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man, I'm quite weak at number properties as well I just found out, trying this one... well lets give it a shot:

xy != 0 means x and y can be +ve or -ve but none is zero

(1) the ^4 neutralizes the sign and says that the absolute value of y > x
The true value cannot be know.
- both could be +ve then y>x
- both could be -ve then y<x
---> INSUFF

(2) rewrite to 1/(x^3) < 1/(x^3), and signs play a role
For this statement to be true, the true values could be:
- both could be +ve then x>y
- both could be -ve then x<y
---> INSUFF

(1)&(2)
On the premise, that the absolute value of y > x from (1),
in order for (2) to be true, x must be -ve.
Hence, y>x
SUFF.

(1) x^4<y^4 --> x^4-y^4<0 ---> (x^2-y^2) ( x^2+y^2) <0
since x^2+y^2> 0 ( coz no x , y equal to 0 so the sum can't be 0)---> x^2-y^2<0 ---> (x-y)*(x+y) <0
We have to cases:
case1 : x-y<0 and x+y>0 ---> -y<x<y
case2: x-y>0 and x+y<0 ---> y<x<-y
---> insuff

(2) x^(-3)< y^(-3) --> 1/x^3< 1/y^3 --> 1/x^3-1/y^3<0
--> ( 1/x-1/y) * (1/x^2+ 1/xy+ 1/y^2) <0
1/x^2+ 1/y^2 >= | 2/xy|>= 2/xy> 1/xy, thus 1/x^2+1/xy+1/y^2>0
---> 1/x-1/y < 0 --> (y-x)/xy<0
we have two cases:
case1 : y-x<0 and xy>0 ---> y<x and xy>0
for example: x=3, y=2 we have 1/3-1/2<0
case2: y-x>0 and xy<0 ---> y>0>x
--->insuff

Last edited by laxieqv on 04 Nov 2005, 09:42, edited 1 time in total.

(1) x^4<y^4 --> x^4-y^4<0 ---> (x^2-y^2) ( x^2+y^2) <0 since x^2+y^2> 0 ( coz no x , y equal to 0 so the sum can't be 0)---> x^2-y^2<0 ---> (x-y)*(x+y) <0 We have to cases: case1 : x-y<0 and x+y>0 ---> -y<x<y case2: x-y>0 and x+y<0 ---> y<x<-y ---> insuff

(2) x^(-3)< y^(-3) --> 1/x^3< 1/y^3 --> 1/x^3-1/y^3<0 --> ( 1/x-1/y) * (1/x^2+ 1/xy+ 1/y^2) <0 1/x^2+ 1/y^2 >= | 2/xy|>= 2/xy> 1/xy, thus 1/x^2+1/xy+1/y^2>0 ---> 1/x-1/y < 0 --> (y-x)/xy<0 we have two cases: case1 : y-x<0 and xy>0 ---> y<x and xy>0 for example: x=3, y=2 we have 1/3-1/2<0 case2: y-x>0 and xy<0 ---> y>0>x --->insuff

(1) x^4<y^4 --> x^4-y^4<0 ---> (x^2-y^2) ( x^2+y^2) <0 since x^2+y^2> 0 ( coz no x , y equal to 0 so the sum can't be 0)---> x^2-y^2<0 ---> (x-y)*(x+y) <0 We have to cases: case1 : x-y<0 and x+y>0 ---> -y<x<y case2: x-y>0 and x+y<0 ---> y<x<-y ---> insuff

(2) x^(-3)< y^(-3) --> 1/x^3< 1/y^3 --> 1/x^3-1/y^3<0 --> ( 1/x-1/y) * (1/x^2+ 1/xy+ 1/y^2) <0 1/x^2+ 1/y^2 >= | 2/xy|>= 2/xy> 1/xy, thus 1/x^2+1/xy+1/y^2>0 ---> 1/x-1/y < 0 --> (y-x)/xy<0 we have two cases: case1 : y-x<0 and xy>0 ---> y<x and xy>0 for example: x=3, y=2 we have 1/3-1/2<0 case2: y-x>0 and xy<0 ---> y>0>x --->insuff

good explanation; but the bold part does not satisfy statement 2. pick some numbers and try
Answer is good: B is suff

(1) x^4<y^4 --> x^4-y^4<0 ---> (x^2-y^2) ( x^2+y^2) <0 since x^2+y^2> 0 ( coz no x , y equal to 0 so the sum can't be 0)---> x^2-y^2<0 ---> (x-y)*(x+y) <0 We have to cases: case1 : x-y<0 and x+y>0 ---> -y<x<y case2: x-y>0 and x+y<0 ---> y<x<-y ---> insuff

(2) x^(-3)< y^(-3) --> 1/x^3< 1/y^3 --> 1/x^3-1/y^3<0 --> ( 1/x-1/y) * (1/x^2+ 1/xy+ 1/y^2) <0 1/x^2+ 1/y^2 >= | 2/xy|>= 2/xy> 1/xy, thus 1/x^2+1/xy+1/y^2>0 ---> 1/x-1/y < 0 --> (y-x)/xy<0 we have two cases: case1 : y-x<0 and xy>0 ---> y<x and xy>0 for example: x=3, y=2 we have 1/3-1/2<0 case2: y-x>0 and xy<0 ---> y>0>x --->insuff

good explanation; but the bold part does not satisfy statement 2. pick some numbers and try Answer is good: B is suff

Actually there are three questions I think have a wrong OA. Questions are from Princeton +48 Gmat math.

OE: 1 is insuf because negative values for the variables give you the opposite value than negative values.
In statement 2, negative numbers retain their sign but we need to look closely at how fractions react. If you plug in x=2 and y=1 that gives you a "no". If you plug in x=1/2 and y=1/4 you also get a "no". Therefore statement 2 is sufficient and answer is B.

But if you plug in x=-1 y=1 you get a yes with statement 2, thus OA/OE is wrong.