It is currently 22 Nov 2017, 00:48

### 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^4 + y^4 > z^4?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 03 Feb 2010
Posts: 67

Kudos [?]: 149 [0], given: 4

Is x^4 + y^4 > z^4? [#permalink]

### Show Tags

19 Sep 2010, 23:51
25
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

43% (01:29) correct 57% (01:25) wrong based on 640 sessions

### HideShow timer Statistics

Is x^4 + y^4 > z^4?

(1) x^2 + y^2 > z^2
(2) x + y > z
[Reveal] Spoiler: OA

Last edited by Bunuel on 03 Jul 2013, 01:32, edited 2 times in total.

Kudos [?]: 149 [0], given: 4

Math Expert
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133003 [14], given: 12402

Re: Gmat Prep exponent [#permalink]

### Show Tags

20 Sep 2010, 00:03
14
KUDOS
Expert's post
11
This post was
BOOKMARKED
Is $$x^4+y^4>z^4$$?

The best way to deal with this problem is plugging numbers. Remember on DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

(1) $$x^2+y^2>z^2$$
It's clear that we get YES answer very easily with big x and y (say 10 and 10), and small z (say 0).

For NO answer let's try numbers from Pythagorean triples:
$$x^2=3$$, $$y^2=4$$ and $$z^2=5$$ ($$x^2+y^2=7>5=z^2$$) --> $$x^4+y^4=9+16=25=z^4$$, so we have answer NO ($$x^4+y^4$$ is NOT more than $$z^4$$, it's equal to it).

Not sufficient.

(2) $$x+y>z$$. This one is even easier: again we can get YES answer with big x and y, and small z.

As for NO try to make z some big enough negative number: so if $$x=y=1$$ and $$z=-5$$, then $$x^4+y^4=1+1=2<25=z^4$$.

Not sufficient.

(1)+(2) As we concluded YES answer is easily achievable. For NO try the case of $$x^2=3$$, $$y^2=4$$ and $$z^2=5$$ again: $$x+y=\sqrt{3}+\sqrt{4}>\sqrt{5}$$ ($$\sqrt{3}+2$$ is more than 3 and $$\sqrt{5}$$ is less than 3), so statement (2) is satisfied, we know that statement (1) is also satisfied ($$x^2+y^2=7>5=z^2$$) and $$x^4+y^4=9+16=25=z^4$$. Not sufficient.

Hope it's clear.
_________________

Kudos [?]: 133003 [14], given: 12402

Manager
Joined: 04 Aug 2010
Posts: 151

Kudos [?]: 33 [0], given: 15

Re: Gmat Prep exponent [#permalink]

### Show Tags

20 Sep 2010, 19:23
Thanks for the explanation. I keep forgetting the unknowns could be negative numbers.

Kudos [?]: 33 [0], given: 15

Retired Moderator
Joined: 02 Sep 2010
Posts: 793

Kudos [?]: 1211 [1], given: 25

Location: London
Re: Gmat Prep exponent [#permalink]

### Show Tags

10 Oct 2010, 03:39
1
KUDOS
1
This post was
BOOKMARKED
utin wrote:
Hi Bunuel,

Can it be done using algebra???

Yes ! and it is fairly simple as well ...

(1) $$x^2+y^2>z^2$$
Both sides are positive, so we can square them up
$$x^4+y^4+2x^2y^2>z^4$$
For statement to be true or false, we need to know the magnitude of $$2x^2y^2$$. So insufficient

(2) $$x+y>z$$
Let me take the special case, x,y,z>0. If we cannot answer for that case, we can certainly not answer in general.
Now simply take both sides to the fourth power :
$$x^4+y^4+4x^3y+6x^2y^2+4xy^3>z^4$$
Again unless we know the extra terms, we cannot say for sure. So insufficient

(1+2) Again consider the special case for positive numbers only. Even then we don't have enough information to say anything. Do the same manipulation as above and it is easy to see that inequality two has the same terms on the LHS as inequality 1 and some extra terms on top. So clearly its enough to look at 1, which we know is not sufficient to give us an answer, so insufficient

_________________

Kudos [?]: 1211 [1], given: 25

Math Expert
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133003 [1], given: 12402

Re: Gmat Prep exponent [#permalink]

### Show Tags

10 Oct 2010, 04:45
1
KUDOS
Expert's post
utin wrote:
Hi Bunuel,

Can it be done using algebra???

