It is currently 20 Oct 2017, 20:27

### 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 June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 06 Apr 2010
Posts: 141

Kudos [?]: 936 [4], given: 15

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

### Show Tags

30 Aug 2010, 09:38
4
KUDOS
11
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

45% (01:33) correct 55% (01:12) wrong based on 290 sessions

### HideShow timer Statistics

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

(1) $$\sqrt{x} > y$$
(2) x > y
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Dec 2012, 03:58, edited 2 times in total.
Edited the question.

Kudos [?]: 936 [4], given: 15

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129046 [6], given: 12187

### Show Tags

30 Aug 2010, 10:31
6
KUDOS
Expert's post
2
This post was
BOOKMARKED
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.
_________________

Kudos [?]: 129046 [6], given: 12187

Manager
Joined: 17 Mar 2010
Posts: 173

Kudos [?]: 210 [1], given: 9

### Show Tags

31 Aug 2010, 04:07
1
KUDOS
When there are squaring or cubing is involved, always remember -ve values as well as the values between -1 and 1

Kudos [?]: 210 [1], given: 9

Manager
Joined: 05 Jan 2011
Posts: 171

Kudos [?]: 188 [0], given: 8

### Show Tags

18 Mar 2011, 07:11
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

Kudos [?]: 188 [0], given: 8

SVP
Joined: 16 Nov 2010
Posts: 1597

Kudos [?]: 592 [0], given: 36

Location: United States (IN)
Concentration: Strategy, Technology

### Show Tags

18 Mar 2011, 21:25
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

Kudos [?]: 592 [0], given: 36

Manager
Joined: 23 May 2013
Posts: 191

Kudos [?]: 108 [1], given: 42

Location: United States
Concentration: Technology, Healthcare
Schools: Stanford '19 (M)
GMAT 1: 760 Q49 V45
GPA: 3.5
Re: If x and y are positive, is x^3 > y? [#permalink]

### Show Tags

30 May 2014, 07:10
1
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.

Last edited by eaze on 30 May 2014, 15:37, edited 1 time in total.

Kudos [?]: 108 [1], given: 42

Intern
Joined: 06 Nov 2013
Posts: 1

Kudos [?]: [0], given: 1

### Show Tags

30 May 2014, 14:46
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 ??

Kudos [?]: [0], given: 1

Manager
Joined: 23 May 2013
Posts: 191

Kudos [?]: 108 [0], given: 42

Location: United States
Concentration: Technology, Healthcare
Schools: Stanford '19 (M)
GMAT 1: 760 Q49 V45
GPA: 3.5

### Show Tags

30 May 2014, 15:40
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.

Kudos [?]: 108 [0], given: 42

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16637

Kudos [?]: 273 [0], given: 0

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

### Show Tags

21 Feb 2016, 08:45
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.
_________________

Kudos [?]: 273 [0], given: 0

VP
Joined: 14 Nov 2016
Posts: 1161

Kudos [?]: 1182 [0], given: 415

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

### Show Tags

14 Feb 2017, 00:26
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 444 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."

Kudos [?]: 1182 [0], given: 415

Re: If x and y are positive, is x^3 > y?   [#permalink] 14 Feb 2017, 00:26
Display posts from previous: Sort by