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

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Q: Is x > y ? (1) x(x-1) > xy (2) (x-y)^3 > 0 [#permalink] New post 09 Mar 2007, 19:48
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.

Please check.
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Re: Algebra 2 [#permalink] New post 09 Mar 2007, 22:57
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.

Please check.


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?? :wink:
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 [#permalink] New post 10 Mar 2007, 01:22
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
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Re: Algebra 2 [#permalink] New post 10 Mar 2007, 01:45
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)
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 [#permalink] New post 10 Mar 2007, 01:58
Agree with (B).... Same approach as Dred :)
  [#permalink] 10 Mar 2007, 01:58
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