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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

If x and y are positive, is \(x^3\) > y? (1) \sqrt{x} > y (2) x > y

My approach: 1.\sqrt{x}>y ....>x>\(y^2\)....NS 2. x>y ....NS

Combine...\(y^2\)>x>y....> \(x^3\)>y. This can be verified by taking no like 4>3>2..where y=2 and x=3

Any other apporach.

If x and y are positive, is x^3>y?

NUMBER PLUGGING:

(1) \(\sqrt{x}>y\) --> if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient.

(2) \(x>y\) --> if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient.

(1)+(2) Both examples are valid for combined statements, so we still have two answers. Not sufficient.

Answer: E.

ALGEBRAIC APPROACH:

For \(1\leq{x}\): ------\(\sqrt{x}\)----\(x\)----\(x^3\), so \(1\leq{\sqrt{x}}\leq{x}\leq{x^3}\) (the case \(\sqrt{x}=x=x^3\) is when \(x=1\)). \(y\) is somewhere in green zone (as \(y<\sqrt{x}\) and \(y<x\)), so if we have this case answer is always YES: \(y<x^3\).

But:

For \(0<x<1\): \(0\)----\(x^3\)----\(x\)----\(\sqrt{x}\)----\(1\), so \(0<x^3<x<\sqrt{x}\). \(y\) is somewhere in green or red zone (as \(y<\sqrt{x}\) and \(y<x\)), so if we have this case answer is sometimes YES: \(y<x^3\) (if \(y\) is in green zone), and sometimes NO: \(x^3<y\) (if \(y\) is in red zone). In fact in this case \(y=x^3\) is also possible, for example when \(x=\frac{1}{2}\) and \(y=\frac{1}{8}\)

If x and y are positive, is \(x^3\) > y? (1) \sqrt{x} > y (2) x > y

My approach: 1.\sqrt{x}>y ....>x>\(y^2\)....NS 2. x>y ....NS

Combine...\(y^2\)>x>y....> \(x^3\)>y. This can be verified by taking no like 4>3>2..where y=2 and x=3

Any other apporach.

If x and y are positive, is x^3>y?

NUMBER PLUGGING:

(1) \(\sqrt{x}>y\) --> if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient.

(2) \(x>y\) --> if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient.

(1)+(2) Both examples are valid for combined statements, so we still have two answers. Not sufficient.

Answer: E.

ALGEBRAIC APPROACH:

For \(1\leq{x}\): ------\(\sqrt{x}\)----\(x\)----\(x^3\), so \(1\leq{\sqrt{x}}\leq{x}\leq{x^3}\) (the case \(\sqrt{x}=x=x^3\) is when \(x=1\)). \(y\) is somewhere in green zone (as \(y<\sqrt{x}\) and \(y<x\)), so if we have this case answer is always YES: \(y<x^3\).

But:

For \(0<x<1\): \(0\)----\(x^3\)----\(x\)----\(\sqrt{x}\)----\(1\), so \(0<x^3<x<\sqrt{x}\). \(y\) is somewhere in green or red zone (as \(y<\sqrt{x}\) and \(y<x\)), so if we have this case answer is sometimes YES: \(y<x^3\) (if \(y\) is in green zone), and sometimes NO: \(x^3<y\) (if \(y\) is in red zone). In fact in this case \(y=x^3\) is also possible, for example when \(x=\frac{1}{2}\) and \(y=\frac{1}{8}\)

Re: If x and y are positive, is x^3 > y? [#permalink]

Show Tags

30 May 2014, 06:10

1

This post received KUDOS

udaymathapati wrote:

If x and y are positive, is x^3 > y?

(1) \(\sqrt{x} > y\) (2) x > y

My method without number plugging:

1) \(\sqrt{x}>y\)

Square both sides. Since x & y are both positive, we know this doesn't change the sign at all (because \(f(x) = x^2\) is a strictly increasing function for \(x \geq 0\)). Thus \(x > y^2\). We're trying to get this in a form that resembles our original question, so we multiply each side by \(x^2\)(which we know is positive). Thus, \(x^3 > x^2y^2\), or that\(x^3 > y (x^2 y).\) Therefore, our question is TRUE if and only if \(x^2y\geq 1\). Since we don't know anything about the behavior of\(x^2y\), we mark 1) as INSUFFICIENT.

2) \(x>y.\)

Multiplying both sides by \(x^2\), we get that \(x^3 > yx^2\). Thus, if \(x^2 \geq 1,\) we get TRUE, otherwise, we get FALSE. Since we have no information about \(x^2\), we mark this as INSUFFICIENT.

Taking 1) and 2) together, we still have no information about \(x^2\) or \(x^2y\), so 1) & 2) are INSUFFICIENT.

Answer: E.

Last edited by speedilly on 30 May 2014, 14:37, edited 1 time in total.

If x and y are positive, is \(x^3\) > y? (1) \sqrt{x} > y (2) x > y

My approach: 1.\sqrt{x}>y ....>x>\(y^2\)....NS 2. x>y ....NS

Combine...\(y^2\)>x>y....> \(x^3\)>y. This can be verified by taking no like 4>3>2..where y=2 and x=3

Any other apporach.

If x and y are positive, is x^3>y?

NUMBER PLUGGING:

(1) \(\sqrt{x}>y\) --> if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient.

(2) \(x>y\) --> if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient.

(1)+(2) Both examples are valid for combined statements, so we still have two answers. Not sufficient.

Answer: E.

ALGEBRAIC APPROACH:

For \(1\leq{x}\): ------\(\sqrt{x}\)----\(x\)----\(x^3\), so \(1\leq{\sqrt{x}}\leq{x}\leq{x^3}\) (the case \(\sqrt{x}=x=x^3\) is when \(x=1\)). \(y\) is somewhere in green zone (as \(y<\sqrt{x}\) and \(y<x\)), so if we have this case answer is always YES: \(y<x^3\).

But:

For \(0<x<1\): \(0\)----\(x^3\)----\(x\)----\(\sqrt{x}\)----\(1\), so \(0<x^3<x<\sqrt{x}\). \(y\) is somewhere in green or red zone (as \(y<\sqrt{x}\) and \(y<x\)), so if we have this case answer is sometimes YES: \(y<x^3\) (if \(y\) is in green zone), and sometimes NO: \(x^3<y\) (if \(y\) is in red zone). In fact in this case \(y=x^3\) is also possible, for example when \(x=\frac{1}{2}\) and \(y=\frac{1}{8}\)

Answer: E.

Hope it's clear.

What if the question says x and y are positive INTEGERS ??

What if the question says x and y are positive INTEGERS ??

If they were both integers, then the answer would be D: either of them are sufficient, since we'd know that\(x^2y \geq 1\) and \(x^2 \geq 1\)(see my solution above). The question doesn't state that, however, so we must consider all cases.

Re: If x and y are positive, is x^3 > y? [#permalink]

Show Tags

21 Feb 2016, 07:45

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...