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.

OA is ..please explain

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.

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.
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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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GT

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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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.

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
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compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

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