Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 23 Aug 2016, 14:07

### 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

# Is x > y ? (1) x^(1/2)>y (2) x^3>y

Author Message
TAGS:

### Hide Tags

Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 100

Kudos [?]: 850 [7] , given: 25

Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink]

### Show Tags

08 Sep 2010, 01:38
7
KUDOS
48
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

37% (02:25) correct 63% (01:14) wrong based on 1237 sessions

### HideShow timer Statistics

I thought this was a really tough question !

Is $$x > y$$ ?

(1) $$\sqrt{x} > y$$
(2) $$x^3 > y$$
[Reveal] Spoiler: OA

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 34393
Followers: 6244

Kudos [?]: 79303 [68] , given: 10016

Re: Is x > y ? [#permalink]

### Show Tags

08 Sep 2010, 05:10
68
KUDOS
Expert's post
27
This post was
BOOKMARKED
shrouded1 wrote:
I thought this was a really tough question !

Is $$x > y$$ ?

(1) $$\sqrt{x} > y$$
(2) $$x^3 > y$$

Is x>y?

(1) $$\sqrt{x}>y$$ --> if $$x=4$$ and $$y=1$$ then the answer will be YES but if $$x=\frac{1}{4}$$ and $$y=\frac{1}{3}$$ then the answer will be NO. Two different answers, hence not sufficient.

Note that from this statement we can derive that $$x\geq{0}$$ because an expression under the square root cannot be negative.

(2) $$x^3>y$$ --> if $$x=4$$ and $$y=1$$ then the answer will be YES but if $$x=2$$ and $$y=3$$ then the answer will be NO. Two different answers, hence not sufficient.

(1)+(2) From (1) we have that $$x\geq{0}$$. Now, $$\sqrt{x}$$, $$x$$, $$x^3$$ can be positioned on a number line only in 2 ways:

1. 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 $$y<x^3$$), so if we have this case answer is always YES: $$y<x$$.

2. For $$0\leq{x}<1$$: $$0$$----$$x^3$$----$$x$$----$$\sqrt{x}$$----$$1$$, so $$0\leq{x^3}\leq{x}\leq{\sqrt{x}}$$. $$y$$ is somewhere in green zone (as $$y<\sqrt{x}$$ and $$y<x^3$$), so if we have this case answer is always YES: $$y<x$$.

So in both cases $$y<x$$. Sufficient.

Hope it's clear.
_________________
Manager
Status: Keep fighting!
Joined: 31 Jul 2010
Posts: 236
WE 1: 2+ years - Programming
WE 2: 3+ years - Product developement,
WE 3: 2+ years - Program management
Followers: 4

Kudos [?]: 374 [0], given: 104

Re: Is x > y ? [#permalink]

### Show Tags

08 Sep 2010, 23:50
1
This post was
BOOKMARKED
Bunuel.... you seem to love number theory! I really am amazed at your patience. Good job. You already have enough Kudos.
Senior Manager
Joined: 20 Jul 2010
Posts: 269
Followers: 2

Kudos [?]: 66 [4] , given: 9

Re: Is x > y ? [#permalink]

### Show Tags

09 Sep 2010, 08:25
4
KUDOS
interesting question.......forgot to consider fractions first.......now its clear.....
_________________

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

VP
Status: Current Student
Joined: 24 Aug 2010
Posts: 1345
Location: United States
GMAT 1: 710 Q48 V40
WE: Sales (Consumer Products)
Followers: 108

Kudos [?]: 418 [4] , given: 73

Re: Is x > y ? [#permalink]

### Show Tags

16 Sep 2010, 19:08
4
KUDOS
That solution is so simple yet I couldn't figure it out. I knew that each statement alone was insufficient, but I couldn't figure out whether both together were sufficient. I kept trying to plug in numbers and it just never worked. Never did it occur to me to put both scenarios on a number line and just use the statements to prove it. I swear the math on the GMAT really makes you think in different ways and the solutions are so easy we make it harder than it needs to be.
_________________

The Brain Dump - From Low GPA to Top MBA (Updated September 1, 2013) - A Few of My Favorite Things--> http://cheetarah1980.blogspot.com

Intern
Joined: 15 Sep 2010
Posts: 1
Followers: 0

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

Re: Is x > y ? [#permalink]

### Show Tags

16 Sep 2010, 22:24
shrouded1 wrote:
I thought this was a really tough question !

Is $$x > y$$ ?

