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Is x^4 + y^4 > z^4?

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Source: — Data Sufficiency |
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by sivaelectric » Sun Jun 05, 2011 7:53 am
My choice E
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by Frankenstein » Sun Jun 05, 2011 8:05 am
Hi,
From(1): x^2 + y^2 > z^2
if x=1,y=3,z=2 then x^4 + y^4 > z^4
if x=1,y=1,z=1.4 then x^4 + y^4 < z^4
Insufficient
From(2): x+y>z
if x=1,y=3,z=2 then x^4 + y^4 > z^4
if x=1,y=1,z=1.4 then x^4 + y^4 < z^4
Insufficient
Both(1)&(2): I have considered the same values in both cases
Insufficient

Hence, E
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by cans » Sun Jun 05, 2011 8:06 am
x^4 + y^4 > z^4?
a) x^2 + y^2 > z^2
squaring both sides, x^4 + y^4 + 2x^2*y^2> z^4
insufficient. (2x^2*y^2 >=0)
b)x+y>z
x=.5,y=.5z=.99
x^4=y^4=.0625 z^4 is nearly equal to 1.
x^4 + y^4 =.125 thus lhs<rhs
x=10,y=10,z=1 ->lhs>rhs
insufficient.
a&b together) x^2 + y^2 + 2xy > z^2
also x^2 + y^2 > z^2
thus xy>0
Insufficient.
IMO E
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by [email protected] » Sun Jun 05, 2011 8:09 am
Is x^4 + y^4 > z^4?


1) x^2 + y^2 > z^2

2) x + y > z


Well according to me the answer is D.

Explanation:

Statement1:
x^2 + y^2 > z^2

if you square on both sides then,

you will get 2x^2 y^2 as extra. so the step would be

x^4 + y^4 + 2 x^2 y^2 > z^4 and we know that x^2 + y^2 > z^2 so, x^4 + y^4 > z^4 this has to be true.

Statement 2:

statement 2 alone is sufficient.

If x + y > z, then (x + y)^4 > z^4 this also will hold as we are given that x + y > z.


I am not an expert but according to me answer is D.
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by cans » Sun Jun 05, 2011 8:29 am
Frankenstein wrote:Hi,
From(1): x^2 + y^2 > z^2
if x=1,y=3,z=2 then x^4 + y^4 > z^4
if x=1,y=1,z=1.4 then x^4 + y^4 < z^4
Insufficient
From(2): x+y>z
if x=1,y=3,z=2 then x^4 + y^4 > z^4
if x=1,y=1,z=1.4 then x^4 + y^4 < z^4
Insufficient
Both(1)&(2): I have considered the same values in both cases
Insufficient

Hence, E
Last edited by cans on Sun Jun 05, 2011 8:45 am, edited 1 time in total.
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by Frankenstein » Sun Jun 05, 2011 8:37 am
Hi,
I will explain "I have considered the same values in both cases"
It means I have used the same set of values of (1) alone and (2) alone. So, for (1)&(2) together, my set of values is same and the result is still same.
Last edited by Frankenstein on Sun Jun 05, 2011 8:54 am, edited 1 time in total.
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by cans » Sun Jun 05, 2011 8:47 am
sorry for the mistake.
in the line x=1,y=1,z=1.4
I thought '.' represents full stop and thus considered z as 1 instead of 1.4
Have edited my previous post :)
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by Frankenstein » Sun Jun 05, 2011 8:57 am
Hi,
I appreciate your efforts. As you have edited, my post of your quote doesn't make sense. So, I edited my post.
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by cans » Sun Jun 05, 2011 4:45 pm
[email protected] wrote:Is x^4 + y^4 > z^4?


1) x^2 + y^2 > z^2

2) x + y > z


Well according to me the answer is D.

Explanation:

Statement1:
x^2 + y^2 > z^2

if you square on both sides then,

you will get 2x^2 y^2 as extra. so the step would be

x^4 + y^4 + 2 x^2 y^2 > z^4 and we know that x^2 + y^2 > z^2 so, x^4 + y^4 > z^4 this has to be true.

Statement 2:

statement 2 alone is sufficient.

If x + y > z, then (x + y)^4 > z^4 this also will hold as we are given that x + y > z.


I am not an expert but according to me answer is D.
x^4 + y^4 + 2 x^2 y^2 > z^4
x^2y^2 is always >=0
let x^4 + y^4 = a and let 2x^2y^2 = b and let z^4 = c
thus a+b>c (a,b,c all are greater than or equal to 0)
a>c-b
now let a=5,b=2,c=6
5+2=7>6 true
but 5 is not greater than 6 or a is not greater than c.
Thus A is insufficient.
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