Great question. Yes you can use algebra, but as shown by shrouded1 above you'd better not. Below is great note from ManhattanGMAT tutor Ron Purewal about this problem:

"this problem should serve as a nice wake-up call to any and all students who don't like "plug-in methods", or who abjure such methods so that they can keep searching ... and searching ... and searching for the elusive "textbook" method.
this problem is pretty much ONLY soluble with plug-in methods. therefore, you MUST make plug-in methods part of your arsenal if you want a fighting chance at all quant problems you'll see.

this is the case for a great many difficult inequality problems, by the way: the most difficult among those problems will often require some sort of plug-ins, or, at the very least, they will be hell on earth if you try to use theory."
_________________

Kudos [?]: 133003 [1], given: 12402

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7745

Kudos [?]: 17843 [3], given: 235

Location: Pune, India
Re: Is x^4 + y^4 > z^4? [#permalink]

### Show Tags

04 Dec 2010, 19:59
3
KUDOS
Expert's post
1
This post was
BOOKMARKED
gmatbull wrote:
Is x^4 + y^4 > z^4?

(1) x^2 + y^2 > z^2
(2) x + y > z

What do you think is the fastest approach to solving this question?

This is the approach I used. It isn't algebraic and it isn't completely based on plugging numbers either (though I am explaining using plugs). It seemed intuitive to me, but your opinion could be different.

Stmnt 1: $$x^2 + y^2 > z^2$$
I do not like $$x^4 + y^4 > z^4$$, powers of 4 since I do not know how to work with them in multiple ways. I prefer squares.
So I look at the question in this way:
Stmnt 1: X + Y > Z (X, Y and Z are all positive)
Is $$X^2 + Y^2 > Z^2$$?
Now it is intuitive to say "Yes, it is." because if X = 8, Y = 9 and Z = 1, it is.
The problem is, is there a case in which I would answer "No".
$$X^2 + Y^2 > Z^2$$ reminds me of pythagorean triplets where $$X^2 + Y^2 = Z^2$$. I check the easiest one 3, 4, 5. I get "No". Hence this is not sufficient.
The toughest part of the question is over.

Stmnt 2: x + y > z
Again, intuitive to say "Yes" because if X = 8, Y = 9 and Z = 1, the answer is yes.
And very easy to say "No" because if x = root(3), y = root(4) and z = root(5), x^4 + y^4 is not greater than z^4

