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Re: On the coordinate plane, is point (u,v) closer to point [#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: On the coordinate plane, is point (u,v) closer to point [#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: On the coordinate plane, is point (u,v) closer to point [#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: On the coordinate plane, is point (u,v) closer to point [#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: On the coordinate plane, is point (u,v) closer to point [#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: On the coordinate plane, is point (u,v) closer to point [#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: On the coordinate plane, is point (u,v) closer to point [#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: On the coordinate plane, is point (u,v) closer to point [#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: On the coordinate plane, is point (u,v) closer to point [#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.
Re: On the coordinate plane, is point (u,v) closer to point [#permalink]
24 Aug 2014, 04:51
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Re: On the coordinate plane, is point (u,v) closer to point [#permalink]
24 Aug 2014, 05:23
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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.
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