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What is the value of x^2 + y^2 ?

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What is the value of x^2 + y^2 ? [#permalink]

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What is the value of \(x^2 + y^2\) ?

(1) \(x^2 + y^2 = 2xy + 1\)
(2) \(x^2 + y^2 = 4 - 2xy\)

[Reveal] Spoiler:
I don't agree with the OA.
If we add (1) and (2):

(1) \(x^2 -2xy + y^2 = 1\)
(2) \(x^2 +2xy+ y^2 = 4\)
-------------------------------------
Total: \(2x^2 +2y^2 = 5\)

Then: \(x^2 + y^2 = \frac{5}{2}\)

Please, your comments.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 01 Dec 2014, 03:57, edited 1 time in total.
Edited the OA.

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Re: What is the value of x^2 + y^2 ? [#permalink]

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New post 26 Aug 2012, 10:41
When trying to solve for x and y, it doesn't appear that we can find a solution for x^2 + y^2
2xy + 1 = 4 - 2xy

4xy = 3
xy = 3/4

However, the approach of canceling out 2xy DOES remove the xy variable and leaves us cleanly with x^2 + y^2. We indeed have the value as you mentioned...perhaps the source for the OA is incorrect.

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What is the value of x^2 + y^2 ? [#permalink]

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New post 23 Nov 2012, 07:29
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What is the value of x^2 + y^2 ?

(1) x^2 + y^2 = 2xy+1
(2) x^2 + y^2 = 4 - 2xy

I have an official answer but I was thinking that it is wrong.
This is a question from GMATHacks Algebra Challange set.


OA SOLUTION

[Reveal] Spoiler:
77. E
Explanation: This question rigorously tests your familiarity with common
binomials. The only way to do anything with statement (1) is to subtract 2xy
from both sides:
x^2 - 2xy + y^2 = 1
(x - y)(x - y) = 1
x - y = 1 or x - y = -1
There are an in…nite number of possible solutions for x and y, so there’s no
way to determine a speci…c value for x^2 + y^2.
Statement (2) is insufficient for similar reasons:
x^2 + y^2 = 4 - 2xy
x^2 + 2xy + y^2 = 4
(x + y)(x + y) = 4
x + y = 2 or x + y = -2
Again, there are an in…nite number of possibilities for x and y.
Taken together, you still don’t have enough information. If you had exactly
two equations (such as x-y = 1 and x+y = 2), you could solve, but you have
two diferent pairs of possible equations, which is insufficient to …find the values
of the variables. Choice (E) is correct.

Last edited by Bunuel on 24 Nov 2014, 01:44, edited 6 times in total.
Edited the OA.

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What is the value of x^2+y^2 ?

(1) x^2+y^2=2xy+1 --> \(x^2-2xy+y^2=1\) --> \((x-y)^2=1\). If \(x=1\) and \(y=0\), then \(x^2+y^2=1\) but if \(x=2\) and \(y=1\), then \(x^2+y^2=5\). Not sufficient.

(2) x^2+y^2=4-2xy --> \(x^2+2xy+y^2=4\) --> \((x+y)^2=4\). If \(x=2\) and \(y=0\), then \(x^2+y^2=4\) but if \(x=1\) and \(y=1\), then \(x^2+y^2=2\). Not sufficient.

(1)+(2) Sum the equations: \(2(x^2+y^2)=5\) --> \(x^2+y^2=2.5\). Sufficient.

Answer: C. OA must be wrong.

Hope it's clear.
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Re: What is the value of x^2 + y^2 ? [#permalink]

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New post 23 Nov 2012, 07:46
Bunuel wrote:
What is the value of x^2+y^2 ?

(1) x^2+y^2=2xy+1 --> \(x^2-2xy+y^2=1\) --> \((x-y)^2=1\). If \(x=1\) and \(y=0\), then \(x^2+y^2=1\) but if \(x=2\) and \(y=1\), then \(x^2+y^2=5\). Not sufficient.