We have used the same examples in both the cases to get a "Yes" and a "No" hence using both together, we will not get the answer.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17843 [3], given: 235 CEO Joined: 17 Nov 2007 Posts: 3583 Kudos [?]: 4670 [1], given: 360 Concentration: Entrepreneurship, Other Schools: Chicago (Booth) - Class of 2011 GMAT 1: 750 Q50 V40 Re: Gmat Prep exponent [#permalink] ### Show Tags 04 Dec 2010, 20:31 1 This post received KUDOS Expert's post Nice question. Here is my reasoning that allowed me to solve the problem pretty fast. The inequity true for X=Y=Z=1 --> 1+1>1 Now, let's increase Z (X=Y=1). Z^2 will grow faster than Z but slower than Z^4. So, at some point Z^4 will be greater than 2 but Z^2 and Z will be lesser than 2 and still both conditions will be satisfied. So E. _________________ HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame Kudos [?]: 4670 [1], given: 360 Manager Joined: 19 Dec 2010 Posts: 135 Kudos [?]: 32 [0], given: 12 Re: Gmat Prep exponent [#permalink] ### Show Tags 17 Mar 2011, 23:17 i worked only with positives on this one. Plug and chug approach... Stmnt 1: x^2 + y^2 > z^2 plug in 1 for each, you will see that the statement holds. if you plug in 2 for x and y...3 for z the statement falls apart Same logic for statement 2 Answer: E Kudos [?]: 32 [0], given: 12 Retired Moderator Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL Joined: 04 Oct 2009 Posts: 1628 Kudos [?]: 1124 [0], given: 109 Location: Peru Schools: Harvard, Stanford, Wharton, MIT & HKS (Government) WE 1: Economic research WE 2: Banking WE 3: Government: Foreign Trade and SMEs Re: Gmat Prep exponent [#permalink] ### Show Tags 27 Jan 2012, 11:26 shrouded1 wrote: utin wrote: Hi Bunuel, Can it be done using algebra??? Yes ! and it is fairly simple as well ... (1) $$x^2+y^2>z^2$$ Both sides are positive, so we can square them up $$x^4+y^4+2x^2y^2>z^4$$ For statement to be true or false, we need to know the magnitude of $$2x^2y^2$$. So insufficient (2) $$x+y>z$$ Let me take the special case, x,y,z>0. If we cannot answer for that case, we can certainly not answer in general. Now simply take both sides to the fourth power : $$x^4+y^4+4x^3y+6x^2y^2+4xy^3>z^4$$ Again unless we know the extra terms, we cannot say for sure. So insufficient (1+2) Again consider the special case for positive numbers only. Even then we don't have enough information to say anything. Do the same manipulation as above and it is easy to see that inequality two has the same terms on the LHS as inequality 1 and some extra terms on top. So clearly its enough to look at 1, which we know is not sufficient to give us an answer, so insufficient Answer is (e) I have doubts about this method. In statement 1, although we don't know the mangnitude of $$2x^2y^2$$, we know that this expression cannot take any arbitrary value; its value depends on the values of $$x^2$$ and $$y^2$$. I have picked different numbers, and I have not found a combination in which $$z^4$$ is less than $$x^4+y^4+2x^2y^2$$. What do you think? I solved the question, picking numbers directly. It is a 700+ question, isn't it? _________________ "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 Kudos [?]: 1124 [0], given: 109 Intern Joined: 06 Jun 2012 Posts: 18 Kudos [?]: [0], given: 9 Re: Is x^4+y^4>z^4? (1) x^2+y^2>z^2 (2) x+y>z [#permalink] ### Show Tags 14 Jun 2012, 11:01 Damn i felt stupid when i saw this question. Should you have the knowledge to solve this kind of questions to hit a 700+ score, before starting your prep to gmat? Kudos [?]: [0], given: 9 Math Expert Joined: 02 Sep 2009 Posts: 42302 Kudos [?]: 133003 [0], given: 12402 Re: Is x^4+y^4>z^4? (1) x^2+y^2>z^2 (2) x+y>z [#permalink] ### Show Tags 14 Jun 2012, 11:41 Shadow88 wrote: Damn i felt stupid when i saw this question. Should you have the knowledge to solve this kind of questions to hit a 700+ score, before starting your prep to gmat? This is definitely a hard question, but some practice should help to deal with such kind of questions in the future. Try the following inequality questions to practice: PS: search.php?search_id=tag&tag_id=184 PS: search.php?search_id=tag&tag_id=189 Hard inequality and absolute value questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html Hope it helps. _________________ Kudos [?]: 133003 [0], given: 12402 Manager Joined: 12 Feb 2012 Posts: 130 Kudos [?]: 63 [0], given: 28 Re: Gmat Prep exponent [#permalink] ### Show Tags 19 Aug 2012, 17:55 shrouded1 wrote: utin wrote: Hi Bunuel, Can it be done using algebra??? Yes ! and it is fairly simple as well ... (1) $$x^2+y^2>z^2$$ Both sides are positive, so we can square them up $$x^4+y^4+2x^2y^2>z^4$$ For statement to be true or false, we need to know the magnitude of $$2x^2y^2$$. So insufficient (2) $$x+y>z$$ Let me take the special case, x,y,z>0. If we cannot answer for that case, we can certainly not answer in general. Now simply take both sides to the fourth power : $$x^4+y^4+4x^3y+6x^2y^2+4xy^3>z^4$$ Again unless we know the extra terms, we cannot say for sure. So