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Is 2x - 3y < x^2 ?

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Is 2x - 3y < x^2 ? [#permalink]

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Is 2x - 3y < x^2 ?

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


OPEN DISCUSSION OF THIS QUESTION IS HERE: is-2x-3y-x-100596.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 23 Mar 2015, 03:11, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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Re: Is 2x - 3y < x^2 ? [#permalink]

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New post 05 Dec 2005, 21:03
mand-y wrote:
:) Is 2x-3y<x^3?

ST 1 2x-3y=-2

ST x>2 and y>0


Plz explain your approach ....


THANKS


My answer is B

Is 2x-3y<x^3?

(1) 2x - 3y = -2 ==> -2 < x^3 ==> no info about x. Insufficient
(2) x > 2 and y > 0 ==> Sufficient

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Re: Is 2x - 3y < x^2 ? [#permalink]

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New post 05 Dec 2005, 23:19
Stmt 1 not suff. Since we do not know the value of x.
Stmt 2. True for any values of x and y. Hence B

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Re: Is 2x - 3y < x^2 ? [#permalink]

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New post 10 Dec 2005, 10:54
You have a typo in your question....x^3 should read 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 22 Mar 2015, 22:06
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^3. You can 'rewrite' the question to ask if 2X < X^3 + 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^3?

The answer to this question depends completely on the value of X.
IF....
X = 0, then the answer to the question is YES
X = -2, then the answer to the question is NO
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^3. We also know that Y > 0, so (X^3 + 3Y) will get larger as Y gets larger. This all serves as evidence that....

2X is ALWAYS < X^3 + 3Y.
Fact 2 is SUFFICIENT

Final Answer:
[Reveal] Spoiler:
B


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Re: Is 2x - 3y < x^2 ? [#permalink]

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New post 22 Mar 2015, 22:33
Quick way to solve this:
St1 - depends on unknown variable X - henice can't be determined.
St2 - all possible values can be tested and hence sufficient.

Hence B.

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Re: Is 2x - 3y < x^2 ? [#permalink]

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mand-y 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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-2x-3y-x-100596.html
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Re: Is 2x - 3y < x^2 ?   [#permalink] 23 Mar 2015, 03:12
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