Last visit was: 24 Apr 2024, 22:31 It is currently 24 Apr 2024, 22:31

Close
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
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.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
User avatar
Retired Moderator
Joined: 02 Sep 2010
Posts: 615
Own Kudos [?]: 2930 [419]
Given Kudos: 25
Location: London
 Q51  V41
Send PM
Most Helpful Reply
Math Expert
Joined: 02 Sep 2009
Posts: 92900
Own Kudos [?]: 618848 [273]
Given Kudos: 81588
Send PM
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64905 [51]
Given Kudos: 426
Location: Pune, India
Send PM
General Discussion
User avatar
Manager
Manager
Joined: 31 Jul 2010
Status:Keep fighting!
Affiliations: IIT Madras
Posts: 150
Own Kudos [?]: 1246 [4]
Given Kudos: 104
 Q49  V34 GMAT 2: 720  Q50  V37
WE 1: 2+ years - Programming
WE 2: 3+ years - Product developement,
WE 3: 2+ years - Program management
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
2
Kudos
2
Bookmarks
Bunuel.... you seem to love number theory! I really am amazed at your patience. Good job. You already have enough Kudos. :P
User avatar
Manager
Manager
Joined: 20 Jul 2010
Posts: 137
Own Kudos [?]: 247 [6]
Given Kudos: 9
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
6
Kudos
interesting question.......forgot to consider fractions first.......now its clear.....
User avatar
VP
VP
Joined: 24 Aug 2010
Status:Current Student
Posts: 1332
Own Kudos [?]: 447 [8]
Given Kudos: 74
Location: United States
GMAT 1: 710 Q48 V40
WE:Sales (Consumer Products)
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
8
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.
User avatar
Retired Moderator
Joined: 02 Sep 2010
Posts: 615
Own Kudos [?]: 2930 [5]
Given Kudos: 25
Location: London
 Q51  V41
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
4
Kudos
1
Bookmarks
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

Answer is (C)
User avatar
SVP
SVP
Joined: 12 Oct 2009
Status:<strong>Nothing comes easy: neither do I want.</strong>
Posts: 2279
Own Kudos [?]: 3594 [16]
Given Kudos: 235
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
12
Kudos
4
Bookmarks
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.
User avatar
Retired Moderator
Joined: 02 Sep 2010
Posts: 615
Own Kudos [?]: 2930 [2]
Given Kudos: 25
Location: London
 Q51  V41
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
2
Kudos
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
Math Expert
Joined: 02 Sep 2009
Posts: 92900
Own Kudos [?]: 618848 [9]
Given Kudos: 81588
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
1
Kudos
8
Bookmarks
Expert Reply
Math Expert
Joined: 02 Sep 2009
Posts: 92900
Own Kudos [?]: 618848 [11]
Given Kudos: 81588
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
5
Kudos
6
Bookmarks
Expert Reply
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.
User avatar
Manager
Manager
Joined: 13 Jun 2011
Status:Do till 740 :)
Posts: 61
Own Kudos [?]: 31 [0]
Given Kudos: 19
Concentration: Strategy, General Management
GMAT 1: 460 Q35 V20
GPA: 3.6
WE:Consulting (Computer Software)
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
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!
Please help me with this1
Thanks
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64905 [3]
Given Kudos: 426
Location: Pune, India
Send PM
Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
2
Kudos
Expert Reply
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!
Please help me with this1
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.

Originally posted by KarishmaB on 18 Apr 2012, 03:28.
Last edited by KarishmaB on 08 Oct 2022, 21:33, edited 1 time in total.
User avatar
Manager
Manager
Joined: 22 Dec 2011
Posts: 175
Own Kudos [?]: 1042 [0]
Given Kudos: 32
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
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 ??
User avatar
Director
Director
Joined: 22 Mar 2011
Posts: 520
Own Kudos [?]: 2136 [1]
Given Kudos: 43
WE:Science (Education)
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
1
Kudos
Jp27 wrote:
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 ??



\(\sqrt{x} > y\) -> cannot square this but I can always cube both sides - YES (although it won't help much, the square root will stay)
\(y > \sqrt{x}\) - 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 ?? YES, but the right hand side always non-negative (it can be 0)
User avatar
VP
VP
Joined: 06 Sep 2013
Posts: 1345
Own Kudos [?]: 2391 [0]
Given Kudos: 355
Concentration: Finance
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
gurpreetsingh wrote:
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.


Yeah kinda tough indeed

Hey gupreet or anyone how did you get to this? If y is +ve then x>y^2 and x^3>y => x^6 > y^2

Like from both statements to the x^6 > y^2?

Thanks

Cheers!
J :)

Kudos rain!
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64905 [0]
Given Kudos: 426
Location: Pune, India
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
Expert Reply
jlgdr wrote:
gurpreetsingh wrote:
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.


Yeah kinda tough indeed

Hey gupreet or anyone how did you get to this? If y is +ve then x>y^2 and x^3>y => x^6 > y^2

Like from both statements to the x^6 > y^2?

Thanks

Cheers!
J :)

Kudos rain!


If y is positive and x^3>y, it means x^3 is also positive (since it is greater than y which is positive).
If both sides of an inequality are positive, you can square the inequality.

\(x^3>y\)
\((x^3)^2>y^2\)
\(x^6 > y^2\)
User avatar
Senior Manager
Senior Manager
Joined: 17 Apr 2013
Status:Verbal Forum Moderator
Posts: 361
Own Kudos [?]: 2197 [0]
Given Kudos: 298
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
Bunuel wrote:
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.

Answer: C.

Hope it's clear.


when if x=1/4 and y=1/3 then \sqrt{X} = 1/2, and 1/2 > 1/3. You have chosen wrong plugin data here.
Math Expert
Joined: 02 Sep 2009
Posts: 92900
Own Kudos [?]: 618848 [0]
Given Kudos: 81588
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
Expert Reply
honchos wrote:
Bunuel wrote:
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.

Answer: C.

Hope it's clear.


when if x=1/4 and y=1/3 then \sqrt{X} = 1/2, and 1/2 > 1/3. You have chosen wrong plugin data here.


Nope.

If \(x=\frac{1}{4}\) and \(y=\frac{1}{3}\), then \((\sqrt{x}=\frac{1}{2})>(y=\frac{1}{4})\) (statement is satisfied) but \((x=\frac{1}{4})<(y=\frac{1}{3})\) giving a NO answer to the question.
avatar
Intern
Intern
Joined: 27 Nov 2014
Posts: 4
Own Kudos [?]: 6 [0]
Given Kudos: 118
Send PM
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
gurpreetsingh wrote:
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.



Hi gurpreetsingh, could you please elaborate on:
gurpreetsingh wrote:
If y is +ve then x>y^2 and x^3>y => x^6 > y^2
. Thank you!
GMAT Club Bot
Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
 1   2   
Moderator:
Math Expert
92900 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne