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# Q: Is x > y ? (1) x(x-1) > xy (2) (x-y)^3 > 0

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Senior Manager
Joined: 20 Feb 2007
Posts: 257
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Q: Is x > y ? (1) x(x-1) > xy (2) (x-y)^3 > 0 [#permalink]  09 Mar 2007, 18:48
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Q: Is x > y ?

(1) x(x-1) > xy
(2) (x-y)^3 > 0

Solution:
(1) x(x-1) > xy
(x-1) > xy / x
x-1 > y
x > y+1 so x > y (A is SUFF)

(2) (x-y)^3 > 0

now x-y can not be negative because cube-root of negative is -ve. So x is either more than y or equal to y. x>=y

If x = y then e.g. (3-3)^3 > 0 => 0 > 0 which is NOT TRUE so B is NOT SUFF.

My answer is A alone SUFF.

Director
Affiliations: FRM Charter holder
Joined: 02 Dec 2006
Posts: 736
Schools: Stanford, Chicago Booth, Babson College
Followers: 9

Kudos [?]: 38 [0], given: 4

Re: Algebra 2 [#permalink]  09 Mar 2007, 21:57
Summer3 wrote:
Q: Is x > y ?

(1) x(x-1) > xy
(2) (x-y)^3 > 0

Solution:
(1) x(x-1) > xy
(x-1) > xy / x
x-1 > y
x > y+1 so x > y (A is SUFF)

(2) (x-y)^3 > 0

now x-y can not be negative because cube-root of negative is -ve. So x is either more than y or equal to y. x>=y

If x = y then e.g. (3-3)^3 > 0 => 0 > 0 which is NOT TRUE so B is NOT SUFF.

My answer is A alone SUFF.

Hey Summer, why don't u please post just the question and then compare responses with your answer later? That way people will respond. If you post answers, how will we have fun in solving??
Intern
Joined: 13 Jun 2005
Posts: 29
Followers: 0

Kudos [?]: 14 [0], given: 0

I'd go with B.

(1)
x(x - 1) > x.y
x [ x - y - 1] > 0

if x > 0, x > y + 1 => x > y
if x < 0, x - y- 1 < 0 => x < y + 1 => x < y
(insuff)

(2)
(x - y) ^ 3 > 0
=> x - y needs to be +ve => x > y
Intern
Joined: 02 Mar 2007
Posts: 30
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Kudos [?]: 0 [0], given: 0

Re: Algebra 2 [#permalink]  10 Mar 2007, 00:45
Summer3 wrote:
Q: Is x > y ?

(1) x(x-1) > xy
(2) (x-y)^3 > 0

Solution:
(1) x(x-1) > xy
(x-1) > xy / x
x-1 > y

dred is right. Summer3, u divide by x without knowing whether it is positive or negative. If it is negative, the sign of the inequality will change.
I go for B)
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

Agree with (B).... Same approach as Dred
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