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

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

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Updated on: 20 Feb 2017, 22:39
1
9
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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.
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Re: Is (|x^(-1)*y^(-1)|)^(-1)> xy?  [#permalink]

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20 Feb 2017, 23:17
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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Re: Is (|x^(-1)*y^(-1)|)^(-1)> xy?  [#permalink]

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24 Mar 2017, 22:17
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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Is (|x^(-1)*y^(-1)|)^(-1)> xy?  [#permalink]

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16 Apr 2018, 04:59
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

If negative, answer is Yes

If positive, answer is No

Insuffcient

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

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16 Apr 2018, 07:59
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.

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

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15 Feb 2019, 19:48
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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Re: Is (|x^(-1)*y^(-1)|)^(-1)> xy?   [#permalink] 15 Feb 2019, 19:48
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# Is (|x^(-1)*y^(-1)|)^(-1)> xy?

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