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Is x^4 + y^4 > z^4?
Source: Beat The GMAT — Data Sufficiency |
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sivaelectric
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My choice E
If I am wrong correct me 
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.
Chitra Sivasankar Arunagiri
, If my post helped let me know by clicking the Thanks button .Chitra Sivasankar Arunagiri
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
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
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
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
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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Cans!!
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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.
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.
IT IS TIME TO BEAT THE GMAT
LEARNING, APPLICATION AND TIMING IS THE FACT OF GMAT AND LIFE AS WELL... KEEP PLAYING!!!
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LEARNING, APPLICATION AND TIMING IS THE FACT OF GMAT AND LIFE AS WELL... KEEP PLAYING!!!
Whenever you feel that my post really helped you to learn something new, please press on the 'THANK' button.
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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Cans!!
Contact me about long distance tutoring!
[email protected]
Cans!!
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.
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.
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
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
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
If my post helped you- let me know by pushing the thanks button
Contact me about long distance tutoring!
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Cans!!
Contact me about long distance tutoring!
[email protected]
Cans!!
Hi,
I appreciate your efforts. As you have edited, my post of your quote doesn't make sense. So, I edited my post.
I appreciate your efforts. As you have edited, my post of your quote doesn't make sense. So, I edited my post.
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
x^4 + y^4 + 2 x^2 y^2 > z^4[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^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.
If my post helped you- let me know by pushing the thanks button
Contact me about long distance tutoring!
[email protected]
Cans!!
Contact me about long distance tutoring!
[email protected]
Cans!!