(2) x^2+y^2=4-2xy --> \(x^2+2xy+y^2=4\) --> \((x+y)^2=4\). If \(x=2\) and \(y=0\), then \(x^2+y^2=4\) but if \(x=1\) and \(y=1\), then \(x^2+y^2=2\). Not sufficient.

(1)+(2) Sum the equations: \(2(x^2+y^2)=5\) --> \(x^2+y^2=2.5\). Sufficient.

Answer: C. OA must be wrong.

Hope it's clear.



Thanks Bunuel!!! (I feel like a VIP having you respond to one of my questions!) haha

"C" is what I thought! I got worried and couldn't see what I was doing wrong even after reading the OA description.

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New post 23 Nov 2012, 07:48
maxLRok wrote:
Bunuel wrote:
What is the value of x^2+y^2 ?

(1) x^2+y^2=2xy+1 --> \(x^2-2xy+y^2=1\) --> \((x-y)^2=1\). If \(x=1\) and \(y=0\), then \(x^2+y^2=1\) but if \(x=2\) and \(y=1\), then \(x^2+y^2=5\). Not sufficient.

(2) x^2+y^2=4-2xy --> \(x^2+2xy+y^2=4\) --> \((x+y)^2=4\). If \(x=2\) and \(y=0\), then \(x^2+y^2=4\) but if \(x=1\) and \(y=1\), then \(x^2+y^2=2\). Not sufficient.

(1)+(2) Sum the equations: \(2(x^2+y^2)=5\) --> \(x^2+y^2=2.5\). Sufficient.

Answer: C. OA must be wrong.

Hope it's clear.



Thanks Bunuel!!! (I feel like a VIP having you respond to one of my questions!) haha

"C" is what I thought! I got worried and couldn't see what I was doing wrong even after reading the OA description.


Can you please post their solution?
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New post 23 Nov 2012, 08:50
Bunuel wrote:
maxLRok wrote:
Bunuel wrote:
What is the value of x^2+y^2 ?

(1) x^2+y^2=2xy+1 --> \(x^2-2xy+y^2=1\) --> \((x-y)^2=1\). If \(x=1\) and \(y=0\), then \(x^2+y^2=1\) but if \(x=2\) and \(y=1\), then \(x^2+y^2=5\). Not sufficient.

(2) x^2+y^2=4-2xy --> \(x^2+2xy+y^2=4\) --> \((x+y)^2=4\). If \(x=2\) and \(y=0\), then \(x^2+y^2=4\) but if \(x=1\) and \(y=1\), then \(x^2+y^2=2\). Not sufficient.

(1)+(2) Sum the equations: \(2(x^2+y^2)=5\) --> \(x^2+y^2=2.5\). Sufficient.

Answer: C. OA must be wrong.

Hope it's clear.



Thanks Bunuel!!! (I feel like a VIP having you respond to one of my questions!) haha

"C" is what I thought! I got worried and couldn't see what I was doing wrong even after reading the OA description.


Can you please post their solution?



Just put it in the main post

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New post 23 Nov 2012, 08:56
maxLRok wrote:
Bunuel wrote:
maxLRok wrote:

Thanks Bunuel!!! (I feel like a VIP having you respond to one of my questions!) haha

"C" is what I thought! I got worried and couldn't see what I was doing wrong even after reading the OA description.


Can you please post their solution?



Just put it in the main post


Their solution is not correct.

Four pairs of (x,y) satisfy (x-y)^2=1 and (x+y)^2=4: (-3/2, -1/2), (-1/2, -3/2), (3/2, 1/2) and (1/2, 3/2). For each pair x^2+y^2=10/2.
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Re: What is the value of x^2 + y^2 ? [#permalink]

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New post 06 Jun 2014, 07:02
Hello Everyone,

This DS problem is from the Jeffrey Sackmann Challenge Algebra set, Quadratics. Not sure about the difficulty level.

