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Is 2x-3y<x^2? 1)2x-3y=-2 2)x>2 and y>0

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Is 2x-3y<x^2? 1)2x-3y=-2 2)x>2 and y>0 [#permalink] New post 01 Mar 2005, 17:31
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Is 2x-3y<x^2?
1)2x-3y=-2
2)x>2 and y>0
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Re: DS problem-2 [#permalink] New post 01 Mar 2005, 20:48
aks720 wrote:
Rupstar wrote:
Is 2x-3y<x^2?
1)2x-3y=-2
2)x>2 and y>0


D???


1- -2<x^2 , x^2 is always >0, sufficient.
2 - x^2>2x is true for x>2 so x^2>2x. Since y>0 then -3y<0 is true
Therefore 2x-3y will always be less than x^2, sufficient.

Answer = D
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 [#permalink] New post 02 Mar 2005, 07:40
D it is ...everyone is a DS genius now
:)
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Re: DS problem-2 [#permalink] New post 02 Mar 2005, 10:38
Rupstar wrote:
Is 2x-3y<x^2?
1)2x-3y=-2
2)x>2 and y>0


From I) -2<x^2
Squares are always +ve, thus we know that is true.Suff
From II) x>2 and we know for x>2, 2x<x^2
thus 2x-3y will always be lass than x^2. Suff

D.
Re: DS problem-2   [#permalink] 02 Mar 2005, 10:38
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