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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From Statement 1, we know that 2x - 3y is negative, so it certainly must be less than x^2, which cannot be negative.

From Statement 2, we know that 2 < x. Multiplying by x on both sides, that means 2x < x^2. Now, if y is positive, certainly 2x - 3y < 2x, and since 2x < x^2, then 2x - 3y is definitely less than x^2.

The answer is D.
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Thanxs Bunuel!, you rule!
don't you ever think in writing a book about GMAT?
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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

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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Thanks Bunuel.. Now it is clear

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?

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.

Great Manipulation for Statement 2!

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:

GMAT assassins arne't born, they're made,
Rich

Here is a visual that should help.

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

Hi Bunuel,

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

No. All numbers on the GMAT are by default real numbers.
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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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Asked: Is 2x - 3y < x^2 ?


(1) 2x - 3y = -2
2x - 3y = - 2 < 0 < = x^2
SUFFICIENT

(2) x > 2 and y > 0
For x > 2 ; 2x < x^2
-3y < 0
2x - 3y < 2x < x^2
SUFFICIENT

IMO D

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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Here is a simple solution:

St(1) -: 2x-3y=-2, substituting in the inequality: (2x-3y<x^2)....> -2<x^2 or Is X^2>-2 ? , Ofcourse Yes! Square of any number is >=0 (Sufficient)

St(2) -: x>2 & Y>0

Original inequality : 2x-3y<x^2
or.... 2x-x^2<3y
or.... x(2-x)<3y

Now if we plug in the above inequality x>2 & y>0, we know that LHS will always be Negative and RHS will always be positive, Hence Sufficient

Answer : D

Hope it helps..!