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1) x^2 + y^2 > Z^2 X^2, Y^2, Z^2 are always positive [#permalink]
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23 Jul 2009, 20:29
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1) x^2 + y^2 > Z^2 X^2, Y^2, Z^2 are always positive numbers, so squaring on both sides does not have any effect on the inequality i.e (x^2 + y^2)^2 > (Z^2)^2 X^4 + Y^4 + 2X^2Y^2 > Z^4 All the values mentioned above are positive numbers. So obviously X^4 + Y^4 > Z^4 2) X + Y > Z In this scenario nothing is mentioned about X, Y & Z. They might be either positive or negative numbers. Lets suppose X= 1, Y = 1, Z = 50 2> 50 squaring both sides, 4 Z^4 holds true. So 1 alone is sufficient but 2 is not required and the answer is "A". == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



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Re: DS Inequalities Problem for Math Wiz [#permalink]
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23 Jul 2009, 23:08
Aleehsgonji wrote: 1) x^2 + y^2 > Z^2
X^2, Y^2, Z^2 are always positive numbers, so squaring on both sides does not have any effect on the inequality i.e (x^2 + y^2)^2 > (Z^2)^2 X^4 + Y^4 + 2X^2Y^2 > Z^4
All the values mentioned above are positive numbers. So obviously X^4 + Y^4 > Z^4
2) X + Y > Z
In this scenario nothing is mentioned about X, Y & Z. They might be either positive or negative numbers.
Lets suppose X= 1, Y = 1, Z = 50
2> 50
squaring both sides, 4< 2500. Squaring again will retain the same inequality as both the sides have positive numbers
X^4 + Y^4 < Z^4
but if X, Y, Z are +ve then X^4 + Y^4 > Z^4 holds true.
So 1 alone is sufficient but 2 is not required and the answer is "A". No offence but neither is the approach to this problem correct nor is the solution. Lets start with the first statement: I. x^2+y^2>z^2Squaring both sides we get: x^4+y^4+2x^2y^2 > z^4 Since 2x^2y^2 is a positive quantity, on it's own this expression is insufficient to determine whether x^4+y^4>z^4 Alternatively, Let's try and plugin certain values in this expression: a. x=1; y=0; z=0 x^2+y^2>z^2 Clearly, x^4+y^4>z^4 holds true here since [((0+1)^2)^2]>[(0)^2]^2 TRUEb. x=1; y=3; z=2 z^2=4 x^2+y^2=(1+9)=10 Clearly x^2+y^2>z^2 x^4=1; y^4=81; z^4=16 x^4+y^4 > z^4 TRUEc. x=4; y=5; z=6 z^2=36 x^2+y^2=(16+25)=41 Clearly x^2+y^2>z^2 x^4=256; y^4=625; z^4=1296 x^4+y^4 < z^4 FALSEHence, Statement I is INSUFFICIENTLet's look at the second statement: II. x+y>zSquaring both sides (x+y)^2>z^2 x^2+y^2+2xy > z^2 Squaring again we get: x^4+4x^3y+6x^2y^2+4xy^3+y^4 > z^4 This expression is Insufficient on it's own to determine whether x^4+y^4 > z^4 since don't know whether x,y,z are positive or negative integers. Alternatively, Lets plugin certain values for x,y and z: a. x=1; y=1; z=0 (1^4 + 1^4) > 0^4 Hence, x^4+y^4 > z^4 TRUEb. x=2; y=3; z=4 (2^4+3^4)= (16+81)=97 4^4=256 x^4+y^4 < z^4 FALSEc. x=1; y=1; z=10 (x^4+y^4)=2 z^4= 10000 x^4+y^4 < z^4 FALSEHence, Statement II is INSUFFICIENTCombining both the statements doesn't help either as there is no common ground between the two statements which could lead to a solution to the question. Hence the Answer is E



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Re: DS Inequalities Problem for Math Wiz [#permalink]
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24 Jul 2009, 02:19
x=3, y=3, z=4 x+y>z x^2+y^2>z^2 But x^4+y^4<z^4 x=10, y=10, z=1 x+y>z x^2+y^2>z^2 And x^4+y^4>z^4 So E



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Re: DS Inequalities Problem for Math Wiz [#permalink]
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24 Jul 2009, 02:27
christinee wrote: Please solve:
Is x^4 + y^4 > z^4?
1) x^2 + y^2 > z^2
2) x+y>z 2) x+y>z we have: x^2+y^2>= [(x+y)^2] . 1/2 > z^2 So in this problem (1) is relatively stronger than (2). So if statement 1 is not suff, st 2 is too. Statement 1: x^2 + y^2 > z^2 x^4+y^4 >= [(x^2+y^2)^2] . 1/2 > z^4 . 1/2 So statement 1 just determines that x^4 + y^4 > z^4 . 1/2 so insuf E for me.



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Re: DS Inequalities Problem for Math Wiz [#permalink]
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24 Jul 2009, 05:33
christinee wrote: Please solve:
Is x^4 + y^4 > z^4? rephrase is x^4+y^4z^4>0
1) x^2 + y^2 > z^2
2) x+y>z from 1 ( square) x^4+2x^2*y^2+y^4 > z^4 x^4+y^4 z^4> 2x^2*y^2 thus the question became (WHETHER 2x^2*y^2 >0 we dont have enough info as x or y could = 0.......insuff from 2 x+y>z......insuff both still insuff.....E



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Re: DS Inequalities Problem for Math Wiz [#permalink]
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24 Jul 2009, 06:17
If you want to choose numbers here, you might notice that S1 looks a lot like the Pythagorean Theorem. In any right triangle with hypotenuse c, and legs a and b, we always have a + b > c, but a^2 + b^2 is not greater than c^2 (they're equal). So if we make x^2, y^2 and z^2 the lengths of three sides of a right triangle, we'll get numbers that demonstrate that x^4 + y^4 does not need to be larger than z^4. If you let, say, x^2 = 3, y^2 = 4 and z^2 = 5 (so x = root(3), y root(4) = 2, and z = root(5)), then x+y > z, x^2 + y^2 > z^2, but x^4 + y^4 = z^4.
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Re: DS Inequalities Problem for Math Wiz [#permalink]
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24 Jul 2009, 11:12
IanStewart wrote: If you want to choose numbers here, you might notice that S1 looks a lot like the Pythagorean Theorem. In any right triangle with hypotenuse c, and legs a and b, we always have a + b > c, but a^2 + b^2 is not greater than c^2 (they're equal). So if we make x^2, y^2 and z^2 the lengths of three sides of a right triangle, we'll get numbers that demonstrate that x^4 + y^4 does not need to be larger than z^4. If you let, say, x^2 = 3, y^2 = 4 and z^2 = 5 (so x = root(3), y root(4) = 2, and z = root(5)), then x+y > z, x^2 + y^2 > z^2, but x^4 + y^4 = z^4. Thanks Ian..Its all about identifying the pattern == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.




Re: DS Inequalities Problem for Math Wiz
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