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# Is 2x-3y< x^2

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Intern
Joined: 20 Jul 2008
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10 Aug 2008, 12:40
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Is 2x-3y< x^2?

(1) 2x-3y=-2
(2) x>2 and y>0

Please explain if you can. I will post the OA after I see some answers

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Director
Joined: 10 Sep 2007
Posts: 933

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10 Aug 2008, 13:39
Rephrased question Is x(2-x) < 3y.

1) Sufficient. x^2 is always greater than or equal to zero. Given 2x-3y = -ve number.
So -ve number < 0 or some +ve number is true.

2) Sufficient. As x > 2, x(2-x) will be a -ve term. and as y > 0 so 3y will be greater than zero.
So again -ve number < +ve number is true.

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SVP
Joined: 28 Dec 2005
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10 Aug 2008, 13:42
i would say D. i can think of two ways to do this, algebraically or just by picking numbers if youre in a crunch.

2x-x^2 < 3y
x(2-x)< 3y.

we are told in stat 2 that x>2, so the left side will give you a negative number, and since y>0, the RS is positive ... since RS is positive and LS is negative, this is sufficient.

for stat 1, since we have x^2 on the RS, it will always be positive.

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Manager
Joined: 15 Jul 2008
Posts: 205

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10 Aug 2008, 18:29
D.

Stmt 1 is obvious, RHS is a square so always +ve. so sufficient.

stmt 2 is sufficient because x>2. General rule is that for any x>2, x^2 > 2x. So if 2x < x^2, then 2x-3y < x^2 for every y>0

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Re: DS: Algebra Inequalities   [#permalink] 10 Aug 2008, 18:29
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