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anuramm
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Alex_NL
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anuramm
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sorry for the confusion, there is no minus sign.
the two parts read as
1. x^4 < y^4
2. x^-3 < Y^-3
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smandalika
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I think its still B.
this is because with even power the relevance of the sign is lost
ie. consider -2 ^ 4 = 16 and -3 ^ 4 = 81 and
2^4 = 16 3^4 =81
so in the former case x > y and the latter case x<y

but the 2nd condition keeps the sign parity and is sufficient since the power is odd

(-2)^3 > (-3)^3 => -2 > -3
and (4)^3 > (3)^3 => 4 > 3
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B.

I does not say anything about the sign of Y. It might be negative or positive.
II. X^3>Y^3 =>(X^2)X > (Y^2)Y
X^2 and Y^2 does not have any sign. So X is greater than Y.

Just for discussion, lets call this problem #2, what would be the solution for:
If xy != 0, is x < y?
1. - x^4 < y^4
2. - x^-3 < Y^-3.
(there is minus sign before the x in both statements)
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intr3pid
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First things first, if xy! here stands for the factorial of xy, then it can't be true. A factorial is a non-zero number.

Assuming that's a typo, the answer to the original question would be B for the reasons you guys mentioned.

The answer to Problem # 2 will be E. We don't have enough information from the givens to determine the answer.
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My apologies that I came up with the signs for x and y.
However I mentioned this because I try to calculate problems within certain time limits. This question was not sure for me whether I have to use minus or not.

Back to the question: My answer is B.

Regards,

Alex
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anuramm
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The OA is B.

Could anyone explain what is wrong with my reasoning.

Statement 2 says x^-3 y.

Let x = -3 and y =2.

-1/27 < 1/8, but x < y.

So 2 is insufficient.

So answer is E.

Thanks.



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