What is the value of x^2 + y^2 ?
1) x^2 + y^2 = 2xy + 1
2) x^2 + y^2 = 4 - 2xy

The answer is (E) But I'm not sure why we didn't just use the two statements together > add them > 2xy cancels out and the answer would be: x^2 + y^2 = 5/2 ?

Thanks in advance :)

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New post 06 Jun 2014, 07:08
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Mfz83 wrote:
Hello Everyone,

This DS problem is from the Jeffrey Sackmann Challenge Algebra set, Quadratics. Not sure about the difficulty level.

What is the value of x^2 + y^2 ?
1) x^2 + y^2 = 2xy + 1
2) x^2 + y^2 = 4 - 2xy

The answer is (E) But I'm not sure why we didn't just use the two statements together > add them > 2xy cancels out and the answer would be: x^2 + y^2 = 5/2 ?

Thanks in advance :)


Merging similar topics. please refer to the discussion above.

You are right the OA must be C, instead of E.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rule 3. Thank you.
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New post 11 Jun 2014, 04:41
What is the value of \(x^2\) +\(y^2\)?

(1)\(x^2\) +\(y^2\) = \(2xy + 1\)

(1) \(x^2 +y^2 = 4 - 2xy\)

Last edited by Gnpth on 11 Jun 2014, 04:58, edited 1 time in total.
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irda wrote:
What is the value of \(x^2\) +\(y^2\)?

(1)\(x^2\) +\(y^2\) = \(2xy + 1\)

(1) \(x^2 +y^2 = 4 - 2xy\)



The question is already discussed here. Also the OA is wrong.

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Re: What is the value of x^2 + y^2 ? [#permalink]

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New post 30 Nov 2014, 13:20
metallicafan wrote:
What is the value of \(x^2 + y^2\) ?
(1) \(x^2 + y^2 = 2xy + 1\)
(2) \(x^2 + y^2 = 4 - 2xy\)

I don't agree with the OA.
If we add (1) and (2):

(1) \(x^2 -2xy + y^2 = 1\)
(2) \(x^2 +2xy+ y^2 = 4\)
-------------------------------------
Total: \(2x^2 +2y^2 = 5\)

Then: \(x^2 + y^2 = \frac{5}{2}\)

Please, your comments.




+1 for C. I did the same as you did.

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New post 30 Nov 2014, 23:32
metallicafan wrote:
What is the value of \(x^2 + y^2\) ?
(1) \(x^2 + y^2 = 2xy + 1\)
(2) \(x^2 + y^2 = 4 - 2xy\)

I don't agree with the OA.
If we add (1) and (2):

(1) \(x^2 -2xy + y^2 = 1\)
(2) \(x^2 +2xy+ y^2 = 4\)
-------------------------------------
Total: \(2x^2 +2y^2 = 5\)

Then: \(x^2 + y^2 = \frac{5}{2}\)

Please, your comments.



well metallicafan,
i see nothing wrong with your approach, but i solved this the following way...
(1) x^2 + y^2 = 2.x.y + 1
=> (x - y)^2 = 1
=> x - y = 1 or -1 insuff

(2) x^2 + y^2 = 4 - 2.x.y
=> (x + y)^2 = 4
=> x + y = 2 or -2 insuff

combining the two,
since the above equations can have multiple values when we solve for x and y, hence, insufficient.
answer is E. is this a GMATPrep question?? because i dont think the real gmat would confuse students like this.

happy to help, a kudos if you like. :)

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Re: What is the value of x^2 + y^2 ? [#permalink]

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New post 01 Dec 2014, 01:40
arnabs wrote:
metallicafan wrote:
What is the value of \(x^2 + y^2\) ?
(1) \(x^2 + y^2 = 2xy + 1\)
(2) \(x^2 + y^2 = 4 - 2xy\)

I don't agree with the OA.
If we add (1) and (2):

(1) \(x^2 -2xy + y^2 = 1\)
(2) \(x^2 +2xy+ y^2 = 4\)
-------------------------------------
Total: \(2x^2 +2y^2 = 5\)

