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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< 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".



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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




Re: DS Inequalities Problem for Math Wiz
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