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

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Is x^4 + y^4 > z^4?  [#permalink]

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Updated on: 03 Jul 2013, 01:32
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Is x^4 + y^4 > z^4?

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

Originally posted by ksharma12 on 19 Sep 2010, 23:51.
Last edited by Bunuel on 03 Jul 2013, 01:32, edited 2 times in total.
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20 Sep 2010, 00:03
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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.
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20 Sep 2010, 19:23
Thanks for the explanation. I keep forgetting the unknowns could be negative numbers.
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10 Oct 2010, 03:39
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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

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10 Oct 2010, 04:45
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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."
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Re: Is x^4 + y^4 > z^4?  [#permalink]

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04 Dec 2010, 19:59
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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.
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04 Dec 2010, 20:31
2
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.
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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
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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

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?
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Re: Is x^4+y^4>z^4? (1) x^2+y^2>z^2 (2) x+y>z  [#permalink]

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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?
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Re: Is x^4+y^4>z^4? (1) x^2+y^2>z^2 (2) x+y>z  [#permalink]

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14 Jun 2012, 11:41
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.
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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

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?
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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

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.
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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 $$x^4+y^4=z^4$$
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Re: Is x^4 + y^4 > z^4?  [#permalink]

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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.
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Re: Is x^4 + y^4 > z^4?  [#permalink]

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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.
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Is x^4 + y^4 > z^4?  [#permalink]

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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
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Re: Is x^4 + y^4 > z^4?  [#permalink]

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14 Oct 2015, 06:58
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

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

From the original condition, we get 3 variables (x,y,z), and in order to match the numbers, we need 3 equations. We are only given 2, so there is high chance (E) will be our answer. Looking at the conditions together, the answer to the question is ‘yes’ for x=y=z=1, but ‘no’ for x=y=0.6, z=0.8. So the conditions are insufficient, and the answer becomes (E).

For cases where we need 3 more equation, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 3517 Location: United States (CA) Is x^4 + y^4 > z^4? [#permalink] ### Show Tags Updated on: 03 Jan 2018, 07:08 ksharma12 wrote: Is x^4 + y^4 > z^4? (1) x^2 + y^2 > z^2 (2) x + y > z We want to determine whether x^4 + y^4 > z^4. Statement One Alone: x^2 + y^2 > z^2 If we square both sides of the inequality, we will have: x^4 + 2(x^2)(y^2) + y^4 > z^4 x^4 + y^4 > z^4 – 2(x^2)(y^2) We still cannot determine whether x^4 + y^4 > z^4, since 2(x^2)(y^2) is a nonnegative quantity. For example, if x = 0, y = 2, and z = 1, then x^4 + y^4 > z^4, since 0^4 + 2^4 > 1^4. However, if x = 3, y = 3, and z = 4, then x^4 + y^4 is not greater than z^4, since 3^4 + 3^4 is not greater than 4^4. Statement one is not sufficient to answer the question. Statement Two Alone: x + y > z We cannot determine whether x^4 + y^4 is greater than z^4. For example, if x = 0, y = 2, and z = 1, then x^4 + y^4 > z^4, since 0^4 + 2^4 > 1^4. However, if x = 3, y = 3, and z = 4, then x^4 + y^4 is not greater than z^4, since 3^4 + 3^4 is not greater than 4^4. Statement two is not sufficient to answer the question. Statements One and Two Together: Using our two statements, we still cannot determine whether x^4 + y^4 is greater than z^4. Using the same numerical examples used earlier, we have: If x = 0, y = 2, and z = 1, then x^4 + y^4 > z^4. However, if x = 3, y = 3, and z = 4, then x^4 + y^4 is not greater than z^4. Answer: E _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Originally posted by ScottTargetTestPrep on 12 Jul 2017, 16:59. Last edited by ScottTargetTestPrep on 03 Jan 2018, 07:08, edited 2 times in total. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12415 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Is x^4 + y^4 > z^4? [#permalink] ### Show Tags 20 Dec 2017, 18:12 Hi All, This question can be solved by TESTing VALUES. When a DS prompt involves lots of exponents, it's likely that you'll have to consider values OTHER than positive integers though (including 0, negatives, fractions, etc.). We're asked if X^4 + Y^4 > Z^4. This is a YES/NO question. Since Fact 2 is easier to deal with than Fact 1, I'm going to start there (we might also be able to use our work on Fact 2 to more easily deal with Fact 1).... 2) X + Y > Z IF.... X = 1, Y = 0 and Z = 0, then the answer to the question is YES. X = 1/2, Y = 1/2 and Z = .7, then the answer to the question is NO. Fact 2 is INSUFFICIENT 1) X^2 + Y^2 > Z^2 The TESTs that we used in Fact 2 also "fit" the information in Fact 1... IF.... X = 1, Y = 0 and Z = 0, then the answer to the question is YES. X = 1/2, Y = 1/2 and Z = .7, then the answer to the question is NO. Fact 1 is INSUFFICIENT Combined, we have two sets of numbers that fit both Facts and produce different answers to the given question (one YES and one NO). Combined, INSUFFICIENT. Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: Is x^4 + y^4 > z^4? &nbs [#permalink] 20 Dec 2017, 18:12

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