Then: \(x^2 + y^2 = \frac{5}{2}\)

Please, your comments.



well metallicafan,
i see nothing wrong with your approach, but i solved this the following way...
(1) x^2 + y^2 = 2.x.y + 1
=> (x - y)^2 = 1
=> x - y = 1 or -1 insuff

(2) x^2 + y^2 = 4 - 2.x.y
=> (x + y)^2 = 4
=> x + y = 2 or -2 insuff

combining the two,
since the above equations can have multiple values when we solve for x and y, hence, insufficient.
answer is E. is this a GMATPrep question?? because i dont think the real gmat would confuse students like this.

happy to help, a kudos if you like. :)


But don't you get an answer of 5/2 for\(x^2 + y^2\) if you take any of the combinations?, Note that we are asked to find \(x^2 + y^2\)

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New post 01 Dec 2014, 02:36
santorasantu wrote:
arnabs wrote:
metallicafan wrote:
What is the value of \(x^2 + y^2\) ?
(1) \(x^2 + y^2 = 2xy + 1\)
(2) \(x^2 + y^2 = 4 - 2xy\)

I don't agree with the OA.
If we add (1) and (2):

(1) \(x^2 -2xy + y^2 = 1\)
(2) \(x^2 +2xy+ y^2 = 4\)
-------------------------------------
Total: \(2x^2 +2y^2 = 5\)

Then: \(x^2 + y^2 = \frac{5}{2}\)

Please, your comments.



well metallicafan,
i see nothing wrong with your approach, but i solved this the following way...
(1) x^2 + y^2 = 2.x.y + 1
=> (x - y)^2 = 1
=> x - y = 1 or -1 insuff

(2) x^2 + y^2 = 4 - 2.x.y
=> (x + y)^2 = 4
=> x + y = 2 or -2 insuff

combining the two,
since the above equations can have multiple values when we solve for x and y, hence, insufficient.
answer is E. is this a GMATPrep question?? because i dont think the real gmat would confuse students like this.

happy to help, a kudos if you like. :)


But don't you get an answer of 5/2 for\(x^2 + y^2\) if you take any of the combinations?, Note that we are asked to find \(x^2 + y^2\)


hmmm, yes you do. yeah should be C. my bad :)

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What is the value of x^2 + y^2 ?

(1) x^2 + y^2 = 2xy + 1
(2) x^2 + y^2 = 4 - 2xy

----------------------------------------------------------------------------------------------------------------------------------------
Good tricky question almost carried away and started finding (x+y) ^2

but finally marked correct ans C

1. x^2 + y^2 = 2xy+1 --> value of 2XY is not known hence not sufficient
2. x^2 + y^2 = 4-2xy --> value of 2XY is not known hence not sufficient

Combining (1+2)
2x^2+2y^2 = 5
i.e x^2 + y^2 = 5/2 hence sufficient

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Re: What is the value of x^2 + y^2 ? [#permalink]

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New post 02 Dec 2014, 00:56
St1. x^2+y^2=2xy+1 is INSUFFICIENT

St2. x^2+y^2=4-2xy is INSUFFICIENT

St1+St2
means 2xy+1=4-2xy
4xy=3
xy=3/4
putting it in the first equation gives 2*3/4+1=5/2

C

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Re: What is the value of x^2 + y^2 ? [#permalink]

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Re: What is the value of x^2 + y^2 ? [#permalink]

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New post 06 Jul 2016, 06:39
maxLRok wrote:
What is the value of x^2 + y^2 ?

(1) x^2 + y^2 = 2xy+1
(2) x^2 + y^2 = 4 - 2xy



Statement 1 and 2 both do not give us the value of xy and hence are not sufficient to arrive at the answer

adding both statements

2(x^2 + y^2)= 5

x^2 + y^2= 2.5

C is the answer
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Re: What is the value of x^2 + y^2 ?   [#permalink] 06 Jul 2016, 06:39
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