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:

Re: On the coordinate plane, is point (u,v) closer to point [#permalink]

Show Tags

21 Nov 2009, 08:41

gmat620 wrote:

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

1. V + u^2 = -1

2 V< 0.

Is the answer 'A'

1. Statement 1

v+u^2= -1

u^2 = -1-v

since square of any number is always positive so v is negative and\(|v|>1\)

Now distant between (u,v) and (0,0) = \(u^2+v^2\) since\(|v|>1\) \(u^2+v^2\) >1 and Distance between (u,v) and (u,v+1) = 1 so (u,v) is closer to (u,v+1) than to (0,0)

2. This statement doesn't say anything about 'u'. Not Sufficient

Last edited by swatirpr on 21 Nov 2009, 09:36, edited 1 time in total.

Re: On the coordinate plane, is point (u,v) closer to point [#permalink]

Show Tags

21 Nov 2009, 09:10

How is it even A? The distance between (u,v) and (0,0) is sqrt(v squared + u squared), consider this value-1.

The distance between (u, v) and (u, v+1) is 1, consider this value-2.

The question asks us is if value-1 < value-2.

Statement-1: -v = 1 + u squared. Square on both sides and then substitute in the equation of the question, u powered 4 + 3(u squared) + 1 < 1 . Hence, either u squared < 0 or u squared < -3 which is not possible.

Statement-2: Makes no difference.

Statement-3: Combine both statements, still same as above.

I think it should be E. Both statements are insufficient to come up with a solid answer. _________________

Re: On the coordinate plane, is point (u,v) closer to point [#permalink]

Show Tags

21 Nov 2009, 09:44

SensibleGuy wrote:

How is it even A? The distance between (u,v) and (0,0) is sqrt(v squared + u squared), consider this value-1.

The distance between (u, v) and (u, v+1) is 1, consider this value-2.

The question asks us is if value-1 < value-2.

Statement-1: -v = 1 + u squared. Square on both sides and then substitute in the equation of the question, u powered 4 + 3(u squared) + 1 < 1 . Hence, either u squared < 0 or u squared < -3 which is not possible.

Statement-2: Makes no difference.

Statement-3: Combine both statements, still same as above.

I think it should be E. Both statements are insufficient to come up with a solid answer.

Please check the explanation and let me know if I am doing it wrong.

1. Statement 1

\(v+u^2= -1\)

\(u^2 = -1-v\)

since square of any number is always positive so v is negative and|v|>1

Now distant between (u,v) and (0,0) = \(u^2+v^2\) since\(|v|>1\) then \(u^2+v^2 >1\) and Distance between (u,v) and (u,v+1) = 1

so (u,v) is closer to (u,v+1) than to (0,0)

2. This statement doesn't say anything about 'u'. Not Sufficient

Re: On the coordinate plane, is point (u,v) closer to point [#permalink]

Show Tags

22 Nov 2009, 05:14

1

This post received KUDOS

Quote:

I don't know I encountered this question on gmat club test and OA given there was A. Even I reached D.

Question Stem : Which is greater between \(\sqrt{u^2+v^2}\) and \(\sqrt{u^2+v^2+1+2v}\)

St. (1) : \(u^2 = - v - 1\) Applying this to the question stem we get --> Which is greater between \(\sqrt{v^2 - v - 1}\) and \(\sqrt{v^2+v}\) It is obvious that \(\sqrt{v^2-v-1}\) will always be greater and therefore always be further away from the origin. Hence Sufficient.

St. (2) : \(v<0\) This just tells us that v is negative. However, it is not mentioned that v is an integer. It can be a fraction as well. Thus, we will get different solutions for v < -0.5, v = -0.5 and v > -0.5. (\(\sqrt{u^2+v^2}\) will be greater than, equal to and less than \(\sqrt{u^2+v^2+1+2v}\) respectively). Hence Insufficient.

Answer : A

Thus Statement is insufficient. _________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html

Last edited by sriharimurthy on 22 Nov 2009, 15:51, edited 1 time in total.

Re: On the coordinate plane, is point (u,v) closer to point [#permalink]

Show Tags

22 Nov 2009, 15:26

sriharimurthy wrote:

Quote:

I don't know I encountered this question on gmat club test and OA given there was A. Even I reached D.

Question Stem : Which is greater between \(\sqrt{u^2+v^2}\) and \(\sqrt{u^2+v^2+1+2v}\)

St. (1) : \(u^2 = - v - 1\) Applying this to the question stem we get --> Which is greater between \(\sqrt{v^2 - v - 1}\) and \(\sqrt{v^2+v}\) It is obvious that \(\sqrt{v^2+v}\) will always be greater and therefore always be further away from the origin. Hence Sufficient.

St. (2) : \(v<0\) This just tells us that v is negative. However, it is not mentioned that v is an integer. It can be a fraction as well. Thus, we will get different solutions for v < -0.5, v = -0.5 and v > -0.5. (\(\sqrt{u^2+v^2}\) will be greater than, equal to and less than \(\sqrt{u^2+v^2+1+2v}\) respectively). Hence Insufficient.

