Is |x^-1 y^-1| ^ -1 xy ? (1) xy 1 (2) x^2 y^2

Question

Is |x^-1 y^-1| ^ -1 > xy ?

(1) xy > 1
(2) x^2 > y^2

bewakoof · Answer

Is it A

solving the LHS I get 

|xy| >xy

and that holds false for all cases when xy>0.. 

x^2 > y^2.. in this case |xy| > xy and |xy| = xy

FN · Answer

I get A...

basically you can reduce the eq. to be |xy| > xy?

(1) says that the XY>1, then that means both X and Y have the same sign and are not fractions...in that case...|xy|=xy....Sufficient..

(2) X^2>y^2, well we dont know if X is positive or not..same about Y...

insuff....

duttsit · Answer

Great job! OA is A.

We can rephrase the question by manipulating it algebraically:

(|x-1 * y-1|)-1 > xy

(|1/x * 1/y|)-1 > xy

(|1/xy|)-1 > xy

1/(|1/(xy)|) > xy

|xy| > xy

The question can be rephrased as â€œIs the absolute value of xy greater than xy?â€

krisrini · Answer

This is A.

From the question we need to evaluate if |xy| > xy

From stmt 1 we see xy to be > 1 implying xy is a positive number in which case |xy| = xy, hence sufficient to answer.

From stmt 2 x and y can be + ve or -ve and will not yield a definite result.

Hence A.

Is |x^-1 y^-1| ^ -1 > xy ? (1) xy > 1 (2) x^2 > y^2

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