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 350,000 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:

Re: wats this mean? [#permalink]
28 Aug 2008, 08:45

1

This post received KUDOS

1

This post was BOOKMARKED

arjtryarjtry wrote:

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

1.v + u^2 = -1 2. v < 0

OA is A. but sorry, i didnt understand this ...

Boss don't post OA .. GIVE A CHANCE TO OTHERS.

Distance between (0, 0) and (u, v) = \(sqrt {u^2+v^2 }\) Distance between (0, 0) and (u, v+1) = \(sqrt {u^2+v^2 +2v+1}\)

1.v + u^2 = -1 --> u^2 = -1-v ( u^2 is alwasy positive.. So.. v must be -ve and <-1) 2v+1 --> -ve \(sqrt {u^2+v^2 +2v+1}\) is alwasy less than \(sqrt {u^2+v^2 }\) _________________

Your attitude determines your altitude Smiling wins more friends than frowning

Last edited by x2suresh on 29 Aug 2008, 05:27, edited 1 time in total.

Re: wats this mean? [#permalink]
28 Aug 2008, 10:33

x2suresh wrote:

arjtryarjtry wrote:

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

1.v + u^2 = -1 2. v < 0

OA is A. but sorry, i didnt understand this ...

Boss don't post OA .. GIVE CHANCE TO OTHERS.

Distance between (0, 0) and (u, v) = \(sqrt {u^2+v^2 }\) Distance between (0, 0) and (u, v+1) = \(sqrt {u^2+v^2 +2v+1}\)

1.v + u^2 = -1 --> u^2 = -1-v ( u^2 is alwasy positive.. So.. v must be -ve and <-1) 2v+1 --> -ve \(sqrt {u^2+v^2 +2v+1}\) is alwasy less than \(sqrt {u^2+v^2 }\)

Why will S2 not work? What is the OA? _________________

To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton

Re: wats this mean? [#permalink]
28 Aug 2008, 10:50

leonidas wrote:

x2suresh wrote:

arjtryarjtry wrote:

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

1.v + u^2 = -1 2. v < 0

OA is A. but sorry, i didnt understand this ...

Boss don't post OA .. GIVE CHANCE TO OTHERS.

Distance between (0, 0) and (u, v) = \(sqrt {u^2+v^2 }\) Distance between (0, 0) and (u, v+1) = \(sqrt {u^2+v^2 +2v+1}\)

1.v + u^2 = -1 --> u^2 = -1-v ( u^2 is alwasy positive.. So.. v must be -ve and <-1) 2v+1 --> -ve \(sqrt {u^2+v^2 +2v+1}\) is alwasy less than \(sqrt {u^2+v^2 }\)

Re: wats this mean? [#permalink]
29 Aug 2008, 00:44

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

Let's consider each statement carefully:

a) On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ? First of all, u does not influence on answer. Therefore, we can restate: On the coordinate line is point (0) closer to point (v) than to point (v + 1) ? Now, we can translate it to language of formulas: |v|<|v+1| Eventually, we can write: v>-0.5

Re: wats this mean? [#permalink]
29 Aug 2008, 02:44

thanks , but how did u get 2 from 1?? could not quite understand...

walker wrote:

Let's consider each statement carefully:

a) On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ? First of all, u does not influence on answer. Therefore, we can restate: On the coordinate line is point (0) closer to point (v) than to point (v + 1) ? Now, we can translate it to language of formulas: |v|<|v+1|......(1) Eventually, we can write: v>-0.5....(2)

Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]
13 May 2014, 23:51

1

This post received KUDOS

Expert's post

arjtryarjtry wrote:

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).

So basically the question asks whether the distance between the points \((0, 0)\) and \((u, v)\) is less than the distance between the points \((0, 0)\) and \((u, v + 1)\): is \(\sqrt{(u-0)^2+(v-0)^2}<\sqrt{(u-0)^2+(v+1-0)^2}\)? --> is \(\sqrt{u^2+v^2}<\sqrt{u^2+(v+1)^2}\)? --> is \(u^2+v^2<u^2+v^2+2v+1\)? --> is \(v>-\frac{1}{2}\)?

(1) \(v + u^2 = -1\) --> \(v=-1-u^2\leq{-1}\) --> so the answer to the question is NO. Sufficient.

Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]
04 Dec 2014, 02:27

1

This post was BOOKMARKED

Bunuel wrote:

arjtryarjtry wrote:

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).

So basically the question asks whether the distance between the points \((0, 0)\) and \((u, v)\) is less than the distance between the points \((0, 0)\) and \((u, v + 1)\): is \(\sqrt{(u-0)^2+(v-0)^2}<\sqrt{(u-0)^2+(v+1-0)^2}\)? --> is \(\sqrt{u^2+v^2}<\sqrt{u^2+(v+1)^2}\)? --> is \(u^2+v^2<u^2+v^2+2v+1\)? --> is \(v>-\frac{1}{2}\)?

(1) \(v + u^2 = -1\) --> \(v=-1-u^2\leq{-1}\) --> so the answer to the question is NO. Sufficient.

Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]
04 Dec 2014, 03:40

Expert's post

arshu27 wrote:

Bunuel wrote:

arjtryarjtry wrote:

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).

So basically the question asks whether the distance between the points \((0, 0)\) and \((u, v)\) is less than the distance between the points \((0, 0)\) and \((u, v + 1)\): is \(\sqrt{(u-0)^2+(v-0)^2}<\sqrt{(u-0)^2+(v+1-0)^2}\)? --> is \(\sqrt{u^2+v^2}<\sqrt{u^2+(v+1)^2}\)? --> is \(u^2+v^2<u^2+v^2+2v+1\)? --> is \(v>-\frac{1}{2}\)?

(1) \(v + u^2 = -1\) --> \(v=-1-u^2\leq{-1}\) --> so the answer to the question is NO. Sufficient.

As for your question, we need to find whether \(v>-\frac{1}{2}\).

(1) gives \(v=-1-u^2\). Since u^2 (the square of a number) must be non-negative, then we have that \(v=-1-(nonnegative)\leq{-1}\), therefore v is NOT greater than -1/2, so we have a definite NO answer to the question.

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

McCombs Acceptance Rate Analysis McCombs School of Business is a top MBA program and part of University of Texas Austin. The full-time program is small; the class of 2017...

My swiss visa stamping is done and my passport was couriered through Blue dart. It took 5 days to get my passport stamped and couriered to my address. In...