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

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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]

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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]

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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]

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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]

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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]

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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]

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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]

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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]

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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

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24 Aug 2014, 05:51

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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.

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