(1) $$\sqrt{x} > y$$
(2) $$x^3 > y$$

（1）x^2>y^4
(2) x^3 > y

if both 1and 2 are true,
then x>0,
if y<=0,
then x>y;
if y>0,
we make x,y have the same power by (1)*（2）
x^5>y^5,
so x>y;
so x>y.
Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 100

Kudos [?]: 850 [3] , given: 25

Re: Is x > y ? [#permalink]

### Show Tags

09 Oct 2010, 09:10
3
KUDOS
Again there is a wonderfully simple explanation for this one using graphs.

x>y, sqrt(x)>y, x^3>y all three represents the region below the graph for all three cases.

We need to answer is x>y

(1) sqrt(x)>y
You are below the yellow line does not imply you are below the purple line. Insufficient

(2) x^3>y
Now x need not be just positive, but looking at the graph is enough to conclude this is not sufficient. Being below the blue line does not imply being below the purple line

(1+2) Now x>0 since we are using sqrt(x)
You are below the blue line and the yellow line both
To satisfy both, you must always be below the purple line

_________________
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 220

Kudos [?]: 1508 [5] , given: 235

Re: Is x > y ? [#permalink]

### Show Tags

09 Oct 2010, 10:02
5
KUDOS
Statement 1: $$\sqrt{x} > y$$ ....take x = 4, y = 1 (yes); take x = 1/4 and y = 1/3 (no) hence not sufficient.

Statement 2: $$x^3 > y$$ take x = 2 and y = 3 (no) , take x = 2, y = 1 (yes) hence not sufficient.

take statement 1 and 2 together. Now the answer could be either C or E.

Either y is -ve or postive. If y is -ve then x >y always holds true.

If y is +ve then x>y^2 and x^3>y => x^6 > y^2

divide both the inequalities: x^5 >1 => x >1. Since square root of x (> 1) is greater than y => x>y.

Hence C.

The graphical approach is good, but its essential to understand that how the inequalities behaves in different domains.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 100

Kudos [?]: 850 [0], given: 25

Re: Is x > y ? [#permalink]

### Show Tags

09 Oct 2010, 11:28
Agreed

It's just that the way I was taught algebra, there was a lot of focus on graphs. I find it way more intuitive than algebraic manipulation, more straight forward than plugging values and in almost all cases faster especially for very simple functions like these

Posted from my mobile device
_________________
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 220

Kudos [?]: 1508 [0], given: 235

Re: Is x > y ? [#permalink]

### Show Tags

09 Oct 2010, 12:15
shrouded1 wrote:
Agreed

It's just that the way I was taught algebra, there was a lot of focus on graphs. I find it way more intuitive than algebraic manipulation, more straight forward than plugging values and in almost all cases faster especially for very simple functions like these

Posted from my mobile device

Even I was taught in same way. But post JEE, I never used them. I will work one day on graphs for sure. It is the best approach.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Math Expert
Joined: 02 Sep 2009
Posts: 34393
Followers: 6244

Kudos [?]: 79303 [0], given: 10016

Re: DS question : need help [#permalink]

### Show Tags

28 Oct 2010, 14:30
Expert's post
2
This post was
BOOKMARKED
Merging similar topics. By the way OA fo this question is C, not E.

Also check:
inequality-ds-100086.html?hilit=green%20zone
ds-algebra-101438.html?hilit=green%20zone

Hope it helps.
_________________
Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1712
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Followers: 89

Kudos [?]: 783 [0], given: 109

Re: Is x > y ? [#permalink]

### Show Tags

28 Oct 2010, 15:28
Hi Bunuel!,
do you have similar questions? They would be very helpful to be sure that we have learned this
Thanks!
_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

GMAT Club Premium Membership - big benefits and savings

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6827
Location: Pune, India
Followers: 1912

Kudos [?]: 11906 [12] , given: 221

Re: Is x > y ? [#permalink]

### Show Tags

29 Oct 2010, 06:56
12
KUDOS
Expert's post
1
This post was
BOOKMARKED
shrouded1 wrote:
I thought this was a really tough question !

Is $$x > y$$ ?

(1) $$\sqrt{x} > y$$
(2) $$x^3 > y$$

One of those gorgeous questions that seem so simple at first but surprise you later...
Best way to work on these is to fall back on your drawing skills (Yes, I love diagrams!)

