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goelmohit2002
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Powers of X and Y

by goelmohit2002 Fri Aug 21, 2009 1:05 pm

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.
vrajesh.dave
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Re: Powers of X and Y

by vrajesh.dave Sat Aug 22, 2009 3:38 pm

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.
nitin_prakash_khanna
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Re: Powers of X and Y

by nitin_prakash_khanna Sun Aug 23, 2009 10:36 am

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?
maxpeed
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Re: Powers of X and Y

by maxpeed Sun Aug 23, 2009 3:47 pm

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
RonPurewal
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Re: Powers of X and Y

by RonPurewal Fri Sep 25, 2009 10:30 pm