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

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Manager
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Is 2x-3y<x^2? 1)2x-3y=-2 2)x>2 and y>0 [#permalink]  01 Mar 2005, 16:31
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Is 2x-3y<x^2?
1)2x-3y=-2
2)x>2 and y>0
Intern
Joined: 18 Feb 2005
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: DS problem-2 [#permalink]  01 Mar 2005, 19:48
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.

Manager
Joined: 15 Feb 2005
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D it is ...everyone is a DS genius now
Manager
Joined: 01 Jan 2005
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Re: DS problem-2 [#permalink]  02 Mar 2005, 09:38
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, 09:38
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