insufficient (1+2) Again consider the special case for positive numbers only. Even then we don't have enough information to say anything. Do the same manipulation as above and it is easy to see that inequality two has the same terms on the LHS as inequality 1 and some extra terms on top. So clearly its enough to look at 1, which we know is not sufficient to give us an answer, so insufficient Answer is (e) Having a problem in statement (1) (1) $$x^2+y^2>z^2$$ Both sides are positive, so we can square them up $$x^4+y^4+2x^2y^2>z^4$$ $$2x^2y^2>=0$$. so why isnt this sufficient? If $$2x^2y^2>=0$$ then either X or Y equals 0 (but not both!). which implies that if x=0, $$y^2>z^2$$ ===> $$y^4>z^4$$ or y=0 $$x^2>z^2$$ ===> $$x^4>z^4$$ , in both cases sufficient. What I am doing wrong? What am I not considering? Kudos [?]: 63 [0], given: 28 Intern Joined: 07 Apr 2012 Posts: 11 Kudos [?]: 8 [0], given: 3 Location: United States Concentration: Finance, Economics Schools: Owen '15, Olin '15 GMAT 1: 690 Q50 V31 Re: Gmat Prep exponent [#permalink] ### Show Tags 19 Aug 2012, 19:45 alphabeta1234 wrote: shrouded1 wrote: utin wrote: Hi Bunuel, Can it be done using algebra??? Yes ! and it is fairly simple as well ... (1) $$x^2+y^2>z^2$$ Both sides are positive, so we can square them up $$x^4+y^4+2x^2y^2>z^4$$ For statement to be true or false, we need to know the magnitude of $$2x^2y^2$$. So insufficient (2) $$x+y>z$$ Let me take the special case, x,y,z>0. If we cannot answer for that case, we can certainly not answer in general. Now simply take both sides to the fourth power : $$x^4+y^4+4x^3y+6x^2y^2+4xy^3>z^4$$ Again unless we know the extra terms, we cannot say for sure. So insufficient (1+2) Again consider the special case for positive numbers only. Even then we don't have enough information to say anything. Do the same manipulation as above and it is easy to see that inequality two has the same terms on the LHS as inequality 1 and some extra terms on top. So clearly its enough to look at 1, which we know is not sufficient to give us an answer, so insufficient Answer is (e) Having a problem in statement (1) (1) $$x^2+y^2>z^2$$ Both sides are positive, so we can square them up $$x^4+y^4+2x^2y^2>z^4$$ $$2x^2y^2>=0$$. so why isnt this sufficient? If $$2x^2y^2>=0$$ then either X or Y equals 0 (but not both!). which implies that if x=0, $$y^2>z^2$$ ===> $$y^4>z^4$$ or y=0 $$x^2>z^2$$ ===> $$x^4>z^4$$ , in both cases sufficient. What I am doing wrong? What am I not considering? Hi there: You are right on those calculations, but one thing you seem to have neglect: 2x^2y^2>=0 (yes, and for sure), but do we really need to add this term for x^4+y^4>z^4 to be true, what if it means little to the equality, 1000>10, and 1000+1>10, but you could not logically conclude that just because 1>0 and thus 1000<10. I use a totally different approach for this question: it will be relatively easy to find a confirming result giving either statement, just use ridiculous numbers to prove, there is no need to compute, say x=1000 and y=1, and z=10. Therefore, what matters is that if we can find refuting evidence given those( or either statements), I have not actually computed those equations, but as you can recall from the number properties, a number 0<and <1 will decrease when the power increases. For statement 1, we just need to find if there are x^2<1 and Y^2<1, and Z^2>1 (and possible be very close to the left side so that we can find a countering example)..We surely can, for instance, if x=y=.9, and z=1.6. We know that squaring each term on the left side will decrease the value and squaring the right side will increase the value, so bingo, we found it, and S1 is thus insufficient. For statement 2, the pythagorean is the best approach IMO, 3,4,5 are so handy and could be easily validated or you can use my strategy to stay consistent. Combine those will still be insufficient. Hope that helps. Kudos [?]: 8 [0], given: 3 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7745 Kudos [?]: 17843 [0], given: 235 Location: Pune, India Re: Gmat Prep exponent [#permalink] ### Show Tags 20 Aug 2012, 02:54 alphabeta1234 wrote: Having a problem in statement (1) (1) $$x^2+y^2>z^2$$ Both sides are positive, so we can square them up $$x^4+y^4+2x^2y^2>z^4$$ $$2x^2y^2>=0$$. so why isnt this sufficient? If $$2x^2y^2>=0$$ then either X or Y equals 0 (but not both!). which implies that if x=0, $$y^2>z^2$$ ===> $$y^4>z^4$$ or y=0 $$x^2>z^2$$ ===> $$x^4>z^4$$ , in both cases sufficient. What I am doing wrong? What am I not considering? Ok, say you know that a+b > c and that b is positive. Can you say that a is definitely greater than c? Given that $$x^4+y^4+2x^2y^2>z^4$$ and you know that $$2x^2y^2$$ is positive or 0, can you say that $$x^4+y^4>z^4$$ ? Isn't it possible that $$2x^2y^2$$ part of the left hand side is making the left hand side greater than the right hand side? Assume, $$x = \sqrt{3}$$, $$y = 2$$ and$$z = \sqrt{5}$$ $$x^2+y^2>z^2$$ but $$x^4+y^4=z^4$$ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Kudos [?]: 17843 [0], given: 235

