GMAT Prep CAT Geometry

Question

Please download. My explanation and the OA are below. I got it wrong initially, and then I tried to solve it after the test. I think I am doing it correctly, but I would like to see other thought processes.

robinantony · Answer

Firstly, Your solution is not there in the attachment. 

Secondly, there is no relation between line OP and OQ. Point Q can be anywhere on the circle in Quadrant 1. How would you find point Q when it can be anywhere?

djhouse81 · Answer

Sorry, I posted the problem late last night, and since I tried copying down the information from the CAT to a piece of paper then to GMAT Club, I might have written down some of the information incorrectly. I see your concern, and I will go back and look at the problem again to see what information, if any, is missing.

My apologies.

djhouse81 · Answer

Ok, let's try this again. I checked my GMAT Prep test and the figure that I originally posted did not have 90 degree angle sign.

With that being said, I am scrapping my solution. I thought I had it, but I am making too many assumptions about the coordinates and angles. Although my assumptions might be correct, I would like to see other answers.


Once again, OA is B.

hd54321 · Answer

I think OA is wrong (gasp) and it's C. Here's my opinion.

Since the center of the circle is O and since the center is bisected by the y axis and since POQ = 90 degrees, it's safe to assume OP = OQ.

Since we have coordinates for P (-sqrt(3), 1) and for O (0, 0) we could use the distance formula to calculate that the distance between these two points. The distance between P & O = 2, so OP = 2. 

Then OP = OQ = 2, and we have a 45-45-90 Isosceles Triangle. Using the 45-45-90 proportions  , we arrive at the length of the hypotenuse which is 2*sqrt(2).

Since the y-axis bisects the circle and triangle (evident by the 90 degree angle), we divide 2*sqrt(2)/2 = sqrt(2) = answer C.

Anyone else?

djhouse81 · Answer

OA is B according to GMAT Prep software.

misterJJ2u · Answer

Ah, i remember this one on the gmat prep software. The OA is B and is correct. The key to this question is to think in terms of triangle definitions.

Here goes: 
(1) Point P is to the left of the origin by sqr root 3 and up by 1. Using the right triangle rule you calculate that line segment PO is 2. 

(2) Tricky part: rules of 90 degree triangles. If an angle yields the segments, 1, 2, sqr root 3, what kind of triangle is it? This is the part you need to figure out. It is a 30-60-90 triangle. The angle opposite the segment value of 1 is 30 degrees. 

(3) you know the inside angle (angle POR) is 30, where R is a made up point directly under point P. 30 + 90 (angle POQ) = 120. 180-120 = 60. 

(4) Angle QOS, where S is directly under point Q, is 60 degrees. 

(5) We know that OQ is 2 b/c we already figured out the radius in step 1. Therefore, we need to figure out what the length of the side QS is, given it is opposite a 60 degree angle within a 30-60-90 triangle and the hypotenuse is 2. Line segment across from the 60 degree angle is sqr root 3 (QS) and line segment across from the 30 degree angle is 1 (OS).

Therefore, s = 1.

GK_Gmat · Answer

This question had been bugging me for a long time. Thanks a lot for clarifying it.

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