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

Question

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

(1) xy > 1

(2) x^2 > y^2

AR15J · Answer

Correct answer is A because when xy> 1, (|x−1y−1|)−1>xy(|x−1y−1|)−1>xy can never be true; however, statement 2 can not determine if it would be true or not

nks2611 · Answer

the answer would be A 
 after simplifying the above equations we have two cases 
case 1. xy>xy 
case 2. -xy>xy

now statement 1. xy>1 which means both x and y must be -ve or +ve say x=2 and y=3 after putting values in case 1 and in case 2, we get to know that in both the cases we get a clear no, so sufficient

now statement 2. X^2 - Y^2>0
in this we have two cases say x=3 and y= -2
 x=-3 and y=2 
now put these values in above cases , case 1 . 3*-2>3*-2 which not true now in case 2, -(3*-2)>3*-2 which is true so now no need to check more cases , as we gat one yes and one no so insufficient .

hence A 
thanks

Mo2men · Answer

Let's simplify the LHS in the question stem

(|x^{-1}y^{-1}|)^{-1} = (|\frac{1}{x} * \frac{1}{y}|)^{-1} =|xy|

The question becomes:

Is  |xy|  >  xy ??

This happens when  xy  is negative. So question could be rephrased:

Is  xy  < 0??

(1) xy > 1

Here, we have direct answer NO

Sufficient

(2) x^2 > y^2

Here it gives us that  |x|>|y|

so  xy  could be negative or positive

If negative, answer is Yes

If positive, answer is No

Insuffcient

Answer: A

MathRevolution · Answer

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. (|x^{-1} y^{-1}|)^{-1} > xy ⇔ |1/(xy)|^{-1} = > xy ⇔ |xy| > xy ⇔ xy < 0 Condition 1) : xy > 1 Since xy > 1 , xy is positive. Thus xy < 0 is false and the answer is "no". Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, the condition 1) is sufficient, when used together. Condition 2) x^2 > y^2 If x = 2 , y = -1 , then xy < 0 . The answer is "yes" If x = 2 , y = 1 , then xy > 0 . The answer is "no". Since we don't have a unique solution, the condition 2) is not sufficient. Therefore, A is the answer.

KanishkM · Answer

So lets do manipulation Question becomes |xy| > xy now this can have 2 cases -xy -xy > 0 or 0>0(NA) -2xy > 0 => xy < 0 from 1) xy > 1 Question will always be false from 2) x^2 > y^2 x^2 - y^2 > 0 x-y x+y > 0 x > y & x > - y or x< y & x < - y Now for the above relationship to be true, x can be +ive or -ive or y can be +ive or -ive Give 2 answers to the question, making this statement insufficient A

bumpbot · Answer

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

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