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123) Is x<0?

by ern5231 » Sun May 16, 2010 9:21 pm
x<y. Is x<0?
1) x^2>y^2
2) x ^ 3> y ^ 3

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by ballubalraj » Mon May 17, 2010 1:11 am
Stmt 1: If x < y, then x^2 > y^2 can be true only if x is negative; sufficient

Stmt 2: If x < y,then x^3 > y^3 can be true only if x = 0; sufficient

Ans D.

Note: As no information is mentioned about x and y, I assumed that zero is an acceptable values for x.

Any expert comments on this?

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by mj78ind » Mon May 17, 2010 1:18 am
1 - For any positive x and y. There is no x and y combination where x^2 > y^2 and x < y are both satisfied. If x is negative and y positive we can have both conditions met. If x is negative and y is also negative, y has to be greater than x for both conditions to hold. Thus, 1 is sufficient, as x<0 is always the case.

2 - For x^3 > y^3, means x > y, but the question says x<y. Thus there is no solution.

Hence, my bet is A

OA please .......?

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by ballubalraj » Mon May 17, 2010 1:52 am
I agree with mj78ind for stmt 2 reasoning, which I missed in my previous answer.

OA please!