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Source: — Data Sufficiency |
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by sanju09 » Sat Apr 30, 2011 2:00 am
ruplun wrote:IF X=3Y, IS X^3 > Y^3?

(1). Y+X<Y-X
(2).X^2=9Y^2

If X = 3 Y, then X^3 > Y^3 is possible only if X is positive.

(1) Y + X < Y - X implies that X < 0, hence X^3 is NOT greater than Y^3. Sufficient

(2) X^2 = 9 Y^2 reduces to X = ±3 Y, but we already have X = 3 Y in the stem, which is already insufficient to decide on X^3 > Y^3. Insufficient

[spoiler]A[/spoiler]
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by clock60 » Sat Apr 30, 2011 2:33 am
i hope that in given problem it is assumed that y is not equal to zero.otherwise the immediate answer is no
is x^3>y^3, or is x^3-y^3>0
x^3-y^3=(x-y)*(x^2+xy+y^2) we are given that x=3y, thus
(3y-y)((3y)^2+3y*y+y^2)=2y*(13y^2) ignore 2 and 13, and y^2 as well
and left with y, so the real question
does y>0
(1)x+y<y-x here we can cancel y and left with 2x<0, x<0 and 2*3y<0 strictly y<0.
the answer is no and it is suff
(2) (x-3y)(x+3y)=0
from here x=3y or x=-3y but we are given that x=3y from the problem so it can`t be x=-3y, but this info does not help us to estimate the value of y, insuff
my answer is A
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