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09 Mar 2007, 19:48
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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.
Please check.
(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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09 Mar 2007, 22:57
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Summer3
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??
10 Mar 2007, 01:22
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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
(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
