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Is 2x - 3y < x^2 ?

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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]

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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]

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Re: Inequalities [#permalink]

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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]

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New post 11 Sep 2010, 15:09
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]

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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]

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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]

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New post 13 Sep 2010, 05:58
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.

Thanks

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Re: Inequalities [#permalink]

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New post 13 Sep 2010, 08:21
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]

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New post 13 Sep 2010, 22:44
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]

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New post 09 Apr 2013, 10:04
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]

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New post 09 Apr 2013, 10:16
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]

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New post 04 Jul 2013, 01:43
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: Is 2x - 3y < x^2 ? [#permalink]

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New post 23 Mar 2015, 18:28
Hi All,

In this DS question, you might find that 'rewriting' the question makes it easier to answer. Either way, you'll find that a combination of logic, Number Properties and TESTing VALUES will come in handy.

We're asked if 2X - 3Y < X^2. You can 'rewrite' the question to ask if 2X < X^2 + 3Y. Either way, this is a YES/NO question.

Fact 1: 2X - 3Y = -2

Here, the 'original' version of the question is probably easier to answer, since we now have a value that we can 'substitute' in for (2X - 3Y)....

The question now asks.....

Is -2 < X^2?

X^2 can be 0 or any positive value, so with ALL possible values of X, the answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT

Fact 2: X > 2 and Y > 0

With this Fact, the 'rewritten' version of the question is probably easier to answer.

Since we know that X > 2......2X will ALWAYS be < X^2.

We also know that Y > 0, so (X^2 + 3Y) will get larger as Y gets larger. This all serves as evidence that....

2X is ALWAYS < X^2 + 3Y. The answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT

Final Answer:
[Reveal] Spoiler:
D


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Is 2x - 3y < x^2 ? [#permalink]

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New post 27 Mar 2016, 20:17
Here is a visual that should help.
Attachments

Screen Shot 2016-03-27 at 8.17.47 PM.png
Screen Shot 2016-03-27 at 8.17.47 PM.png [ 476.59 KiB | Viewed 2219 times ]


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Re: Is 2x - 3y < x^2 ? [#permalink]

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New post 28 Mar 2016, 05:53
metallicafan wrote:
Is 2x - 3y < x^2 ?

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


Statement 1)
2x-3y =-2 => 3y =2x+2
The question -> 2x-3y<x^2 => 2x-x^2<3y
Comparing these e equations we get that 3y> 2x-x^2 (x^2 is always positive and hence the value of 2x-x^2 will always be less than 2x+2
Sufficient

Statement 2)
x>2, y>0
Consider, x=4 and y = 2, substituting the values of x and y in 2x-3y<x^2, we get -> 2<16;
or
x=3 and y=4 we get 2*3-3*4<3^2 => 6-12<9 or -6<9
In all the cases we get 2x-3y<x^2

Sufficient.

Answer D

Please suggest if this approach is correct. Thanks a lot :)

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Re: Is 2x - 3y < x^2 ? [#permalink]

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New post 19 Oct 2016, 02:09
Hi Bunuel,

Can we not consider imaginary number such as (\sqrt{(-2)}) that woud result -2 for x^2?

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Re: Is 2x - 3y < x^2 ? [#permalink]

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Re: Is 2x - 3y < x^2 ? [#permalink]

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New post 11 Feb 2017, 00:45
This is a yes/no problem

Statement1:
if 2x - 3y = -2 , then 2x - 3y will always be less than x^2 because x^2 has to be a positive number. Sufficient to answer the question

Statement2:
2x=x^2 when x equals 2. This means that 2x is always less than x^2 if x is greater than 2. 3y is not a concern because we are subtracting that quantity from 2x. 3y would only be a concern if y is less than zero and the "-3y" becomes a positive term. However, statement 2 lists that y>0. This means that the term "-3y" will always be negative, and for possible combinations of x and y, 2x-3y will always be greater than x^2. Sufficient to answer the question.

Both statements are sufficient, the answer should be D.
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Re: Is 2x - 3y < x^2 ?   [#permalink] 11 Feb 2017, 00:45
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