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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".

Re: DS Inequalities Problem for Math Wiz [#permalink]
23 Jul 2009, 22:08

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^2 Squaring 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 plug-in 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 TRUE

b. 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 TRUE

c. 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 FALSE

Hence, Statement I is INSUFFICIENT

Let's look at the second statement: II. x+y>z

Squaring 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 plug-in 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 TRUE

Combining 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.

Re: DS Inequalities Problem for Math Wiz [#permalink]
24 Jul 2009, 01:27

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.

Re: DS Inequalities Problem for Math Wiz [#permalink]
24 Jul 2009, 05:17

Expert's post

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. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Re: DS Inequalities Problem for Math Wiz [#permalink]
24 Jul 2009, 10:12

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

gmatclubot

Re: DS Inequalities Problem for Math Wiz
[#permalink]
24 Jul 2009, 10:12

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