Statement (1): If I can say that $$x >= \sqrt{x}$$ for all values of x, then I can say that x > y. The green line shows me the region where $$x >= \sqrt{x}$$ but the red line shows me the region where it isn't. Then, for the red line region, x MAY NOT be greater than y. Not Sufficient.
I also understand from this statement that x >= 0.

Statement (2): If I can say that $$x >= x^3$$ for all values of x, then I can say that x > y. The green lines show me the region where $$x >= x^3$$ but the red lines show me the region where it isn't. Then, for the red line region, x MAY NOT be greater than y. Not Sufficient.
Attachment:

Ques.jpg [ 9.04 KiB | Viewed 33163 times ]

Using both together, I know x >= 0.
If 0 <= x <= 1, then we know $$x >= x^3$$. Since statement (2) says that $$x^3 > y$$, I can say that x > y.
If x > 1, then we know $$x > \sqrt{x}$$. Since statement (1) says that $$\sqrt{x} > y$$, I can deduce that x > y.
For all possible values of x, we can say x > y. Sufficient. Answer (C).
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 34393 Followers: 6244 Kudos [?]: 79303 [3] , given: 10016 Re: Is x > y ? [#permalink] ### Show Tags 29 Oct 2010, 15:47 3 This post received KUDOS Expert's post 1 This post was BOOKMARKED satishreddy wrote: hey bunnel.....can we also rewrite sq root X>M as X>M2 , by squaring both sides to make things simple,,,,,,,,, We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: $$2<4$$ --> we can square both sides and write: $$2^2<4^2$$; $$0\leq{x}<{y}$$ --> we can square both sides and write: $$x^2<y^2$$; But if either of side is negative then raising to even power doesn't always work. For example: $$1>-2$$ if we square we'll get $$1>4$$ which is not right. So if given that $$x>y$$ then we can not square both sides and write $$x^2>y^2$$ if we are not certain that both $$x$$ and $$y$$ are non-negative. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: $$-2<-1$$ --> we can raise both sides to third power and write: $$-2^3=-8<-1=-1^3$$ or $$-5<1$$ --> $$-5^2=-125<1=1^3$$; $$x<y$$ --> we can raise both sides to third power and write: $$x^3<y^3$$. Hope it helps. _________________ Intern Joined: 25 Sep 2010 Posts: 19 Followers: 0 Kudos [?]: 30 [0], given: 7 Re: Is x > y ? [#permalink] ### Show Tags 29 Oct 2010, 16:06 1 This post was BOOKMARKED Bunuel wrote: satishreddy wrote: hey bunnel.....can we also rewrite sq root X>M as X>M2 , by squaring both sides to make things simple,,,,,,,,, We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: $$2<4$$ --> we can square both sides and write: $$2^2<4^2$$; $$0\leq{x}<{y}$$ --> we can square both sides and write: $$x^2<y^2$$; But if either of side is negative then raising to even power doesn't always work. For example: $$1>-2$$ if we square we'll get $$1>4$$ which is not right. So if given that $$x>y$$ then we can not square both sides and write $$x^2>y^2$$ if we are not certain that both $$x$$ and $$y$$ are non-negative. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: $$-2<-1$$ --> we can raise both sides to third power and write: $$-2^3=-8<-1=-1^3$$ or $$-5<1$$ --> $$-5^2=-125<1=1^3$$; $$x<y$$ --> we can raise both sides to third power and write: $$x^3<y^3$$. Hope it helps. ofcourse, it helps a lot bunnel.....thank you,,,my science background is killing me........ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6827 Location: Pune, India Followers: 1912 Kudos [?]: 11906 [1] , given: 221 Re: Is x > y ? [#permalink] ### Show Tags 30 Oct 2010, 05:52 1 This post received KUDOS Expert's post satishreddy wrote: hey karishma,,,,we can also rewrite sq root X>M as X>M2 , by squaring both sides to make things simple,,,,,,,,, How about providing the solution with your suggested modifications? Is always great to see different takes on the same question! _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Senior Manager
Status: May The Force Be With Me (D-DAY 15 May 2012)
Joined: 06 Jan 2012
Posts: 289
Location: India
Concentration: General Management, Entrepreneurship
Followers: 2

Kudos [?]: 213 [0], given: 16

### Show Tags

12 Apr 2012, 04:56
Hi,

To Prove : X> Y
Statement 1 : sqrt (x) > y
From this statement X is always positive hence X> Y
4 > 2 ,16> 4, etc

Statement 2 : cube root (x) > y
X can be negative & smaller than y eg -8 < -4

Thus A alone is sufficient.

