Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

P and Q are on the circumference of the circle. so OP = OQ and POQ = 90 degrees. As P and Q are on the same side of X axis, that implies P and Q are reflections along Y axis. It is the conclusion we arrived at based on the above conditions.

Is it not possible that OP makes an angle of 30 degree with Y axis and OQ makes an angle of 60 degress then in such a case POQ is 90 degree but P & Q are not reflections and r will not be equal to t.

If P makes 30 and Q makes 60 then P and Q will not be on the circumference of the circle. If P and Q are on the circumference then P, Q make 45 degrees with Y axis.

I am sorry but i don't agree with your point as there can be many points where POQ can be 90 degrees with OP & OQ making angles other than 45degree and still remain on circumference since OP & OQ are radius and they will always be on circumference.

Just try to imagine that you are rotating OP and OQ such that angle between them is 90 and there could be many such points and OP & OQ will still stay on circumference.

I guess it should be C since we know r & s so we can calculate radius and the angle that OP with X & Y axis.So we can then get angle that OQ makes with X & Y axis.NOW OP = OQ and we know the angle that OQ makes with X axis so we can calculate t & u.

it is clear that statement 1 & 2 is not sufficient

on combining both 1&2 => we know the radius of the circle 'x' by distant formula ( we know co ordinate of P (root3 , 1) )

Now think of this as we are at (0,0) => we can draw a circle of radius 'x' got from both eqns => now on the drawn semicircle; we can plot point (root3,1) and hence line from this point to (0,0) => also we can draw perpendicular to his line ending on semicirlcle at Q (t,u) now => if a point Q can be found by construction ; its co-ordinates also can be found out.... hence C

plz consider for kudos if right ...trying to access tests
_________________

Bhushan S. If you like my post....Consider it for Kudos

Yes, it is logical that we can find the coordinates of Q. Can anyone give a calculation on how we can derive the values of coordinates of Q from the given information in 1) and 2) ?

Angle made by OP with Y axis Tan angle = (x coordinate) / (Y coordinate) = |-(squareroot 3)/1| = 60 So angle made by Q with Y axis = 90 - 60 = 30 Tan 30 = (x coordinate) / (Y coordinate) = 1/(square root 3) but as OQ = 2, common factor for the ratio of x and y coordinates of Q is 1. So Q = (1, square root(3))

Use the similar/equilateral triangles to solve this!

Assuming (p,s as positive numbers)

Line with (-p,s) will be perpendicular to the one with ==> (s,p) in First quadrant ==> (-s,-p) in the third quadrant.

Did'nt get your explanation. p,s as positive integers? P is a point and s is a coordinate of P !?

I am saying assume that we have two positive numbers (p and s) because (-p, s) will then be in the second quadrant in that case.

Further regarding reasoning, draw perpendicular from point P and S on X axis. The two resulting triangles, one in first quadrant and other in second quadrant will be equivalent.

Reason for equilateral:

First they are similar: you can easily see that in the figure for angles.

Now as one side is also of same length of two triangles (=radius) ==> they are equilateral

the difference ==> y distance will become x distance and vice versa for the coordinates.
_________________

Consider kudos for the good post ... My debrief : http://gmatclub.com/forum/journey-670-to-720-q50-v36-long-85083.html

Last edited by sudeep on 28 Jul 2009, 14:45, edited 1 time in total.

If you are tranforming by 90 degree that is a transformation across the quadrant. Whenever you transform in that way an original coordinate (r,s) -> (-s, r) (If it was a 180 degree transformation, it would have been (r,s)->(-r, -s), i.e. value will negate but not their position). What is the source of the question btw? IS it OG or a third party vendor?