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

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Director
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is 2x-3y<x^2 ? 1) 2x-3y = -2 2) x>2 and y>0 [#permalink]  22 Sep 2008, 23:39
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is 2x-3y<x^2 ?

1) 2x-3y = -2
2) x>2 and y>0
VP
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Re: from veritas prep [#permalink]  22 Sep 2008, 23:43
bigtreezl wrote:
is 2x-3y<x^2 ?

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

1) 2x-3y =-2. This implies that -2 < x^2 . Sufficient since -2 is negative and x^2 is always positive.

2) x>2 and y>0.
2x-3y<x^2 will always hold good. We have to first understand that for +ve numbers ( apart from those between 0 and 1) x^2 is always bigger than 2x. And here in our equation, we are subtracting a part of 2x in the form of 3y where y is a positive number. So 2x-3y will definitely be smaller than x^2

Some examples:
if x=3 and y=1 => 2(3)-3(1)= 3 which is less than x^2 ie 9
if x=3 and y=3 => 2(3)-3(3)=-3 which is less than x^2 ie 9
if x=1000 y=100 => 2(1000)-3(100)=1700 again less than x^2 which is 1000000

2 also sufficient.

_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Last edited by amitdgr on 22 Sep 2008, 23:54, edited 1 time in total.
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Re: from veritas prep [#permalink]  22 Sep 2008, 23:47
D. X^2 can be 0 or any +#.so st1 is suffi. St2) sufficient.
Director
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Re: from veritas prep [#permalink]  22 Sep 2008, 23:50
OA is D

"in statement 2, x>2 so x^2>2x, and since y>0, 3y is negative, therefore 2x-3y must be less than X^2"
Manager
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Re: from veritas prep [#permalink]  23 Sep 2008, 06:38
when you have X2 it will always be positive.
D
Re: from veritas prep   [#permalink] 23 Sep 2008, 06:38
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