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Manager  B
Joined: 08 Sep 2015
Posts: 62
Is (|x^(-1)*y^(-1)|)^(-1)> xy?  [#permalink]

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1
9 00:00

Difficulty:   35% (medium)

Question Stats: 70% (01:45) correct 30% (01:59) wrong based on 197 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, 22:30.
Last edited by Bunuel on 20 Feb 2017, 22:39, edited 1 time in total.
Renamed the topic and edited the question.
Senior Manager  G
Joined: 21 Aug 2016
Posts: 258
Location: India
GPA: 3.9
WE: Information Technology (Computer Software)
Re: Is (|x^(-1)*y^(-1)|)^(-1)> xy?  [#permalink]

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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
Senior Manager  S
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Re: Is (|x^(-1)*y^(-1)|)^(-1)> xy?  [#permalink]

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

(1) xy > 1

(2) x^2 > y^2

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
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Joined: 26 Mar 2013
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Is (|x^(-1)*y^(-1)|)^(-1)> xy?  [#permalink]

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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

If negative, answer is Yes

If positive, answer is No

Insuffcient

Answer: A
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Re: Is (|x^(-1)*y^(-1)|)^(-1)> xy?  [#permalink]

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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.

Therefore, A is the answer.
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Re: Is (|x^(-1)*y^(-1)|)^(-1)> xy?  [#permalink]

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

(1) xy > 1

(2) x^2 > y^2

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
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If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up. Re: Is (|x^(-1)*y^(-1)|)^(-1)> xy?   [#permalink] 15 Feb 2019, 19:48
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