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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.

(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

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."
_________________

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. Answer (E).
_________________

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.
_________________

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

(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

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?

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:

(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?

(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.

(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?

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.

Answer: E.

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)

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.

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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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

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