The rate of rotation in Statement 1 is enough to answer the question because the image shows the radar bear somewhere in the middle of the 1st Quadrant (near around 45degrees). Statement 2 simply reaffirms the position of the beam within the 2nd quadrant. Can someone confirm the OA?

But no where its mentioned teh diagram is drawn to scale. The OA is A. But I diagree fo rthe same reason I mentioned in my previous sentence.

9. The figure above shows the present position on a radar screen of a sweeping beam that is rotating at a constant rate in a clockwise direction. In which of the four quadrants will the beam lie 30 seconds from now ? (1) In each 30-second period, the beam sweeps through 3.690° (2) r = 40

Please see attached document for teh diagram.

To rephrase this question, is the beam with r deg still in the same quadrant?

Statement (1) alone gives us the speed and degree of the movement of the beam every 30-second period. Sufficient for us to know whethere the beam is. Eliminate B, C and E.

Statement (2) alone tells us r = 40. But not where it can be 30 seconds from now. Not sufficient. Eliminate D.

9. The figure above shows the present position on a radar screen of a sweeping beam that is rotating at a constant rate in a clockwise direction. In which of the four quadrants will the beam lie 30 seconds from now ? (1) In each 30-second period, the beam sweeps through 3.690° (2) r = 40

Please see attached document for teh diagram.

To rephrase this question, is the beam with r deg still in the same quadrant?

Statement (1) alone gives us the speed and degree of the movement of the beam every 30-second period. Sufficient for us to know whethere the beam is. Eliminate B, C and E.

Statement (2) alone tells us r = 40. But not where it can be 30 seconds from now. Not sufficient. Eliminate D.

Ans: A

In which of the four quadrants will the beam lie 30 seconds from now ?

We have no information regarding the current position.

The figure above shows the present position on a radar screen of a sweeping beam that is rotating at a constant rate in a clockwise direction. In which of the four quadrants will the beam lie 30 seconds from now ? (1) In each 30-second period, the beam sweeps through 3.690° (2) r = 40

Not sure if there's a typo in the question, but the OA is A, which I don't think it's right.

Just after looking at the figure think that angle r is 2degrees and then think that angle r is say 10 degrees. When you think that it is 2 degrees then after 30 seconds we the beam will in quadrant 2. If angle r is 10degrees then after 30 seconds it will stay in quadrant 1. All the books says that the figure are not drawn to scale in a DS question. There is no way option A is suff.

The present position of radar: 0 < r < 90 degrees A: during 30 seconds the beam will sweep through 3690 degrees, i.e. 3600 + 90 degrees: it will make 10 complete turns + 90 more degrees. We will have 0+90 < r+90 < 90+90 => 90 < r+90 < 180, which tells us that the beam will be in the 2nd quadrant.

Re: Radar.jpg 9. The figure above shows the present position on [#permalink]
14 Feb 2013, 11:50

Expert's post

Hello Friends,

just read in OG quant 2nd edition that The position of points, angles, regions, etc., exist in the order shown.

so if the given figure showing the present position of radar in the middle of first quadrant, we can assume so and can go with choice A
_________________

Re: The figure above shows the present position on a radar [#permalink]
19 Feb 2014, 10:13

1

This post received KUDOS

Guys,

I think there's nothing wrong with the question other than a typo and it's 3,690°. Which I didn't even realize at first because I'm from Brazil, and here works the opposite way. The decimal "point" is "," and the "," is the decimal point.

So, the present position of the sweeping beam it's the one in the figure. "The figure above shows the present position on a radar screen of a sweeping beam" Every 90° the sweeping beam will change quadrants, and every 4 times it will come back to the starting point. If we divide 3690 by 90 we get 41. So, that means it will rotate and go back to the starting point 10 times (10*4 = 40) and one more time (10*4+1=41) So. the answer would be A. It will be in the 4th quadrant.

Re: The figure above shows the present position on a radar [#permalink]
16 Apr 2014, 01:39

Expert's post

PathFinder007 wrote:

Bunnel could you please look into this question. its not clear

The figure above shows the present position on a radar screen of a sweeping beam that is rotating at a constant rate in a clockwise direction. In which of the four quadrants will the beam lie 30 seconds from now ?

(1) In each 30-second period, the beam sweeps through 3690° --> 3690 = 10*360 + 90 = 10 revolution + 90 degrees. Regardless the value of r, the beam will be in the IV quadrant. Sufficient.