I hope you didn't get these problems from the official source because they are much tougher than typical GMAT difficult geometry problems. I spent more than 2 minutes on each.

Problem 1:

We can translate the coordinate values of point P (-sqrt(3), 1) into lengths of a triangle. Draw a verticle line connecting point P perpendicular to the horizontal axis. The triangle would have a horizontal length of sqrt(3), a verical length of 1, and a hypotenuse of 2 (from equation a^2 + b^2 = c^3, where a and b are sides of a right triangle, and c is the hypotenuse).

Right triangle with sides 1, 2, sqrt(3) is a standard right triangle with angles measuring 30, 60, and 90.

From that, we know the angle between the horizontal axis and line OP is 30 degree. The angle that separates line OP and OQ is said to be 90 degree. Thus, the angle between line OP and vertical line at origin O must be 90-30 = 60 degree, and the angle between line OQ and vertical line at origin O must be 90-60 = 30 degree.

Next, draw a line from point Q perpendicular to the vertical line O. This will give you another 30-60-90 right triangle. The hypotenuse of this triangle, line OQ, is actually the radius of this circle (which we already figured out r = 2). Thus, this is another 1 - 2 - sqrt(3) triangle. The coordinate value of point S is the horizontal length of this triangle- the length of the side opposite the 30-degree angle. The answer is 1.

Problem 2:

First, what we know from the given diagram:

a) Angle R + Angle T (let’s just say R+T) = 90
b) Angle TSU + Angle RSQ + Angle x = 180

Statement 1: QR = RS which means angles RQS = RSQ. This doesn’t give us enough to figure out angle x, so it’s insufficient.

Statement 2: ST = TU which means angles TSU = TUS. Again, this doesn’t give us enough to figure out x. Insufficient.

1&2 together:

R = 180 – 2(RSQ) --derived from (1)
T = 180 – 2(TSU) --derived from (2)

From b) TSU + RSQ + x = 180 so

(180 – T)/2 + (180 – R)/2 + x = 180
180 – T + 180 – R + 2x = 360

2x = 360 – 180 – 180 + T + R

From a) we know that T+R = 90, so

2x = 360 – 180 – 180 + 90
X = 45

Sufficient.

There is a much easier way to do the 1st question.

Product of slope of two lines perpendicular to each other is -1

so,

(0-1)/0-(- sq root 3) * (0-t)/ (0/s) = -1

so t/s = sq root 3/1

so s = 1

me4mba: I encountered these questions last night on the GMAC provided practice tests. Needless to say, I was very angry losing as much time as I did trying to figure them out.

ri2007: how do you know that the two slopes are perpendicular?

its given that the angle formed by the 2 lines is 90. So they are perpendicular

I got stuck on this problem (the circle) a few days ago as well. I derived a rule from it, that in questions like that, if the angle is 90 degrees, the points won't me mirrow opposite, but will be reverse mirrow opposite. Maybe this sounds like BS, but if you try to draw it carefully a few times and understand what's going on - you'll be able to figure out the answer in seconds. Calculations are long and questionable. We know that the answer can only either 1 or sqrt3, it's insane if there's something else out there, so the key is to get a way to see this angles and how they project on the xy. Thank you all for the detailed explanations though. The angles, triangles question is totally hard core. "more than two minutes" I would stare at it the whole lunch and just guess...