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goelmohit2002
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Post subject: Powers of X and Y Posted: Fri Aug 21, 2009 1:05 pm |
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Posts: 226
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Source GMATPrep
Is x^4 + y^4 > z^4
1) x^2 + y^2 > z^2 2) x + y > z
OA = E.
Can someone please tell how to solve ?I am able to understand how "2" is insufficient...but could not figure out how "1" is insufficient.
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vrajesh.dave
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Post subject: Re: Powers of X and Y Posted: Sat Aug 22, 2009 3:38 pm |
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This one is interesting.
IMO for stmt 1)
let x = 0.33; y = 0.4; z = 0.5
(0.33)^2 + (0.4)^2 > (0.5)^2 is true
== > 0.1089 + 0.16 > 0.25
(0.33)^4 + (0.4)^4 > (0.5)^4 == > 0.01185 + 0.0256 is NOT greater then 0.0625
Hence stmt 1) is not sufficient
Again, it took me lot of time to figure this one out.
Can someone give us a simple and quicker explanation for this problem.
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nitin_prakash_khanna
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Post subject: Re: Powers of X and Y Posted: Sun Aug 23, 2009 10:36 am |
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Statement 2 can be made insufficient by choosing
x= 3/2 ,,,, y=1 ,,,,z=2
X+y >2
buy X^4 + y^4 < z^4
And hence insufficient.
But still looking for right approach to attack this. Ron?
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maxpeed
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Post subject: Re: Powers of X and Y Posted: Sun Aug 23, 2009 3:47 pm |
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hi as for 1): let's say x,y, z>2 if x^2+y^2=c, we can choose z^2 = c-1 -this condition complies with 1)-
c^2=x^4 + y^4 +2*x^2*y^2 => x^4 + y^4 = c^2 - 2*x^2*y^2 z^4= (c-1)^2 => z^4 = c^2 -2*c+1
let's check now: x^4 + y^4 > z^4 c^2 - 2*x^2*y^2 > c^2 -2*c+1 2*(c-x^2*y^2)>1 ----but c=x^2+y^2 x^2+y^2 - x^2*y^2 > 1/2 x^2*y^2*(1/y^2+1/x^2-1) > 1/2 since both x and y are >2, the part in brackets is < 0, therefore this conditions is never true. 1) Not sufficient.
ex x=3, y=4, z=(24)^1/2 : 25>24 but 337<576 ex x=6, y=8, z=(99)^1/2 : 100>99 but 5392<9801
hope it helps. though i dont know how they pretend students to find an answer within a couple of minutes... mp
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RonPurewal
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Post subject: Re: Powers of X and Y Posted: Fri Sep 25, 2009 10:30 pm |
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Posts: 7146
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