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# R = 1 + 2xy + x^2y^2, what is the value of xy?

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Intern
Joined: 25 Sep 2010
Posts: 12
R = 1 + 2xy + x^2y^2, what is the value of xy?  [#permalink]

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Updated on: 03 Sep 2015, 07:37
3
25
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35% (medium)

Question Stats:

69% (01:22) correct 31% (01:51) wrong based on 555 sessions

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$$R = 1 + 2*x*y + x^2*y^2$$, what is the value of $$x*y$$?

(1) R = 0
(2) x > 0

Originally posted by satishreddy on 14 Oct 2010, 20:56.
Last edited by ENGRTOMBA2018 on 03 Sep 2015, 07:37, edited 3 times in total.
Formatted the question.
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Re: R = 1 + 2xy + x^2y^2, what is the value of xy?  [#permalink]

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15 Oct 2010, 03:30
5
7
satishreddy wrote:
R=1+2xy+x^2*y^2, what is the value of xy

1) R=0
2) x>0

OA : A

Question should be:

R=1+2xy+x^2*y^2, what is the value of xy?

(1) R=0 --> $$1+2xy+x^2*y^2=0$$ --> $$(xy+1)^2=0$$ --> $$xy=-1$$. Sufficient.
(2) x>0 --> Clearly insufficient.

 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

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R = 1 + 2xy + x^2y^2, what is the value of xy?  [#permalink]

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04 Sep 2015, 04:37
Forget conventional ways of solving math questions. In DS, Variable approach is

the easiest and quickest way to find the answer without actually solving the

problem. Remember equal number of variables and equations ensures a solution.

R=1+2xy+(xy)^2, what is the value of x∗y ?

(1) R = 0
(2) x > 0

==> transforming the original condition and the question by variable approach method, we have, R=(1+xy)^2, 1+xy=R, -R, then xy=R-1, xy=-R-1. if we know R we can find xy
In case of 1), if R=0 then xy=-1, a unique answer. Therefore it is sufficient. Thus A is the answer.
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Re: R = 1 + 2xy + x^2y^2, what is the value of xy?  [#permalink]

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04 Sep 2015, 04:58
1
1
MathRevolution wrote:
Forget conventional ways of solving math questions. In DS, Variable approach is

the easiest and quickest way to find the answer without actually solving the

problem. Remember equal number of variables and equations ensures a solution.

R=1+2xy+(xy)^2, what is the value of x∗y ?

(1) R = 0
(2) x > 0

==> transforming the original condition and the question by variable approach method, we have, R=(1+xy)^2, 1+xy=R, -R, then xy=R-1, xy=-R-1. if we know R we can find xy
In case of 1), if R=0 then xy=-1, a unique answer. Therefore it is sufficient. Thus A is the answer.

If you know our own innovative logics to find the answer, you don’t need to

actually solve the problem.
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Also, put the quoted post in proper formatting, so that your actual post is clearly visible.

Thanks
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Re: R = 1 + 2xy + x^2y^2, what is the value of xy?  [#permalink]

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10 Nov 2016, 13:05
R=1+2xy+x^2*y^2, what is the value of x∗y ?
(1) R = 0
(2) x > 0
R=1+2xy+x^2*y^2=(1+xy)^2
1- R=0 ⇒ 1+xy=0 ⇒ xy=-1 -- SUFFICIENT
2- x>0 -- Irrelevant
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Re: R = 1 + 2xy + x^2y^2, what is the value of xy?  [#permalink]

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28 Jan 2017, 14:07

R=1+2xy+x^2y^2 => (1+xy)^2

Statement I

R=0, therefore (1+xy)^2=0
xy=-1

Statement II
x>0

It does not gives us any specific value of x or y or R

Not Sufficient.

Hence A.
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Re: R = 1 + 2xy + x^2y^2, what is the value of xy?  [#permalink]

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07 Jun 2018, 10:29
1st identify what is being asked, then looked for familiar structures (eg Quadratic equation when moving 1 to other side)
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Re: R = 1 + 2xy + x^2y^2, what is the value of xy?  [#permalink]

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09 Oct 2018, 23:31
1
satishreddy wrote:
$$R = 1 + 2*x*y + x^2*y^2$$, what is the value of $$x*y$$?

(1) R = 0
(2) x > 0

$$R = 1 + 2xy + (x^2)(y^2)$$
$$R = 1 + (xy)^2$$
$$xy = \sqrt{R}-1$$

Statement 1: R=0
Surely it gives xy =-1
Sufficient

Statement 2: x>0
Irrelevant . Not Sufficient

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Re: R = 1 + 2xy + x^2y^2, what is the value of xy?  [#permalink]

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16 Oct 2018, 20:10
Solving this question from statement 1 we get xy= -1 as (xy+1)^2 =0 which is sufficient
Statement 2 x>0, we cant find any value hence not suffice

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Re: R = 1 + 2xy + x^2y^2, what is the value of xy?  [#permalink]

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12 Jun 2019, 13:42
shashankism wrote:
satishreddy wrote:
$$R = 1 + 2*x*y + x^2*y^2$$, what is the value of $$x*y$$?

(1) R = 0
(2) x > 0

$$R = 1 + 2xy + (x^2)(y^2)$$
$$R = 1 + (xy)^2$$
$$xy = \sqrt{R}-1$$

Statement 1: R=0
Surely it gives xy =-1
Sufficient

Statement 2: x>0
Irrelevant . Not Sufficient

Please what happened to the 2xy in the question? I'm confused

Posted from my mobile device
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Joined: 02 Sep 2009
Posts: 58427
Re: R = 1 + 2xy + x^2y^2, what is the value of xy?  [#permalink]

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12 Jun 2019, 13:56
Peachie wrote:
shashankism wrote:
satishreddy wrote:
$$R = 1 + 2*x*y + x^2*y^2$$, what is the value of $$x*y$$?

(1) R = 0
(2) x > 0

$$R = 1 + 2xy + (x^2)(y^2)$$
$$R = 1 + (xy)^2$$
$$xy = \sqrt{R}-1$$

Statement 1: R=0
Surely it gives xy =-1
Sufficient

Statement 2: x>0
Irrelevant . Not Sufficient

Please what happened to the 2xy in the question? I'm confused

Posted from my mobile device

There is a typo in that post. Should be:
$$R = 1 + 2xy + x^2y^2$$
$$R = 1 + 2xy + (xy)^2$$
$$R = (1 + xy)^2$$
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Re: R = 1 + 2xy + x^2y^2, what is the value of xy?   [#permalink] 12 Jun 2019, 13:56
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