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

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

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New post Updated on: 03 Sep 2015, 07:37
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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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New post 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.

Answer: A.


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

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New post 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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New post 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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Your actual solution is very difficult to find in the quoted post and your advertisement. Can you please reduce the size of your advertisement text? or better put it in your signatures. This way your posts will be useful for people in this forum.

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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New post 10 Nov 2016, 13:05
Answer:A
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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New post 28 Jan 2017, 14:07
The answer is indeed A

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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New post 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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New post 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

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

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New post 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

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

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New post 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

Answer A


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

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

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New post 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

Answer A


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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