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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

OA is A. but sorry, i didnt understand this ...

Boss don't post OA .. GIVE A CHANCE TO OTHERS.

Distance between (0, 0) and (u, v) = \(sqrt {u^2+v^2 }\)
Distance between (0, 0) and (u, v+1) = \(sqrt {u^2+v^2 +2v+1}\)

1.v + u^2 = -1
--> u^2 = -1-v ( u^2 is alwasy positive.. So.. v must be -ve and <-1)
2v+1 --> -ve
\(sqrt {u^2+v^2 +2v+1}\) is alwasy less than \(sqrt {u^2+v^2 }\)
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thanks , but
how did u get 2 from 1??
could not quite understand...


walker
Let's consider each statement carefully:

a) On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?
First of all, u does not influence on answer. Therefore, we can restate: On the coordinate line is point (0) closer to point (v) than to point (v + 1) ?
Now, we can translate it to language of formulas: |v|<|v+1|......(1)
Eventually, we can write: v>-0.5....(2)

b) v + u^2 = -1 --> v=-1-u^2 --> v<-1

Now, we can restate our problem as following:

Does v>-0.5 ?
1. v<-1
2. v<0

Answer is obviously A.
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Full solution:

|v|<|v+1|

1) v<-1: -v<-v-1 --> 0<-1 --> always false
2) -1<=v<=0: -v<v+1 --> v>-0.5 --> -0.5<v<=0
3) v>0: --> v<v+1 --> 0<1 always true.

Therefore, inequality is true when v>-0.5

Fast solution:

large negative v: inequality is false
large positive v: inequality is true
switch point v=-0.5 --> v>-0.5
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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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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.
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How we could have represented this equation on graph ..... and solved by using graph technique

Please help
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GmatDestroyer2013
How we could have represented this equation on graph ..... and solved by using graph technique

Please help

This question is not a good candidate for graphic approach.
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