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Re: Is 2x - 3y < x^2 ? (1) 2x - 3y = -2 (2) x > 2 and y > 0 [#permalink]
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Thanxs Bunuel!, you rule!
don't you ever think in writing a book about GMAT?
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Re: Is 2x - 3y < x^2 ? (1) 2x - 3y = -2 (2) x > 2 and y > 0 [#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: Is 2x - 3y < x^2 ? (1) 2x - 3y = -2 (2) x > 2 and y > 0 [#permalink]
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: Is 2x - 3y < x^2 ? (1) 2x - 3y = -2 (2) x > 2 and y > 0 [#permalink]
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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: Is 2x - 3y < x^2 ? (1) 2x - 3y = -2 (2) x > 2 and y > 0 [#permalink]
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 > 0 [#permalink]
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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:

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Re: Is 2x - 3y < x^2 ? (1) 2x - 3y = -2 (2) x > 2 and y > 0 [#permalink]
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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 33498 times ]

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Re: Is 2x - 3y < x^2 ? (1) 2x - 3y = -2 (2) x > 2 and y > 0 [#permalink]
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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 ? (1) 2x - 3y = -2 (2) x > 2 and y > 0 [#permalink]
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metallicafan wrote:
Is 2x - 3y < x^2 ?


(1) 2x - 3y = -2

(2) x > 2 and y > 0


Statement 1: \(2x - 3y = -2\)
Substituting \(2x-3y=-2\) into \(2x - 3y < x^2\), we can rephrase the question stem as follows:
Is \(-2 < x^2\)?
Since the least possible value for \(x^2\) is 0, the answer to the rephrased question stem is YES.

Statement 2: \(x > 2\) and \(y > 0\)
If \(x=2\), then \(2x=x^2\).
For all values of \(x\) greater than 2, \(2x < x^2\).
Since \(y>0,\) \(2x-3y < 2x.\)
Linking together \(2x-3y<2x\) and \(2x< x^2\), we get:
\(2x-3y<2x<x^2\)
\(2x-3y<x^2\)
Thus, the answer to the question stem is YES.
SUFFICIENT.

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Re: Is 2x - 3y < x^2 ? (1) 2x - 3y = -2 (2) x > 2 and y > 0 [#permalink]
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metallicafan wrote:
Is 2x - 3y < x² ?

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


Target question: Is 2x - 3y < x² ?

Statement 1: 2x - 3y = -2
The target question becomes: Is -2 < x² ?
Since x² is greater than or equal to 0 for all values of x, we can be certain that -2 is less than x²
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x > 2 and y > 0
Let's first take the target question, Is 2x - 3y < x²?, and subtract 2x from both sides to get: Is -3y < x² - 2x?
Factor to get: Is -3y < x(x - 2)?

If x > 2 then (x - 2) is POSITIVE, and x is POSITIVE
Conversely, if y > 0, then -3y is NEGATIVE.
The rephrased target question becomes, Is some NEGATIVE number < (POSITIVE)(POSITIVE)?
The answer to the rephrased target question is YES, some NEGATIVE number is less than (POSITIVE)(POSITIVE)
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
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Re: Is 2x - 3y < x^2 ? (1) 2x - 3y = -2 (2) x > 2 and y > 0 [#permalink]
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