Answer : A

Thus Statement is insufficient.

Agreed. v could be a -ve fraction and if it is too close to 0, then (u, v+1) will be reversed from where (u, v) was. _________________

Re: On the coordinate plane, is point (u,v) closer to point [#permalink]

Show Tags

23 Nov 2009, 16:54

Hello, I feel like the answer here is D If u draw a graph, u will notice that the 2points are on a vertical line, parallel to the Y axis --> u does not change.

(1) V + U^2 = -1 u^2 = -1 - v this means that (-1-v) > 0 thus v< -1 v is always negative for all v<0, v+1 > v then (u;v) is NOT closer to (0;0) than (u;v+1)

(2) v<0 for all v<0, v+1 > v then (u;v) is NOT closer to (0;0) than (u;v+1) Up here is what i previously did... After thinking about it more... I should admit that I misunderstood the question. Sorry

Last edited by johjok on 23 Nov 2009, 19:20, edited 1 time in total.

Re: On the coordinate plane, is point (u,v) closer to point [#permalink]

Show Tags

23 Nov 2009, 17:08

swatirpr wrote:

Please check the explanation and let me know if I am doing it wrong.

Ms Perfect, +1 awarded. I learnt not to blindly square on both sides to be able to substitute into the question. Values that variables can take sometimes get totally tricky!!!!

I hate DS, deadly mother of all causes that evoke agony!!!! _________________

Re: On the coordinate plane, is point (u,v) closer to point [#permalink]

Show Tags

24 Nov 2009, 18:02

swatirpr wrote:

opal258 wrote:

for those of you who said A, have you tried (0,-1)?

E seems more correct to me.

For (0, -1) question stem would look like On the coordinate plane, is point (0,-1) closer to point (0,0) than to point (0,0) ?

and that doesn't make any sense to me.

when someone chooses A, he/she usually concludes that "(u,v) is closer to (u,v+1) than to (0,0)." However, if (u,v) = (0,-1), the distance from (u,v) to (u,v+1) would be the same with that from (u,v) to (0,0).

Re: On the coordinate plane, is point (u,v) closer to point [#permalink]

Show Tags

25 Nov 2009, 03:11

1

This post received KUDOS

opal258 wrote:

swatirpr wrote:

opal258 wrote:

for those of you who said A, have you tried (0,-1)?

E seems more correct to me.

Guys, In my earlier post, I had misread the question as On the coordinate plane, is point (u,v) closer to point (0,0) than is point (u,v + 1) ?

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

Thus the distances would be as follows: (a) Between (u,v) and (0,0) ----> \(\sqrt{u^2+v^2}\) (b) Between (u,v) and (u,v+1) --> \(\sqrt{(v+1-v)^2}\) = 1

Therefore, in order to answer the question stem, point \(\sqrt{u^2+v^2}\) must either be less than 1 [point (u,v) is closer to (0,0) than to (u,v+1)] or greater than equal to 1 [point (u.v) is not closer to (0,0) that to (u,v+1)].

Now, let us evaluate the statements:

St. (1) : \(u^2 = - v - 1\)

This implies that \(v\leq-1\)

Therefore, distance between (u,v) and (0,0) becomes : \(\sqrt{v^2-v-1}\) where \(v\leq-1\)

Thus, we can see that for no value of \(v\leq-1\) will \(\sqrt{u^2+v^2}\) be less than 1. (Since \(u^2\) will always be greater than 0 and \(v^2\) will always be greater than or equal to 1).

Even when v = -1, the point (0,0) will be the same distance away from (u,v) as (u,v+1) but not closer.

Thus we can see that if the values of u and v satisfy statement 1, the question stem will always be proved false. That is, (u,v) will never be closer to (0,0).

Hence statement 1 is sufficient.

St. (2) : \(v<0\) Since it is not mentioned that either u or v have to be integers, we can have the following cases : (a) \(\sqrt{u^2+v^2}\) less than 1. Point (0,0) will be closer to (u,v). Eg. u = 0.5 and v = -0.5 This would prove the question stem true. (b) \(\sqrt{u^2+v^2}\) greater than equal to 1. Point (0,0) will not be closer to (u,v). Eg. u = 1 and v = -1 This would prove the question stem false.

Since this statement gives us contradicting solutions, it cannot be sufficient.

Answer : A _________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html

Re: On the coordinate plane, is point (u,v) closer to point [#permalink]

Show Tags

24 Aug 2014, 05:51

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: On the coordinate plane, is point (u,v) closer to point [#permalink]

Show Tags

24 Aug 2014, 06:23

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

gmat620 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.

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

I barely remember taking decent rest in the last 60 hours. It’s been relentless with submissions, birthday celebration, exams, vacating the flat, meeting people before leaving and of...

Rishabh from Gyan one services, India had a one to one interview with me where I shared my experience at IMD till now. http://www.gyanone.com/blog/life-at-imd-interview-with-imd-mba/ ...