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

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maxLRok
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What is the value of x^2 + y^2 ? [#permalink] New post   Updated on: 24 Nov 2014, 01:44
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E

Difficulty:

35% (medium)

Question Stats:

66% (01:30) correct 34% (01:46) wrong based on 246 sessions
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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

Show 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.
ShowHide Answer
Official Answer
C

Originally posted by maxLRok on 23 Nov 2012, 07:29.
Last edited by Bunuel on 24 Nov 2014, 01:44, edited 6 times in total.
Edited the OA.
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Re: What is the value of x^2 + y^2 ? [#permalink] New post   23 Nov 2012, 07:41
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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] 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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Re: What is the value of x^2 + y^2 ? [#permalink] New post   23 Nov 2012, 07:48
Expert Reply
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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Re: What is the value of x^2 + y^2 ? [#permalink] 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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Re: What is the value of x^2 + y^2 ? [#permalink] New post   23 Nov 2012, 08:56
Expert Reply
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? (1) x^2 + y^2 = 2xy +1 (2) [#permalink] New post   07 Oct 2022, 08:44
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Re: What is the value of x^2 + y^2? (1) x^2 + y^2 = 2xy +1 (2) [#permalink]
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