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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

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. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

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