Math Expert
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133003 [0], given: 12402

Re: Is x^4 + y^4 > z^4? [#permalink]

### Show Tags

08 Jul 2013, 00:54
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

To find DS questions by Kudos, sort by Kudos here: gmat-data-sufficiency-ds-141/
To find PS questions by Kudos, sort by Kudos here: gmat-problem-solving-ps-140/

_________________

Kudos [?]: 133003 [0], given: 12402

Current Student
Joined: 06 Sep 2013
Posts: 1970

Kudos [?]: 743 [0], given: 355

Concentration: Finance
Re: Gmat Prep exponent [#permalink]

### Show Tags

05 Feb 2014, 13:00
Bunuel wrote:
Is $$x^4+y^4>z^4$$?

The best way to deal with this problem is plugging numbers. Remember on DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

(1) $$x^2+y^2>z^2$$
It's clear that we get YES answer very easily with big x and y (say 10 and 10), and small z (say 0).

For NO answer let's try numbers from Pythagorean triples:
$$x^2=3$$, $$y^2=4$$ and $$z^2=5$$ ($$x^2+y^2=7>5=z^2$$) --> $$x^4+y^4=9+16=25=z^4$$, so we have answer NO ($$x^4+y^4$$ is NOT more than $$z^4$$, it's equal to it).

Not sufficient.

(2) $$x+y>z$$. This one is even easier: again we can get YES answer with big x and y, and small z.

As for NO try to make z some big enough negative number: so if $$x=y=1$$ and $$z=-5$$, then $$x^4+y^4=1+1=2<25=z^4$$.

Not sufficient.

(1)+(2) As we concluded YES answer is easily achievable. For NO try the case of $$x^2=3$$, $$y^2=4$$ and $$z^2=5$$ again: $$x+y=\sqrt{3}+\sqrt{4}>\sqrt{5}$$ ($$\sqrt{3}+2$$ is more than 3 and $$\sqrt{5}$$ is less than 3), so statement (2) is satisfied, we know that statement (1) is also satisfied ($$x^2+y^2=7>5=z^2$$) and $$x^4+y^4=9+16=25=z^4$$. Not sufficient.

