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hmm. i didnt quite understand this...

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Current Student
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hmm. i didnt quite understand this... [#permalink]

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New post 11 Sep 2008, 19:48
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

hmm.
i didnt quite understand this....
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Re: PR-D1 [#permalink]

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New post 11 Sep 2008, 20:19
a = -ve
b = -ve.

so (-a, b) and (-b, a) are in iv quadrant.

1: x and y bother be either -ve or +ve. so insuff..
2: if ax > 0, x is also -ve. but what about y? do not know.

togather, x aND Y BOTH -VE. suff.

yes, (-x, y) is also in the same quadrant..
C.
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Re: PR-D1 [#permalink]

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New post 11 Sep 2008, 20:33
IMO C.

Given (-a,b) and (-b,a) are in the same quadrant.. that means :

a ) if a,b > 0 => both the points are in quadrant 2

b) if a,b < 0 => both the points are in quadrant 4

if and b has deiiferent signs, then these 2 points can not be in the same quadrant.

now from options

1 ) xy > 0 i.e X and Y has same sign.. one condition satisfied... but not sure about all 4 a,b,x and Y has same sign.

2) ax> 0 , i.e. a and x has same sign => not sufficient alone , as it does not say about Y

both, x,y has same sign and a,x has same sign , so a,b.x and y has same sign => sufficient

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Re: PR-D1 [#permalink]

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New post 12 Sep 2008, 04:50
arjtryarjtry wrote:
hmm.
i didnt quite understand this....


The question is realy asking whether a and b have the same sign (+ve or -ve) and x has the same sign as a and y has the same sign as b.

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Re: PR-D1 [#permalink]

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New post 12 Sep 2008, 06:43
Let a=1 and b=2. Then (-a, b) = (-1, 2) and (-b, a) = (-2, 1). These are in the same quadrant.

If a and b are negative the two points will also be in the same quadrant. If the signs of a and b or
different, then they won't be in the same quadrant.

Statement 1

xy>0 tells us x and y have the same sign so sufficient. (Note: xy<0 would also be sufficient)


Statement 2

ax>0 just tells us a and x have the same sign but does not tell us anything about the sign of y so
insufficient

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Re: PR-D1 [#permalink]

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New post 12 Sep 2008, 06:49
thanks guys...
definitley helped

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Re: PR-D1   [#permalink] 12 Sep 2008, 06:49
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hmm. i didnt quite understand this...

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