Hope this helps though I'm not 100% sure if I'm correct
_________________

Giving +1 kudos is a better way of saying 'Thank You'.

Manager
Status: Do till 740 :)
Joined: 13 Jun 2011
Posts: 113
Concentration: Strategy, General Management
GMAT 1: 460 Q35 V20
GPA: 3.6
WE: Consulting (Computer Software)
Followers: 1

Kudos [?]: 8 [0], given: 19

Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink]

### Show Tags

13 Apr 2012, 11:12
Hi Karishma,

Quote:
Using both together, I know x >= 0.
If 0 <= x <= 1, then we know x >= x^3. Since statement (2) says that x^3 > y, I can say that x > y.
If x > 1, then we know x > \sqrt{x}. Since statement (1) says that \sqrt{x} > y, I can deduce that x > y.
For all possible values of x, we can say x > y. Sufficient. Answer (C).

from the diagrams you have put up , how can we derive that X>=0 cos from both the diagrams the regions do not over lap!
Thanks
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6827
Location: Pune, India
Followers: 1912

Kudos [?]: 11906 [0], given: 221

Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink]

### Show Tags

18 Apr 2012, 03:28
shankar245 wrote:
Hi Karishma,

Quote:
Using both together, I know x >= 0.
If 0 <= x <= 1, then we know x >= x^3. Since statement (2) says that x^3 > y, I can say that x > y.
If x > 1, then we know x > \sqrt{x}. Since statement (1) says that \sqrt{x} > y, I can deduce that x > y.
For all possible values of x, we can say x > y. Sufficient. Answer (C).

from the diagrams you have put up , how can we derive that X>=0 cos from both the diagrams the regions do not over lap!
Thanks

We know that x >= 0 because statement 1 tells us that $$\sqrt{x} > y$$.
$$\sqrt{x}$$ can be defined only if x is non negative.

Check out this post for a detailed explanation of this solution: http://www.veritasprep.com/blog/2011/08 ... -question/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Senior Manager
Joined: 22 Dec 2011
Posts: 298
Followers: 3

Kudos [?]: 204 [0], given: 32

Re: Is x > y ? [#permalink]

### Show Tags

09 Oct 2012, 05:27
Bunuel wrote:
satishreddy wrote:
hey bunnel.....can we also rewrite sq root X>M as X>M2 , by squaring both sides to make things simple,,,,,,,,,

We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
$$2<4$$ --> we can square both sides and write: $$2^2<4^2$$;
$$0\leq{x}<{y}$$ --> we can square both sides and write: $$x^2<y^2$$;

But if either of side is negative then raising to even power doesn't always work.
For example: $$1>-2$$ if we square we'll get $$1>4$$ which is not right. So if given that $$x>y$$ then we can not square both sides and write $$x^2>y^2$$ if we are not certain that both $$x$$ and $$y$$ are non-negative.

We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
$$-2<-1$$ --> we can raise both sides to third power and write: $$-2^3=-8<-1=-1^3$$ or $$-5<1$$ --> $$-5^2=-125<1=1^3$$;
$$x<y$$ --> we can raise both sides to third power and write: $$x^3<y^3$$.

Hope it helps.

Hi few observations please correct me if Im wrong ->

$$\sqrt{x} > y$$ -> cannot square this but I can always cube both sides

$$y > \sqrt{x}$$ - Can i square this? Rational behind this is right hand side of the inequality is +ve so left must be to, hence this can be written as y^2>x ??
Re: Is x > y ?   [#permalink] 09 Oct 2012, 05:27

Go to page    1   2    Next  [ 39 posts ]

Similar topics Replies Last post
Similar
Topics:
4 Is x^2>y^2? 4 03 Jul 2016, 05:54
5 If x and y are integers, is (x−1)>y? 9 04 Feb 2014, 15:43
3 Is x^3 > y^2? 3 23 May 2010, 00:33
19 If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y 16 09 Jan 2010, 14:17
Y >= 0, What is the value of x? 1) |x-3|>=y 2) |x-3| <=-y 2 14 Mar 2010, 01:03
Display posts from previous: Sort by