Hope it's clear.

Bunuel, is this reasoning correct?

Is x^4 + y^4 > z^4?

So 1st Statement

x^2 + y^2 > z^2

If you square both sides then you have (x^2 + y^2)^2 > z^4
Now since both exponents are even then the total in LHS will be greater than x^4 + y^4
So we cannot really know if by themselves they will be greater than x^4 + y^4 > z^4?

From 2nd Statement

Following the same reasoning
If you elevate both sides to the fourth power then you have (x + y)^4 > z^4
Again this is not sufficient

From both combined

Not sufficient either. We really can't tell

So I think the correct answer for this one should be (E)

Please let us know ok?
Thanks
Cheers
J

Kudos [?]: 743 [0], given: 355

Current Student
Joined: 03 Feb 2013
Posts: 941

Kudos [?]: 1077 [0], given: 548

Location: India
Concentration: Operations, Strategy
GMAT 1: 760 Q49 V44
GPA: 3.88
WE: Engineering (Computer Software)
Re: Is x^4 + y^4 > z^4? [#permalink]

### Show Tags

12 Dec 2014, 12:30
(x^2 + y^2)^2 > z^4 => x^4 + y^4 + 2x^2y^2 > z^4 not sure how much " 2x^2y^2" is compensated.
For example
1 + 1 > 1.99
1 + 1 + 2 > 3.99 but 1 + 1 > 3.99 definitely not.

Not sufficient.

x + y > z same issue - Not sufficient.

Combining the two statements:

Same issue. Hence E)
_________________

Thanks,
Kinjal

My Application Experience : http://gmatclub.com/forum/hardwork-never-gets-unrewarded-for-ever-189267-40.html#p1516961

Please click on Kudos, if you think the post is helpful

Kudos [?]: 1077 [0], given: 548

Intern
Joined: 18 Apr 2015
Posts: 6

Kudos [?]: [0], given: 5

GMAT 1: 600 Q46 V27
Re: Is x^4 + y^4 > z^4? [#permalink]

### Show Tags

02 Aug 2015, 00:06
Brunuel: Is this reasoning right? Bcz we are given with the fact that statement 1 (condition) is true. Then why should we check its validity bringing in Pythagoras theorm.

Kudos [?]: [0], given: 5

Math Expert
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133003 [0], given: 12402

Re: Is x^4 + y^4 > z^4? [#permalink]

### Show Tags

16 Aug 2015, 10:52
josepiusn wrote:
Brunuel: Is this reasoning right? Bcz we are given with the fact that statement 1 (condition) is true. Then why should we check its validity bringing in Pythagoras theorm.

We are not checking whether the first statement is true, it is. We are trying to get an YES and a NO answer to the question (is $$x^4+y^4>z^4$$) to prove its insufficiency.
_________________

Kudos [?]: 133003 [0], given: 12402

Intern
Joined: 04 May 2015
Posts: 25

Kudos [?]: 1 [0], given: 29

Is x^4 + y^4 > z^4? [#permalink]

### Show Tags

13 Oct 2015, 11:00
Let x^4 = m^2
Let y^4 = p ^2
Let z^4 = n ^2

Hence the question is : m^2 + p^2 > n^2
When m^2 + p^2 = n^2 it`s a formula of a circle.
And when m^2 + p^2 > n^2 we need the area outside the circle.

1) Draw a circle with radius z inside the first circle with radius n = z^2 . We are told about the area outside the second circle. But is it still inside the first circle or not ? N/A

2) This is just a plane with a multiple unknown parameters. We can vary z as we want. N/A

1) + 2) Adds nothing N/A

--> E

Kudos [?]: 1 [0], given: 29

Is x^4 + y^4 > z^4?   [#permalink] 13 Oct 2015, 11:00

Go to page    1   2    Next  [ 23 posts ]

Display posts from previous: Sort by

# Is x^4 + y^4 > z^4?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.