GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Jan 2019, 20:59

### GMAT Club Daily Prep

#### 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

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

## Events & Promotions

###### Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### The winning strategy for a high GRE score

January 17, 2019

January 17, 2019

08:00 AM PST

09:00 AM PST

Learn the winning strategy for a high GRE score — what do people who reach a high score do differently? We're going to share insights, tips and strategies from data we've collected from over 50,000 students who used examPAL.
• ### Free GMAT Strategy Webinar

January 19, 2019

January 19, 2019

07:00 AM PST

09:00 AM PST

Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 06 Apr 2010
Posts: 117
If x and y are positive, is x^3 > y?  [#permalink]

### Show Tags

Updated on: 02 Dec 2012, 02:58
5
8
00:00

Difficulty:

85% (hard)

Question Stats:

48% (02:01) correct 52% (01:45) wrong based on 295 sessions

### HideShow timer Statistics

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

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

Originally posted by udaymathapati on 30 Aug 2010, 08:38.
Last edited by Bunuel on 02 Dec 2012, 02:58, edited 2 times in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 52164

### Show Tags

30 Aug 2010, 09:31
6
3
udaymathapati wrote:
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.

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 $$y=\frac{1}{8}$$

Hope it's clear.
_________________
##### General Discussion
Manager
Joined: 16 Mar 2010
Posts: 143

### Show Tags

31 Aug 2010, 03:07
1
When there are squaring or cubing is involved, always remember -ve values as well as the values between -1 and 1
Manager
Joined: 05 Jan 2011
Posts: 124

### Show Tags

18 Mar 2011, 06:11
Bunuel wrote:
udaymathapati wrote:
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.

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 $$y=\frac{1}{8}$$

Hope it's clear.

+1 for u
Great Post Thanks Bunuel
Retired Moderator
Joined: 16 Nov 2010
Posts: 1420
Location: United States (IN)
Concentration: Strategy, Technology

### Show Tags

18 Mar 2011, 20:25
x = 1/4, y = 1/8

From (1)

1/2 > 1/8, but 1/8 = 1/8

but (9)^3 > 2

and sqrt(9) > 2

In (2) also, the same is true.

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Current Student
Joined: 23 May 2013
Posts: 187
Location: United States
Concentration: Technology, Healthcare
GMAT 1: 760 Q49 V45
GPA: 3.5
Re: If x and y are positive, is x^3 > y?  [#permalink]

### Show Tags

Updated on: 30 May 2014, 14:37
2
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.

Originally posted by eaze on 30 May 2014, 06:10.
Last edited by eaze on 30 May 2014, 14:37, edited 1 time in total.
Intern
Joined: 06 Nov 2013
Posts: 1

### Show Tags

30 May 2014, 13:46
Bunuel wrote:
udaymathapati wrote:
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.

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 $$y=\frac{1}{8}$$

Hope it's clear.

What if the question says x and y are positive INTEGERS ??
Current Student
Joined: 23 May 2013
Posts: 187
Location: United States
Concentration: Technology, Healthcare
GMAT 1: 760 Q49 V45
GPA: 3.5

### Show Tags

30 May 2014, 14:40
ShajibAhmed wrote:

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.
Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1323
Location: Malaysia
Re: If x and y are positive, is x^3 > y?  [#permalink]

### Show Tags

13 Feb 2017, 23:26
Bunuel wrote:
udaymathapati wrote:
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.

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 $$y=\frac{1}{8}$$

Hope it's clear.

You will get a better understanding by drawing the number line and plugging in the value.
Attachments

Untitled.png [ 5.87 KiB | Viewed 1142 times ]

_________________

"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

Non-Human User
Joined: 09 Sep 2013
Posts: 9423
Re: If x and y are positive, is x^3 > y?  [#permalink]

### Show Tags

22 Apr 2018, 12:02
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.
_________________
Re: If x and y are positive, is x^3 > y? &nbs [#permalink] 22 Apr 2018, 12:02
Display posts from previous: Sort by