Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

(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.

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) _________________

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. _________________

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.

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 _________________

If you like my post, consider giving me some KUDOS !!!!! Like you I need them

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)

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.

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

(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! _________________