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

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Director
Joined: 23 May 2008
Posts: 739
is 2x-3y<x^2 ? 1) 2x-3y = -2 2) x>2 and y>0  [#permalink]

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23 Sep 2008, 00:39
1
is 2x-3y2 and y>0

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VP
Joined: 30 Jun 2008
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Updated on: 23 Sep 2008, 00:54
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.

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Originally posted by amitdgr on 23 Sep 2008, 00:43.
Last edited by amitdgr on 23 Sep 2008, 00:54, edited 1 time in total.
Senior Manager
Joined: 28 Feb 2007
Posts: 293

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23 Sep 2008, 00:47
D. X^2 can be 0 or any +#.so st1 is suffi. St2) sufficient.
Director
Joined: 23 May 2008
Posts: 739

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23 Sep 2008, 00:50
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
Joined: 12 Feb 2008
Posts: 175

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23 Sep 2008, 07:38
when you have X2 it will always be positive.
D

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Re: from veritas prep &nbs [#permalink] 23 Sep 2008, 07:38
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