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.

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: Help required on geometry question. Please help [#permalink]
21 Nov 2009, 07:41

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, 08:36, edited 1 time in total.

Re: Help required on geometry question. Please help [#permalink]
21 Nov 2009, 08:10

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: Help required on geometry question. Please help [#permalink]
21 Nov 2009, 08:44

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: Help required on geometry question. Please help [#permalink]
22 Nov 2009, 04:14

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.

Re: Help required on geometry question. Please help [#permalink]
22 Nov 2009, 14:26

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: Help required on geometry question. Please help [#permalink]
23 Nov 2009, 15:54

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, 18:20, edited 1 time in total.

Re: Help required on geometry question. Please help [#permalink]
23 Nov 2009, 16:08

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: Help required on geometry question. Please help [#permalink]
24 Nov 2009, 17:02

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: Help required on geometry question. Please help [#permalink]
25 Nov 2009, 02:11

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.