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# On the coordinate plane, is point (u,v) closer to point

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On the coordinate plane, is point (u,v) closer to point [#permalink]  21 Nov 2009, 07:25
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: on-the-coordinate-plane-is-point-0-0-closer-to-point-u-69437.html
[Reveal] Spoiler: OA
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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.

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.
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Re: On the coordinate plane, is point (u,v) closer to point [#permalink]  21 Nov 2009, 07:55
OA is
[Reveal] Spoiler:
A
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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.
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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

Thanks.
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Re: On the coordinate plane, is point (u,v) closer to point [#permalink]  21 Nov 2009, 23:41
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.

Why OA is A and not D? In both case, u is not known and could +ve or -ve. With the same condition, why statement 1 is suff but not statement 2?

The value of u is irrlavant to this question. So it has to be D.
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Re: On the coordinate plane, is point (u,v) closer to point [#permalink]  22 Nov 2009, 02:53
GMAT TIGER wrote:
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.

Why OA is A and not D? In both case, u is not known and could +ve or -ve. With the same condition, why statement 1 is suff but not statement 2?

The value of u is irrlavant to this question. So it has to be D.

I don't know I encountered this question on gmat club test and OA given there was A. Even I reached D.
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Re: On the coordinate plane, is point (u,v) closer to point [#permalink]  22 Nov 2009, 04:05
The question is asking whether u^2+v^2<u^2+ (v+1)^2 => u^2+v^2<u^2+v^2+1+2v

For this happen 1+2v<0

From stem 1 V=-1-u^2; As u^2>0, V<-1 => 1+2V is less than zero. Hence Suff

From stem 2 V<0; but for 1+2V<0 V should be less than -1/2. Hence inSuff

Hence A
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Re: On the coordinate plane, is point (u,v) closer to point [#permalink]  22 Nov 2009, 04:14
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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.

Thus Statement is insufficient.
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Last edited by sriharimurthy on 22 Nov 2009, 14:51, edited 1 time in total.
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Re: On the coordinate plane, is point (u,v) closer to point [#permalink]  22 Nov 2009, 10:57
Thanks to every one for taking time and interest in the problem.
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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.

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.
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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.
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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!!!!
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Re: On the coordinate plane, is point (u,v) closer to point [#permalink]  24 Nov 2009, 16:27
for those of you who said A, have you tried (0,-1)?

E seems more correct to me.
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Re: On the coordinate plane, is point (u,v) closer to point [#permalink]  24 Nov 2009, 16:56
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.
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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).

Therefore, A is insufficient.
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Re: On the coordinate plane, is point (u,v) closer to point [#permalink]  25 Nov 2009, 02:11
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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.

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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.

(2) $$v<0$$. Not sufficient.

OPEN DISCUSSION OF THIS QUESTION IS HERE: on-the-coordinate-plane-is-point-0-0-closer-to-point-u-69437.html
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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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