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:

A square in the XY co-ordinate system has vertices as (1,1), [#permalink]

Show Tags

15 Sep 2003, 08:01

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A square in the XY co-ordinate system has vertices as (1,1), (1,-1), (-1,-1), (-1,1). If a is the x coordinate of any point and b the y-coordinate of any point in the XY coordinate system, find the probability that a point selected at random from the square region will satisfy x^2+ y^2<1?

I have my doubts about this one...but let me see if y'all can crack this one

A square in the XY co-ordinate system has vertices as (1,1), (1,-1), (-1,-1), (-1,1). If a is the x coordinate of any point and b the y-coordinate of any point in the XY coordinate system, find the probability that a point selected at random from the square region will satisfy x^2+ y^2<1?

I have my doubts about this one...but let me see if y'all can crack this one

Thanks Praetorian

Not sure if this is correct ..

x^2 + y^2 =1 is a circle around (0,0) with a radius of 1. And any point within that circle should have x^2 + y^2 < 1. So, probability of a point selected randomly that will satisfy x^2 + y^2 < 1 is:

Area of Circle/Area of Square = pi/4 = 3.14/4 = 0.785

x^2 + y^2 =1 is a circle around (0,0) with a radius of 1. And any point within that circle should have x^2 + y^2 < 1. So, probability of a point selected randomly that will satisfy x^2 + y^2 < 1 is:

Area of Circle/Area of Square = pi/4 = 3.14/4 = 0.785

See the attachment

Area of Square 1 / Area of Square 2 = [sqrt(2)]^2 /(2^2) = 2/4 =0.5

But here we assumed x^2 + y^2 =1

So any value lesser than 0.5 should be a candidate..

You basically have to draw lines x = 1 - y, x = y - 1 and the same set for y in terms of x. Doing this will give you a square between the four lines. You have to recognize that in order for the original equation (x^2 + y^2 < 1) to be true, you have to consider the points inside the new square.

Now targeting the probability question, you can see that the area of the new square is half that of the original square. So of ALL the points in the bigger square, half has many will fall in the new square. This will give you probability of 1/2. Well, technically the probability is a little less than 1/2 but it's so close that we can just say 1/2.

You basically have to draw lines x = 1 - y, x = y - 1 and the same set for y in terms of x. Doing this will give you a square between the four lines. You have to recognize that in order for the original equation (x^2 + y^2 < 1) to be true, you have to consider the points inside the new square.

Now targeting the probability question, you can see that the area of the new square is half that of the original square. So of ALL the points in the bigger square, half has many will fall in the new square. This will give you probability of 1/2. Well, technically the probability is a little less than 1/2 but it's so close that we can just say 1/2.

Good question.

Correct me if I am wrong, but here is a problem with your solution.

With the new square you created within the given square, consider a point say (1/2, 1/2). It is on the line x = 1 - y or x + y = 1. Now, consider a point, (0.55, 0.55) which is a little above (1/2, 1/2).

Now, (0.55)^2 + (0.55)^2 = 0.605 (which is less than 1).

x^2 + y^2 = 1 is a circle and not a square. Hope this helps.

Correct me if I am wrong, but here is a problem with your solution.

With the new square you created within the given square, consider a point say (1/2, 1/2). It is on the line x = 1 - y or x + y = 1. Now, consider a point, (0.55, 0.55) which is a little above (1/2, 1/2).

Now, (0.55)^2 + (0.55)^2 = 0.605 (which is less than 1).

x^2 + y^2 = 1 is a circle and not a square. Hope this helps.

Watchdog: If you explain your solution, maybe we can tell you if you right.

Wonder : I constructed the square using the pythagorous theorem for each set of points. for eg, for (1,1) ....1^2 + 1^2 = 2
So the Hypotenuse or the side of the square = Sqrt(2)

Edeal : GMAT never tests on Circle formulas...so i was trying to see if anyone can do in a simpler way..
But i guess we can leave it at that...we have to rely on the answer choices for a definite answer.