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Is 2x - 3y < x^2 ?

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Is 2x - 3y < x^2 ? [#permalink] New post 07 Sep 2010, 12:15
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Is 2x - 3y < x^2 ?

(1) 2x - 3y = -2
(2) x > 2 and y > 0
[Reveal] Spoiler: OA

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Re: Inequalities [#permalink] New post 07 Sep 2010, 12:23
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metallicafan wrote:
Is 2x - 3y < x^2?

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


Is 2x-3y<x^2?

(1) 2x-3y=-2 --> question becomes: is -2<x^2? as square of a number is always non-negative (x^2\geq{0}) then x^2\geq{0}>-2. Sufficient.

(2) x>2 and y>0 --> is 2x-3y<x^2 --> is x(x-2)+3y>0 --> as x>2 then x(x-2) is a positive number and as y>0 then 3y is also a positive number --> sum of two positive numbers is more than zero, hence x(x-2)+3y>0 is true. Sufficient.

Answer: D.
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Re: Inequalities [#permalink] New post 07 Sep 2010, 12:32
Thanxs Bunuel!, you rule!
don't you ever think in writing a book about GMAT?
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Re: Inequalities [#permalink] New post 07 Sep 2010, 12:42
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metallicafan wrote:
Is 2x - 3y < x^2 ?

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



1. you should see you can replace the terms of 2x-3y so you get -2<x^2 ? since any number squared is positive this is SUFF
2. You can view this problem in another way:
2x will always be less than x^2 as x >2. And since y >0 that means the left side 2x-3y will be even fewer than 2x so left side will always be less than x^2 SUFF
(brunnel's answer is another approach so i chose this way from another pespective)
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Re: Inequalities [#permalink] New post 11 Sep 2010, 14:09
shaselai wrote:
metallicafan wrote:
Is 2x - 3y < x^2 ?

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



1. you should see you can replace the terms of 2x-3y so you get -2<x^2 ? since any number squared is positive this is SUFF
2. You can view this problem in another way:
2x will always be less than x^2 as x >2. And since y >0 that means the left side 2x-3y will be even fewer than 2x so left side will always be less than x^2 SUFF
(brunnel's answer is another approach so i chose this way from another pespective)

Thanks shaselai, much easier to follow.
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Re: Inequalities [#permalink] New post 11 Sep 2010, 14:42
Expert's post
Dawgie wrote:
shaselai wrote:
metallicafan wrote:
Is 2x - 3y < x^2 ?

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



1. you should see you can replace the terms of 2x-3y so you get -2<x^2 ? since any number squared is positive this is SUFF
2. You can view this problem in another way:
2x will always be less than x^2 as x >2. And since y >0 that means the left side 2x-3y will be even fewer than 2x so left side will always be less than x^2 SUFF
(brunnel's answer is another approach so i chose this way from another pespective)

Thanks shaselai, much easier to follow.


Just a little correction, which makes no difference for this particular question but is very important, as GMAT likes to catch on differences like this: square of a number is non-negative (and not positive) --> x^2\geq{0}, because if x=0 then x^2=0.

Hope it helps.
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Re: Inequalities [#permalink] New post 11 Sep 2010, 14:47
In heat of solving question, I missed to notice that statement 1 is same as inquality question. I concluded with D but after spending time
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Re: Inequalities [#permalink] New post 13 Sep 2010, 04:58
Bunuel,

Your explaination for why second statement alone is sufficient to answer the question
proves that x(x-2)+3y > o. But this does not answer whether ( 2x-3y < x^2)

Can you please explain.

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Re: Inequalities [#permalink] New post 13 Sep 2010, 07:21
Expert's post
prashantbacchewar wrote:
Bunuel,

Your explaination for why second statement alone is sufficient to answer the question
proves that x(x-2)+3y > o. But this does not answer whether ( 2x-3y < x^2)

Can you please explain.

Thanks


Question: "is 2x-3y<x^2?" --> rearrange --> "is 2x-x^2-3y<0" --> and finally the question becomes "is x(x-2)+3y>0?". So 2x-3y<x^2 and x(x-2)+3y>0 are the same, if you prove that x(x-2)+3y>0 is true then you know that 2x-3y<x^2 is also true.

Hope it's clear.
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Re: Inequalities [#permalink] New post 13 Sep 2010, 21:44
Thanks Bunuel.. Now it is clear
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Re: Is 2x - 3y < x^2 ? (1) 2x - 3y = -2 (2) x > 2 and y [#permalink] New post 09 Apr 2013, 09:04
Dear Bunuel
Need some clarification on this question, as i am getting A as an answer
What i did:
2x - 3y < X^2 - since x^2 will be positive number i divided both sides by X^2 - the equation provided became (2x-3y)/x^2<0. later when i plugged in various numbers to test the validity they gave me both a Yes and a No answer.
Where am i going wrong. Can you please correct me?
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Re: Is 2x - 3y < x^2 ? (1) 2x - 3y = -2 (2) x > 2 and y [#permalink] New post 09 Apr 2013, 09:16
mbhussain,

Don't divide but just subtract x2 from both sides of the equation

-x2 +2x - 3y < 0 ==> x2 - 2x +3y > 0 ==> x(x-2)+3y > 0

Since statement II says y > 0 then 3y is positive

Also x > 2 implies x(x-2) is also positive

Positive + Positive will always be > 0. Hence B is sufficient. You can use numbers that meet Statement II and you'll get the same result.

//kudos please, if the above explanation is good.
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Re: Inequalities [#permalink] New post 04 Jul 2013, 00:43
Bunuel wrote:

(2) x>2 and y>0 --> is 2x-3y<x^2 --> is x(x-2)+3y>0 --> as x>2 then x(x-2) is a positive number and as y>0 then 3y is also a positive number --> sum of two positive numbers is more than zero, hence x(x-2)+3y>0 is true. Sufficient.

Answer: D.


Great Manipulation for Statement 2!
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Re: Inequalities   [#permalink] 04 Jul 2013, 00:43
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