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

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Manager
Joined: 08 Sep 2015
Posts: 63

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Updated on: 20 Feb 2017, 21:39
1
4
00:00

Difficulty:

35% (medium)

Question Stats:

70% (01:46) correct 30% (01:58) wrong based on 188 sessions

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

(1) xy > 1

(2) x^2 > y^2

Originally posted by BPHASDEU on 20 Feb 2017, 21:30.
Last edited by Bunuel on 20 Feb 2017, 21:39, edited 1 time in total.
Renamed the topic and edited the question.
Senior Manager
Joined: 21 Aug 2016
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GPA: 3.9
WE: Information Technology (Computer Software)

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20 Feb 2017, 22:17
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
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24 Mar 2017, 21:17
BPHASDEU wrote:
Is $$(|x^{-1}y^{-1}|)^{-1}> xy$$?

(1) xy > 1

(2) x^2 > y^2

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
SVP
Joined: 26 Mar 2013
Posts: 2002

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16 Apr 2018, 03:59
1
BPHASDEU wrote:
Is $$(|x^{-1}y^{-1}|)^{-1}> xy$$?

(1) xy > 1

(2) x^2 > y^2

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

Insuffcient

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16 Apr 2018, 06:59
BPHASDEU wrote:
Is $$(|x^{-1}y^{-1}|)^{-1}> xy$$?

(1) xy > 1

(2) x^2 > y^2

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.

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Re: Is (|x^(-1)*y^(-1)|)^(-1)> xy? &nbs [#permalink] 16 Apr 2018, 06:59
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